GSEB Class 11 Physics Solutions Chapter 6 Work, Energy and Power

Question 1. The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative:
1. Work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.
2. Work done by gravitational force in the above case.
3. Work done by friction on a body sliding down an inclined plane.
4. Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity.
5. Work done by the resistive force of air on a vibrating pendulum in bringing it to rest.
Answer: The work done is given by \( W = F.S = FS \cos \theta \), where \( \theta \) is the smaller angle between the force (F) and displacement (S).
1. To lift the bucket, a force equal to its weight must be applied vertically upwards. The bucket also moves in the same direction, so \( \theta = 0 \). Therefore, \( W = FS \) is positive.
2. In this situation, the bucket moves in a direction opposite to the gravitational force, which always pulls vertically downwards. So, \( \theta = 180^\circ \). This means \( W = FS \cos 180^\circ = F S (-1) = -FS \), which is negative.
3. Work done by friction on a body moving down an inclined plane is negative because friction always opposes the relative motion. Thus, \( \theta = 180^\circ \), making \( W = -FS \).
4. When the body moves in the same direction as the applied force, \( \theta = 0 \). So, \( W = FS \), which is positive.
5. Work done is negative because the resistive force of air acting on the vibrating pendulum moves in the opposite direction to the bob's displacement (its motion).
In simple words: Work done is positive if force and motion are in the same direction, and negative if they are in opposite directions. For lifting, the man's work is positive, but gravity's work is negative. Friction and air resistance always do negative work.

Exam Tip: Remember that the sign of work depends on the angle between the force vector and the displacement vector. If the angle is acute (less than 90°), work is positive; if it's obtuse (greater than 90°), work is negative; and if it's 90°, work is zero.

Question 2. A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Compute the
1. work done by the applied force in 10s.
2. work done by friction in 10s.
3. work done by the net force on the body in 10s.
4. change in kinetic energy of the body in 10s and interpret your results.
Answer:
Given:
Mass \( m = 2 \) kg
Applied force \( F = 7 \) N
Initial velocity \( u = 0 \)
Coefficient of kinetic friction \( \mu_k = 0.1 \)
Time \( t = 10 \) s

First, calculate the force of friction:
Force of friction \( f = \mu_k R = \mu_k mg \)
\( = 0.1 \times 2 \times 9.8 \)
\( = 1.96 \) N

Next, find the net force:
\( F_{\text{net}} = F - f = (7 - 1.96) \) N \( = 5.04 \) N

Now, determine the net acceleration of the body:
\( a = \frac{F_{\text{net}}}{m} = \frac{5.04}{2} = 2.52 \) m/s\(^2\)

Calculate the distance (S) covered by the body in 10 s using the equation of motion:
\( S = ut + \frac{1}{2} at^2 \)
\( = 0 \times 10 + \frac{1}{2} \times 2.52 \times (10)^2 \)
\( = 1.26 \times 100 \)
\( S = 126 \) m

Now, let's compute the required quantities:

1. Work done by the applied force:
\( W_1 = F \times S \)
\( = 7 \times 126 \)
\( = 882 \) J

2. Work done by the force of friction:
\( W_2 = -f \times S \)
\( = -1.96 \times 126 \)
\( = -246.96 \) J (approximately \( -247 \) J)

3. Work done by the net force on the body:
\( W_3 = F_{\text{net}} \times S \)
\( = 5.04 \times 126 \)
\( = 635.04 \) J
Alternatively, \( W_3 = W_1 - W_2 = (882 - 246.96) \) J \( = 635.04 \) J

4. Change in kinetic energy of the body in 10s:
According to the work-energy theorem, the change in kinetic energy is equal to the work done by the net force.
Change in K.E. \( = W_3 = 635.04 \) J

Alternatively, calculate final velocity and then K.E.:
Final velocity \( v = u + at \)
\( = 0 + 2.52 \times 10 = 25.2 \) m/s
Final K.E. \( (K.E.)_f = \frac{1}{2} mv^2 = \frac{1}{2} \times 2 \times (25.2)^2 \)
\( = 635.04 \) J
Initial K.E. \( (K.E.)_i = \frac{1}{2} mu^2 = \frac{1}{2} \times 2 \times (0)^2 = 0 \)
Change in K.E. \( = (K.E.)_f - (K.E.)_i = 635.04 - 0 = 635.04 \) J
This shows that the change in kinetic energy of the body is indeed equal to the work done by the net force on the body.
In simple words: First, we found the push from the friction, then the total push that moved the body. We used this to figure out how fast it sped up and how far it went. Then, we calculated the work done by the main push, by the friction, and by the overall total push. Finally, we showed that the body's energy change matched the work done by the overall push.

Exam Tip: For problems involving friction and motion, always start by calculating the frictional force and then the net force to determine acceleration and displacement before calculating work done and kinetic energy changes.

Question 3. Given below are examples of some potential energy) functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle must have in each case. Think of simple physical contexts for which these potential energy shapes are relevant?
Answer: We know that the total energy (E) of a body is the sum of its kinetic energy (K.E.) and potential energy (P.E.): \( E = K.E. + P.E. \). Therefore, \( K.E. = E - P.E. = \frac{1}{2}mv^2 \). Since kinetic energy can never be negative, the potential energy (P.E.) cannot be greater than the total energy (E). If P.E. > E, then K.E. would be negative, which is not possible, meaning the particle cannot exist in such regions.

1. **For V(x) vs x graph (i)** (potential energy is zero between \( x = 0 \) and \( x = a \), then rises to \( V_0 \) for \( x > a \)):
In the region between \( x = 0 \) and \( x = a \), the P.E. is zero, so the K.E. is positive. In the region \( x > a \), the P.E. is \( V_0 \), which is shown as a value greater than E. If P.E. is greater than E, K.E. would be negative. Therefore, the particle cannot be found in the region \( x > a \). The minimum total energy the particle can have in this case is zero.

2. **For V(x) vs x graph (ii)** (potential energy is \( V_0 \) for \( x < a \) and \( x > b \), and a lower value \( -V_1 \) between \( x = a \) and \( x = b \)):
In regions where \( x < a \) and \( x > b \), the potential energy (P.E.) is \( V_0 \), which is greater than the particle's total energy (E). This means the K.E. would be negative in these regions. Thus, the particle cannot be present in the regions \( x < a \) and \( x > b \). In the region between \( x = a \) and \( x = b \), the P.E. is negative (\( -V_1 \)). This means there is a positive K.E. value, so the particle can exist in the region between \( x = a \) and \( x = b \). The minimum total energy the particle can possess in this situation is \( -V_1 \).

3. **For V(x) vs x graph (iii)** (potential energy is always greater than the total energy E in all regions: \( -\infty < x < a \), \( a < x < c \), \( c < x < b \), \( b < x < d \) and \( d < x < \infty \)):
Here, in all the given regions, the potential energy (P.E.) is greater than the total energy (E). Consequently, the K.E. would be negative. So, the particle cannot be found in any region, meaning \( -\infty < x < \infty \). The minimum total energy that the particle can have in this case is \( V_1 \).

4. **For V(x) vs x graph (iv)** (potential energy is an inverted V-shape, rising from \( -V_1 \) at \( x=0 \) to \( V_0 \) at \( x=\pm b/2 \)):
In this case, the P.E. of the particle is more than the total energy (E) in the regions \( \frac{b}{2} < x < \infty \), \( -\infty < x < -\frac{b}{2} \), and also between \( -\frac{a}{2} < x < \frac{a}{2} \). So, the K.E. of the particle will be negative in these regions. Therefore, it will not be present in the regions \( -\infty < x < -\frac{b}{2} \), \( -\frac{a}{2} < x < \frac{a}{2} \), and \( \frac{b}{2} < x < \infty \). The minimum total energy that the particle can have in this case is \( -V_1 \).
In simple words: A particle can only exist where its total energy is greater than or equal to its potential energy, because its movement energy (kinetic energy) cannot be a negative number. If the potential energy is higher than the total energy in a certain area, the particle simply won't be found there. The lowest energy a particle can have is related to the lowest point of its potential energy.

Exam Tip: When analyzing potential energy diagrams, always remember that K.E. must be non-negative. Any region where P.E. > E (total energy) is a classically forbidden region for the particle.

Question 4. The potential energy function for a particle executing linear simple harmonic motion is given by \( V(x) = \frac{1}{2} kx^2 \) where k is the force constant of the oscillator. For \( k = 0.5 \) Nm\(^{-1} \), the graph V(x) versus x is shown in the figure given below. Show that a particle of total energy 1 J moving under this potential must turn back when it reaches \( x = \pm 2 \)m?
Answer: At any moment, the total energy (E) of an oscillator consists of both kinetic energy (K.E.) and potential energy (P.E.). So, E can be written as the sum: \( E = K.E. + P.E. \)
Given: Total energy \( E = 1 \) J. So, \( 1 = \frac{1}{2} mu^2 + \frac{1}{2} kx^2 \) .................. (i)

An oscillating particle reverses its direction at the point where its velocity or speed becomes zero, i.e., \( u = 0 \) .................. (ii)
At this turning point, the total energy is equal to the potential energy.

From (i) and (ii), we get:
\( 1 = 0 + \frac{1}{2} kx^2 \)
\( x^2 = \frac{2 \times 1}{k} \)
Given \( k = 0.5 \) Nm\(^{-1} \), substitute this value:
\( x^2 = \frac{2 \times 1}{0.5} \)
\( x^2 = 2 \times 1 \times 2 = 4 \)
\( x = \pm 2 \) m
Hence, it is proven that the particle will turn back when it reaches \( x = \pm 2 \) m.
In simple words: For a simple harmonic motion, we know the particle's total energy is a mix of its movement energy and stored energy. When the particle turns around, it stops for a moment, meaning its movement energy is zero. At that exact spot, all its total energy is stored energy. By setting the total energy (1 J) equal to the stored energy formula with the given spring constant (0.5 Nm\(^{-1} \)), we can find the position where it stops and turns, which is found to be \( \pm 2 \)m.

Exam Tip: For simple harmonic motion, the turning points occur where the kinetic energy is zero, and all the total mechanical energy is converted into potential energy. This is a key concept for energy conservation problems.

Question 5. Answer the following:
(a) The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained? The rocket or the atmosphere?
(b) Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet's velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why?
(c) An artificial satellite orbiting the earth in very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth?
(d) In Fig. given below (i) the man walks 2m carrying a mass of 15 kg on his hands. In Fig. (ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of 15 kg hangs at its other end. In which case is the work done greater?
Answer:
(a) The heat energy needed for the rocket's casing to burn during flight is obtained from the rocket itself. This energy comes at the cost of the rocket's mass and its potential energy.
(b) The gravitational force acting on a comet due to the Sun is a conservative force. Work done by a conservative force over any closed path is always zero, regardless of the path's nature. Therefore, the work done by the gravitational force over every complete orbit of the comet is zero.
(c) As a satellite moves nearer to Earth, its potential energy decreases. According to the law of conservation of energy, the sum of kinetic and potential energy must remain constant if no external non-conservative forces are at play. However, in this case, the total energy of the satellite steadily decreases somewhat because of energy loss due to friction (atmospheric resistance). Despite this energy loss, as the potential energy goes down (due to decreasing height), the kinetic energy (and therefore speed) increases to maintain the overall energy balance, even if the total sum is slightly lower. This happens because the decrease in potential energy is larger than the total energy loss.
(d) In case (i), the man applies an upward force on the 15 kg mass but walks horizontally for 2m. The angle between the force (upward) and displacement (horizontal) is 90°. Since \( W = FS \cos \theta \), and \( \cos 90^\circ = 0 \), the work done is zero.

In case (ii), the force is applied to lift the 15 kg mass vertically upwards (equal to its weight), and the mass is also lifted vertically. So, \( \theta = 0^\circ \).
Work done \( W = mgS \)
\( = 15 \times 9.8 \times 2 \)
\( = 294 \) J
Therefore, the work done in the second case is greater.
In simple words: (a) A rocket's outer shell burns up because of friction with the air, and this heat energy comes from the rocket itself, using up some of its mass and stored energy. (b) Gravity is a "conservative" force, meaning if an object returns to its starting point after moving, the total work done by gravity is zero, even if it wasn't zero at every moment. (c) A satellite orbiting Earth in a very thin atmosphere slowly loses total energy to air friction. However, as it gets closer to Earth, its stored energy goes down, which makes its movement energy (and speed) go up, even though the overall energy is less. (d) In the first picture, the man carries the weight horizontally, so no work is done against gravity. In the second picture, he pulls the weight up vertically using a pulley, doing actual work to lift it. So, more work is done in the second situation.

Exam Tip: Remember the distinction between conservative and non-conservative forces. Work done by conservative forces over a closed path is zero. Also, for work to be done, there must be displacement in the direction of the applied force.

Question 6. Underline the correct alternative:
(a) When a conservative force does positive work on a body, the potential energy of the body increases/decreases/remain unaltered.
(b) Work done by a body against friction always results in a loss of its kinetic/potential energy.
(c) The rate of change of total momentum of a many particle system is proportional to the external force/sum of the internal forces on the system.
(d) In an inelastic collision of two bodies, the quantities which don't change after the collision are the total kinetic energy/ total linear momentum/total energy of the system of two bodies.
Answer:
(a) When a conservative force does positive work on a body, it means the force moves the body in its direction. This causes the body to move towards the center of force, which then decreases its position (x). Consequently, its potential energy decreases. So, the correct alternative is **decreases**.
(b) Work is done by a body against friction at the expense of its kinetic energy. Therefore, work done against friction always results in a loss of its **kinetic energy**.
(c) Internal forces cannot change the total or net momentum of a system. Therefore, the rate of change of total momentum of a many-particle system is proportional to the **external force** on the system.
(d) In an inelastic collision of two bodies, the quantities that do not change after the collision are the total linear momentum and the total energy of the system of two bodies (if the system is isolated). The total kinetic energy of the system is not conserved as it may transform into an equivalent amount of energy in some other form. Thus, the correct alternatives are **total linear momentum** and **total energy**.
In simple words: (a) When a force that stores energy (conservative force) does positive work, the stored energy of the object goes down. (b) When an object works against rubbing (friction), it always loses energy from its motion. (c) How quickly the total movement of many objects changes depends on outside forces acting on the group, not forces within the group. (d) In a crash where energy isn't perfectly kept (inelastic collision), the total movement and total energy of the objects stay the same, but the energy of motion itself can change.

Exam Tip: Understand the definitions of conservative forces, work-energy theorem, and conservation laws for momentum and energy in different types of collisions (elastic vs. inelastic).

Question 7. State if each of the following statements are true or false? Give reasons for your answer?
(a) In an elastic collision of two bodies, the momentum and energy of each body is conserved.
(b) Total energy of a system is always conserved, no matter what internal and external forces on the body are present.
(c) Work done in the motion of a body over a closed loop is zero for every force in nature.
(d) In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system.
Answer:
(a) **False**. In an elastic collision, the total momentum and total energy of the *system* are conserved, but the momentum and energy of *each individual body* are not necessarily conserved; they can change during the collision.
(b) **False**. The total energy of a system is only conserved if the system is isolated and no external non-conservative forces are doing work on it. External forces on the system may cause the total energy of the system to increase or decrease.
(c) **False**. Work done by a force over a closed loop is zero only when that force is a conservative force (e.g., gravitational or electrostatic forces). It is not zero for non-conservative forces, such as frictional forces.
(d) **True**. In an inelastic collision, some of the initial kinetic energy of the system is converted into other forms of energy, such as heat, sound, or deformation. Therefore, the final kinetic energy of the system is always less than the initial kinetic energy of the system.
In simple words: (a) It's false to say that each object's momentum and energy are saved in an elastic crash; only the total for all objects is. (b) It's false that total energy is always saved no matter what, because outside forces can change it. (c) It's false that work done in a loop is always zero, as this only happens for special forces like gravity, not for rubbing (friction). (d) It's true that in a messy crash (inelastic collision), the final movement energy is always less than what it started with, because some energy changes form (like into heat).

Exam Tip: Be precise with the terms "system" vs. "individual body" and "conservative" vs. "non-conservative forces" when discussing conservation laws.

Question 8. Answer carefully, with reasons:
(a) In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i.e., when they are in contanct)?
(b) Is the total linear momentum conserved during the short time of an elastic collision of two balls?
(c) What are the answers to (a) and (b) for an inelastic collision ?
(d) Is the potential energy of two billiard balls depends only on the separation distance between their centres, is the collision elastic or inelastic? (Note, we are talking here of potential energy corresponding to the force during collision, not gravitational potential energy).
Answer:
(a) No, the total kinetic energy is not conserved during the brief contact interval of an elastic collision. This is because a portion of the kinetic energy is used to deform the balls during that short period of contact, and it is converted into potential energy. However, the kinetic energy before and after the collision remains the same.
(b) Yes, the total linear momentum is conserved during the brief contact interval of an elastic collision of two balls. This holds true as long as no external forces act on the system during the collision.
(c) In an inelastic collision, the total kinetic energy is not conserved *during* the collision, nor is it conserved *after* the collision. It decreases. However, the total linear momentum is conserved during the collision and also after the collision, assuming the system is isolated.
(d) If the potential energy of two billiard balls depends only on the separation distance between their centers, then the collision is elastic. This is because a potential energy function that depends only on separation distance implies that the internal forces involved are conservative. In such cases, kinetic energy lost during deformation is fully recovered as the balls separate, leading to conservation of total kinetic energy before and after the collision, which defines an elastic collision.
In simple words: (a) In an elastic crash, the total energy of motion (kinetic energy) is not kept the same during the exact moment the balls touch because some of it becomes stored energy as the balls squish a little. (b) Yes, the total straight-line movement (linear momentum) is always kept the same during that short time the balls crash, as long as no outside forces interfere. (c) In a messy crash (inelastic), the energy of motion changes (is not kept), but the total straight-line movement still is kept. (d) If the stored energy of two billiard balls only depends on how far apart their centers are, then the crash is elastic, meaning energy is perfectly conserved overall.

Exam Tip: Distinguish between conservation *during* collision and *before and after* collision for both elastic and inelastic scenarios, especially for kinetic energy. The total mechanical energy and total linear momentum are conserved for an isolated system in both types of collisions, but kinetic energy specifically is only conserved before and after an *elastic* collision.

Question 9. A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time t is proportional to
1. t\(^{1/2}\)
2. t
3. t\(^{3/2}\)
4. t\(^2\)
Answer: (2) t
Let \( m \) be the mass of the body.
Let \( a \) be the acceleration produced in the body.
Let \( v \) be the velocity of the body.
Let \( P \) be the power delivered to the body at time \( t \).

We know that \( F = ma \).
Since acceleration is constant, \( a = \frac{dv}{dt} \).
Also, for constant acceleration, \( v = u + at \). Since the body starts from rest, \( u = 0 \), so \( v = at \).

Power \( P \) is given by \( P = Fv \).
Substitute \( F = ma \) and \( v = at \):
\( P = (ma)(at) \)
\( P = ma^2t \)

Since \( m \) (mass) and \( a \) (constant acceleration) are constant, their product \( ma^2 \) is also a constant.
Therefore, \( P \propto t \).

Alternatively, if \( P = Fv \), where \( F = ma \).
Also, \( v = at \).
So, \( P = ma(at) = ma^2t \).
Thus, power is proportional to \( t \).
In simple words: When a body starts from rest and moves with a steady increase in speed, the power given to it increases directly with time. This means if you double the time, you double the power.

Exam Tip: Remember the relationship between power, force, and velocity \( (P = Fv) \), and the equations of motion for constant acceleration. For a body starting from rest with constant acceleration, velocity is proportional to time, leading power to be also proportional to time.

Question 10. A body is moving uni-directionally under the influence of a source of constant power. Its displacement in time t is proportional to
1. t\(^{1/2}\)
2. t
3. t\(^{3/2}\)
4. t\(^2\)
Answer: (3) t\(^{3/2}\)
Let constant power be \( P \) acting on a body of mass \( m \) for time \( t \) to give it velocity \( v \).
The kinetic energy (K.E.) of the body is equal to the work done, which is power multiplied by time:
\( \frac{1}{2} mv^2 = P \times t \)
\( v^2 = \frac{2Pt}{m} \)
\( v = \sqrt{\frac{2P}{m}} t^{1/2} \) .............. (1)

We also know that velocity is the rate of change of displacement:
\( v = \frac{dx}{dt} \)

Substitute the expression for \( v \) from (1) into this equation:
\( \frac{dx}{dt} = \sqrt{\frac{2P}{m}} t^{1/2} \)
To find displacement \( x \), integrate both sides with respect to \( t \):
\( x = \int \sqrt{\frac{2P}{m}} t^{1/2} dt \)
Since \( P \) and \( m \) are constant, \( \sqrt{\frac{2P}{m}} \) is a constant.
\( x = \sqrt{\frac{2P}{m}} \int t^{1/2} dt \)
\( x = \sqrt{\frac{2P}{m}} \frac{t^{1/2 + 1}}{1/2 + 1} \)
\( x = \sqrt{\frac{2P}{m}} \frac{t^{3/2}}{3/2} \)
\( x = \frac{2}{3} \sqrt{\frac{2P}{m}} t^{3/2} \)

Since \( P \) and \( m \) are constants, \( \frac{2}{3} \sqrt{\frac{2P}{m}} \) is also a constant.
Therefore, \( x \propto t^{3/2} \).

Alternative method:
We know \( P = Fv = (ma)v \).
For constant power, \( P = \text{constant} \).
So, \( v \propto \frac{1}{a} \). This is not correct for constant power as F is not constant.
Let's use \( P = Fv \). Since \( F = m \frac{dv}{dt} \), we have \( P = m \frac{dv}{dt} v \).
\( \int P dt = \int mv dv \)
\( Pt = \frac{1}{2} mv^2 \)
\( v^2 = \frac{2Pt}{m} \Rightarrow v = \sqrt{\frac{2P}{m}} t^{1/2} \).
As \( v = \frac{dx}{dt} \),
\( \frac{dx}{dt} = \sqrt{\frac{2P}{m}} t^{1/2} \).
Integrating, \( x = \int \sqrt{\frac{2P}{m}} t^{1/2} dt = \sqrt{\frac{2P}{m}} \frac{t^{3/2}}{3/2} \).
Thus, \( x \propto t^{3/2} \).
In simple words: When a body moves under a steady power source, its speed increases, but not steadily like acceleration. We find that the distance it travels grows proportionally to time raised to the power of one and a half. So, if time doubles, the distance traveled multiplies by \( 2^{1.5} \).

Exam Tip: This problem requires combining the definitions of power, kinetic energy, and velocity. Remember that power is the rate of work done, and for constant power, kinetic energy increases linearly with time, which then allows you to find the velocity and subsequently the displacement as a function of time.

Question 11. A body constrained to move along z axis of a co-ordinate system is subjected to a constant force \( \vec{F} \) given by \( F = -\hat{i} + 2\hat{j} + 3\hat{k} \) N where \( \hat{i}, \hat{j} \) and \( \hat{k} \) are the unit vector along the x, y and z axis of the system respectively. What is the work done by this force in moving the body a distance of 4 m along the z axis?
Answer: The force vector is given as \( \vec{F} = -\hat{i} + 2\hat{j} + 3\hat{k} \) N.
The body is displaced by 4 m along the z-axis only. Therefore, the displacement vector \( \vec{S} \) can be written as:
\( \vec{S} = 0\hat{i} + 0\hat{j} + 4\hat{k} \) m

The work done (W) by the force is given by the dot product of the force vector and the displacement vector:
\( W = \vec{F} \cdot \vec{S} \)
\( W = (-\hat{i} + 2\hat{j} + 3\hat{k}) \cdot (0\hat{i} + 0\hat{j} + 4\hat{k}) \)
To calculate the dot product, multiply the corresponding components and add them:
\( W = (-1 \times 0) + (2 \times 0) + (3 \times 4) \)
\( W = 0 + 0 + 12 \)
\( W = 12 \) J
So, the work done by this force is 12 J.
In simple words: To find the work done, we use the rule where you multiply the force by the distance moved in the same direction. Here, the force has parts in all three directions, but the body only moves along the z-axis. So, we only care about the z-part of the force multiplied by the distance moved along the z-axis. This calculation gives us a work of 12 joules.

Exam Tip: Remember that work done by a constant force is the dot product of the force vector and the displacement vector. This means only the component of the force parallel to the displacement contributes to the work done.

Question 12. An electron and proton are detected in a cosmic ray experiment, the first with kinetic energy 10 keV, and the second with 100 keV. Which is faster, the electron or the proton? Obtain the ratio of their speeds, (electron mass = \( 9.11 \times 10^{-31} \) kg, proton mass = \( 1.67 \times 10^{-27} \) kg, 1 eV = \( 1.60 \times 10^{-19} \)).
Answer: We need to compare the speeds of the electron and proton. We can use the kinetic energy formula \( K.E. = \frac{1}{2} mv^2 \), which can be rearranged to find velocity \( v = \sqrt{\frac{2 \times K.E.}{m}} \).

First, convert the given kinetic energies from keV to Joules:
Given: \( 1 \) eV = \( 1.60 \times 10^{-19} \) J

For the electron:
Kinetic Energy \( K.E._e = 10 \) keV \( = 10 \times 10^3 \) eV \( = 10^4 \) eV
\( K.E._e = 10^4 \times 1.60 \times 10^{-19} \) J \( = 1.60 \times 10^{-15} \) J
Mass of electron \( m_e = 9.11 \times 10^{-31} \) kg

Calculate the speed of the electron \( (v_e) \):
\( v_e = \sqrt{\frac{2 \times K.E._e}{m_e}} \)
\( v_e = \sqrt{\frac{2 \times 1.60 \times 10^{-15} \text{ J}}{9.11 \times 10^{-31} \text{ kg}}} \)
\( v_e = \sqrt{\frac{3.2 \times 10^{-15}}{9.11 \times 10^{-31}}} \)
\( v_e \approx \sqrt{0.35126 \times 10^{16}} \)
\( v_e \approx 0.5926 \times 10^8 \) m/s \( \approx 5.93 \times 10^7 \) m/s

For the proton:
Kinetic Energy \( K.E._p = 100 \) keV \( = 100 \times 10^3 \) eV \( = 10^5 \) eV
\( K.E._p = 10^5 \times 1.60 \times 10^{-19} \) J \( = 1.60 \times 10^{-14} \) J
Mass of proton \( m_p = 1.67 \times 10^{-27} \) kg

Calculate the speed of the proton \( (v_p) \):
\( v_p = \sqrt{\frac{2 \times K.E._p}{m_p}} \)
\( v_p = \sqrt{\frac{2 \times 1.60 \times 10^{-14} \text{ J}}{1.67 \times 10^{-27} \text{ kg}}} \)
\( v_p = \sqrt{\frac{3.2 \times 10^{-14}}{1.67 \times 10^{-27}}} \)
\( v_p \approx \sqrt{1.916 \times 10^{13}} \)
\( v_p \approx 4.377 \times 10^6 \) m/s \( \approx 4.38 \times 10^6 \) m/s

Comparing the speeds:
\( v_e \approx 5.93 \times 10^7 \) m/s
\( v_p \approx 4.38 \times 10^6 \) m/s
Since \( v_e \) is much larger than \( v_p \), the electron moves faster than the proton.

Now, calculate the ratio of their speeds \( \frac{v_e}{v_p} \):
\( \frac{v_e}{v_p} = \frac{5.93 \times 10^7 \text{ m/s}}{4.38 \times 10^6 \text{ m/s}} \)
\( \frac{v_e}{v_p} \approx 13.5 \)

The electron is faster than the proton, and the ratio of their speeds is approximately 13.5.
In simple words: We used the given energy and mass for both the electron and proton to figure out how fast each was moving. Even though the proton had more energy, its much larger mass meant it moved slower. The electron, being much lighter, traveled a lot faster, specifically about 13.5 times quicker than the proton.

Exam Tip: Remember to convert all energy values to Joules and always use the correct units throughout your calculations. The key is to apply the kinetic energy formula correctly to each particle and then compare the derived speeds.

Question 13. A rain drop of radius 2mm falls from a height of 500m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its originial height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey? What is the work done by the resistive force in the entire journey if its speed oh reaching the ground is 10 ms\(^{-1}\)?
Answer: First, let's gather the given information and necessary constants:
Radius of the rain drop \( r = 2 \) mm \( = 2 \times 10^{-3} \) m
Density of water \( \rho = 10^3 \) kg/m\(^3\)
Total height \( H = 500 \) m
Final speed on reaching the ground \( v = 10 \) m/s
Initial speed \( u = 0 \) (assuming it starts from rest)

Work done by gravitational force:
The distance traveled in the first half of its journey is \( S_1 = \frac{H}{2} = \frac{500}{2} = 250 \) m.
The distance traveled in the second half of its journey is also \( S_2 = \frac{H}{2} = 250 \) m.

First, calculate the mass \( (m) \) of the rain drop:
Volume of the drop \( V = \frac{4}{3} \pi r^3 \)
\( V = \frac{4}{3} \times \pi \times (2 \times 10^{-3})^3 \)
\( V \approx \frac{4}{3} \times 3.14159 \times 8 \times 10^{-9} \)
\( V \approx 3.35 \times 10^{-8} \) m\(^3\)

Mass \( m = \rho \times V \)
\( m = 10^3 \text{ kg/m}^3 \times 3.35 \times 10^{-8} \text{ m}^3 \)
\( m = 3.35 \times 10^{-5} \) kg

Gravitational force \( F_g = mg \)
\( F_g = 3.35 \times 10^{-5} \text{ kg} \times 9.8 \text{ m/s}^2 \)
\( F_g \approx 3.283 \times 10^{-4} \) N

Work done by gravitational force in the first half of its journey:
\( W_{g1} = F_g \times S_1 = (3.283 \times 10^{-4} \text{ N}) \times (250 \text{ m}) \)
\( W_{g1} \approx 0.082075 \) J \( \approx 0.082 \) J

Since the distance traveled in the second half is also the same, the work done by the gravitational force in the second half of its journey is also approximately \( 0.082 \) J.

Work done by resistive force in the entire journey:
The work-energy theorem states that the net work done on an object equals its change in kinetic energy: \( W_{\text{net}} = \Delta K.E. \).
Here, the net work done is the sum of work done by gravity and work done by resistive force: \( W_{\text{net}} = W_g + W_r \).
So, \( W_g + W_r = K.E._f - K.E._i \).

Total work done by gravity over the entire journey (500 m):
\( W_g = F_g \times H = (3.283 \times 10^{-4} \text{ N}) \times (500 \text{ m}) \)
\( W_g \approx 0.16415 \) J

Initial kinetic energy \( K.E._i = \frac{1}{2} mu^2 = \frac{1}{2} m (0)^2 = 0 \)

Final kinetic energy \( K.E._f = \frac{1}{2} mv^2 \)
\( K.E._f = \frac{1}{2} \times (3.35 \times 10^{-5} \text{ kg}) \times (10 \text{ m/s})^2 \)
\( K.E._f = \frac{1}{2} \times 3.35 \times 10^{-5} \times 100 \)
\( K.E._f = \frac{1}{2} \times 3.35 \times 10^{-3} \)
\( K.E._f = 1.675 \times 10^{-3} \) J \( = 0.001675 \) J

Now, calculate work done by resistive force \( (W_r) \):
\( W_r = (K.E._f - K.E._i) - W_g \)
\( W_r = (0.001675 \text{ J} - 0 \text{ J}) - 0.16415 \text{ J} \)
\( W_r = 0.001675 - 0.16415 \)
\( W_r = -0.162475 \) J \( \approx -0.163 \) J

The work done by the resistive force is negative, which is expected as it opposes the motion.
In simple words: We first figured out the raindrop's mass. Then, we calculated how much work gravity did in the first half and second half of its fall; it was the same for both halves. To find the work done by air resistance over the whole fall, we used the change in the raindrop's movement energy (kinetic energy) and the total work done by gravity. Since the raindrop's final speed was low, it showed that air resistance had done a lot of negative work, slowing it down.

Exam Tip: For problems involving both conservative (gravity) and non-conservative (air resistance) forces, apply the work-energy theorem, which states that the total work done by all forces equals the change in kinetic energy. Remember that resistive forces always do negative work.

Question 14. A molecule in a gas container hits a horizontal wall with speed 200 ms\(^{-1}\) and angle 30° with the normal and rebounds with the same speed. Is momentum conserved in the collision? Is the collision elastic or inelastic?
Answer: When a molecule collides with a wall, we need to consider the system carefully. The system includes both the molecule and the wall (and the container it belongs to).

Regarding momentum:
Linear momentum is always conserved for an isolated system. In this collision, if we consider the system to be the molecule *and* the wall (or the entire container), then the total momentum of this combined system *is* conserved. The wall will experience a recoil, however small, balancing the change in the molecule's momentum. If we consider only the molecule, its momentum changes direction, so the momentum of the molecule itself is not conserved. However, the question usually implies the total system. Therefore, considering the molecule and the wall as a system, momentum *is conserved*.

Regarding the collision type:
The collision is elastic. This conclusion is drawn because the speed of the molecule remains exactly the same before and after hitting the wall. In an elastic collision, both kinetic energy and momentum are conserved. Since the speed (and thus kinetic energy) is conserved, and momentum of the system is also conserved, the collision is elastic.
In simple words: When the molecule hits the wall, the total forward push (momentum) of the molecule and the wall together stays the same; the wall moves back a tiny bit to make up for the molecule changing direction. Since the molecule bounces off with the same speed it hit with, no energy of motion is lost, meaning it's a perfectly "elastic" collision.

Exam Tip: An elastic collision is characterized by the conservation of both kinetic energy and linear momentum. If kinetic energy is conserved (as implied by same speed before and after), it's an elastic collision. Momentum of the *system* (molecule + wall) is always conserved in any collision, elastic or inelastic, if no external forces are acting.

Question 15. A pump on the ground floor of a building can pump up water to All a tank of value 30 m³ in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump?
Answer: The mass of water to be pumped up is determined by its volume and the density of water. \( = 30 \times 10^3 \) \( = 3 \times 10^4 \) kg The height of the tank, \( h = 40 \) m. Therefore, the work done by the pump to fill the tank is: \( W = mgh = 3 \times 10^4 \times 9.8 \times 40 \) J \( = 1.176 \times 10^7 \) J The time taken, \( t = 15 \) minutes \( = 15 \times 60 \)s \( = 900 \)s So, the required power is: \( P = \frac{W}{t} = \frac{1.176 \times 10^7}{900} = 13.07 \) kW The percentage efficiency of the pump is \( 30\% \). We know that: \( \eta \% = \frac{\text{Output power}}{\text{Input power}} \times 100 \) Or, Input power is the total power used by the pump. So, \( \text{Input power} = \frac{\text{Output power}}{\eta \%} \times 100 \) \( = \frac{13.07}{30} \times 100 = 43.55 \) kW \( = 43.6 \) kW.In simple words: First, we calculate the mass of the water and the work needed to lift it to the tank. Then, we find the power required using the work and time. Finally, we use the pump's efficiency to figure out the actual electrical power it uses.

Exam Tip: Remember to convert all units to SI units (like minutes to seconds, km/h to m/s) before performing calculations for consistency.

Question 16. Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed V. If the collision is elastic, which of the following is a possible result after collision?
(i) Diagram shows the first ball coming to rest, and the second and third balls moving together at \( V/2 \).
(ii) Diagram shows the first two balls coming to rest, and the third ball moving away with speed \( V \).
(iii) Diagram shows all three balls moving together with speed \( V/3 \).
Answer: (ii) Diagram shows the first two balls coming to rest, and the third ball moving away with speed \( V \).
Let \( m \) be the mass of each ball bearing. The initial kinetic energy (K.E.) of the system is given by the kinetic energy of the incoming ball: \( \text{K.E.}_{\text{initial}} = \frac{1}{2} m V^2 + 0 = \frac{1}{2} m V^2 \) After the collision, the kinetic energy for each case is calculated: For case (i): The first ball stops, and the other two balls (total mass \( 2m \)) move together at \( V/2 \). \( \text{K.E.}_{\text{(i)}} = \frac{1}{2} (2m) \left(\frac{V}{2}\right)^2 = \frac{1}{2} (2m) \left(\frac{V^2}{4}\right) = \frac{1}{4} m V^2 \) For case (ii): The first two balls stop, and the third ball (mass \( m \)) moves at \( V \). \( \text{K.E.}_{\text{(ii)}} = \frac{1}{2} m V^2 \) For case (iii): All three balls (total mass \( 3m \)) move together at \( V/3 \). \( \text{K.E.}_{\text{(iii)}} = \frac{1}{2} (3m) \left(\frac{V}{3}\right)^2 = \frac{1}{2} (3m) \left(\frac{V^2}{9}\right) = \frac{1}{6} m V^2 \) From these calculations, we notice that the kinetic energy is conserved only in case (ii). Therefore, case (ii) is the only possible outcome after an elastic collision.In simple words: When one ball hits two resting balls in an elastic crash, the first two balls stop, and the last ball moves away with the same speed as the first ball. This is because kinetic energy and momentum both stay the same.

Exam Tip: In an elastic collision between identical masses, kinetic energy and momentum are both conserved. Often, objects can simply exchange their velocities.

Question 17. The bob A of a pendulum released from 30° to the vertical hits another bob B of the same mass at rest on a table as shown in fig. given below. How high does the bob A rise after the collision? Neglect the size of the bobs and assume the collision to be elastic?
Answer: In a perfectly elastic head-on collision, when two objects of equal mass crash into each other, they swap their speeds. Here, bob A is moving at a certain speed, and bob B is initially at rest. Therefore, after the collision, bob A will stop moving and bob B will begin to move with the speed that bob A had. Bob A transfers all of its momentum to bob B, which means bob A will not rise at all after the impact.In simple words: When two identical pendulums hit each other head-on in an elastic crash, the moving one stops, and the still one starts moving with the first one's speed. So, bob A will not swing back up.

Exam Tip: For elastic collisions between objects of equal mass, a simple exchange of velocities occurs. This principle is often demonstrated by Newton's cradle.

Question 18. The bob of a pendulum is released from a horizontal position A as shown in the figure. If the length of the pendulum is 1.5 m, what is the speed with which the bob arrives at the lower most point B, given that it dissipated 5% of its initial energy against air resistance?
Answer: At point A, the pendulum's energy is entirely potential energy (P.E.). At the lowest point B, this energy is entirely kinetic energy (K.E.). This means that as the bob moves down from A to B, its potential energy changes into kinetic energy. So, at point B, K.E. is equal to the potential energy minus the energy lost. However, 5% of the initial potential energy is lost due to air resistance. Therefore, K.E. at point B \( = 95\% \) of P.E. at point A. Let \( m \) be the mass of the bob and \( v \) be its speed at point B. The height of point A with respect to B, \( h = 1.5 \) m. From the energy relation: \( \frac{1}{2} mv^2 = \frac{95}{100} mgh \) We can simplify for \( v^2 \): \( v^2 = 2 \times \frac{95}{100} gh \) Substituting the values: \( v^2 = 2 \times \frac{95}{100} \times 9.8 \times 1.5 \) So, \( v = \sqrt{\frac{19 \times 9.8 \times 1.5}{10}} = \sqrt{27.93} \) \( v = 5.285 \text{ ms}^{-1} \) Rounding off, \( v = 5.3 \text{ ms}^{-1} \).In simple words: The pendulum starts with potential energy. As it swings down, this energy changes into kinetic energy. Because of air resistance, 5% of the energy is lost. We calculate the final speed at the bottom using the remaining 95% of the original potential energy.

Exam Tip: When dealing with energy conservation problems involving resistance, always remember to subtract the dissipated energy from the initial total energy before equating it to the final kinetic or potential energy.

Question 19. A trolley of mass 300 kg carrying a sand bag of 25 kg is moving uniformly with a speed of 27 km/hr on a frictionless track. After a while, sand starts leaking out of a hole on the floor of trolley at the rate 0.05 kg s\(^{-1}\). What is the speed of the trolley after the entire sand bag is empty?
Answer: The trolley carrying a sand bag is moving with a speed of 27 km/h. This indicates that no external force acts on the system, which includes the trolley and the sand bag. When sand leaks out of a hole on the trolley's floor, it does not create any external force on the trolley. Consequently, the trolley's speed will not change, even if the sand leaks out or after the entire sand bag is empty. This phenomenon occurs because the sand's weight is balanced by the normal reaction from the floor. Both of these forces are perpendicular to the direction of motion, meaning no work is done on or by the trolley as the sand leaks. Therefore, the speed stays the same. So, the speed \( = 27 \text{ kmh}^{-1} \).In simple words: Since the trolley is on a frictionless track and no outside forces push it, its speed will not change even if sand leaks out. The sand falling down doesn't push the trolley forwards or backwards.

Exam Tip: Remember that momentum is conserved in the absence of external forces. The leaking sand's vertical motion doesn't affect the horizontal motion of the trolley.

Question 20. A body of mass 0.5 kg travels in a straight line with velocity \( v = ax^{3/2} \), where \( a = 5 \text{ m}^{-1/2}\text{s}^{-1} \). What is the work done by the net force during its displacement from \( x = 0 \) to \( x = 2 \)m?
Answer: Given values are: Mass \( m = 0.5 \) kg Velocity \( v = ax^{3/2} \) Constant \( a = 5 \text{ m}^{-1/2}\text{s}^{-1} \) Let \( F \) be the net force acting on the body, creating an acceleration \( 'a' \). Therefore, \( F = ma = m \cdot \frac{dv}{dt} \) If \( dW \) represents the small amount of work done by force \( F \) in moving the body by \( dx \), Then \( dW = Fdx = m \frac{dv}{dt} dx \) This can be rewritten as: \( dW = m dv \cdot \frac{dx}{dt} = m v dv \) If \( W \) is the total work done in moving the body from \( x = 0 \) to \( x = 2 \)m, then by integrating the equation, we get: \( W = \int dW = \int mv dv = \frac{1}{2} m v^2 \) Substitute \( v = ax^{3/2} \): \( W = \frac{1}{2} m (ax^{3/2})^2 = \frac{1}{2} m a^2 x^3 \) Now, putting the given values for \( x = 2 \)m, \( a = 5 \), and \( m = 0.5 \), we get: \( W = \frac{1}{2} \times 0.5 \times (5)^2 \times (2)^3 \) \( W = \frac{1}{2} \times \frac{1}{2} \times 25 \times 8 = 50 \) J.In simple words: We calculate the work done by finding the force and displacement. Since velocity depends on displacement, we use integration to find the total work done as the kinetic energy changes from \( x=0 \) to \( x=2 \)m.

Exam Tip: Remember the work-energy theorem, which states that the net work done on an object equals its change in kinetic energy. Also, carefully handle derivatives and integrals when velocity is a function of position.

Question 21. The blades of a windmill sweep out a circle of area A?
(a) If the wind flows at a velocity \( v \) perpendicular to the circle, what is the mass of the air passing through it in time?
(b) What is the kinetic energy of the air?
(c) Assume that windmill converts 25% of the wind's energy into electrical energy, and that \( A = 30\text{m}^2 \), \( v = 36\text{km/h} \) and the density of air is \( 1.2 \text{ kg m}^{-3} \). What is the electrical power produced?
Answer:
(a) The circular area swept out by the windmill blades is \( A \). The velocity of the wind is \( v \). Let the density of air be \( \rho \). The distance covered by the wind in time \( t \) is \( vt \). So, the volume of air passing through the circular area \( A \) is: \( \text{Volume} = \text{Area} \times \text{length} = A \cdot vt \) Then, the mass \( m \) of air passing through the area \( A \) is: \( m = \text{density} \times \text{volume} = \rho A vt \)
(b) If \( E \) is the kinetic energy (K.E.) of the air passing through area \( A \), then: \( E = \frac{1}{2} mv^2 \) Substitute \( m = \rho A vt \): \( E = \frac{1}{2} (\rho A vt) v^2 = \frac{1}{2} \rho A v^3 t \)
(c) Given values are: Area \( A = 30 \text{ m}^2 \) Velocity \( v = 36 \text{ kmh}^{-1} \) Convert velocity to \( \text{ms}^{-1} \): \( v = 36 \times \frac{5}{18} \text{ ms}^{-1} = 10 \text{ ms}^{-1} \) Density of air \( \rho = 1.2 \text{ kgm}^{-3} \) The wind's energy (Input energy) \( = \frac{1}{2} \rho A v^3 t \) The electrical energy (Output energy) is \( 25\% \) of the wind's energy. \( \eta \% = 25 = \frac{\text{Electrical energy}}{\text{Wind's energy}} \times 100 \) So, electrical energy \( = \frac{25}{100} \times \frac{1}{2} \rho A v^3 t \) Electrical power produced \( = \frac{\text{Electrical energy}}{t} = \frac{1}{4} \times \frac{1}{2} \rho A v^3 \) \( = \frac{1}{8} \times 1.2 \times 30 \times (10)^3 \) \( = \frac{1}{8} \times 1.2 \times 30 \times 1000 \) \( = 4.5 \times 10^3 \text{ W} = 4.5 \text{ kW} \).In simple words: First, we find the mass of air passing through the windmill. Then, we calculate the kinetic energy of this air. Finally, since the windmill converts 25% of this energy into electricity, we use the given numbers to find the total electrical power generated.

Exam Tip: Always ensure unit consistency, especially when converting speed from km/h to m/s. Remember that power is energy per unit time, and efficiency relates output power to input power.

Question 22. A person trying to lose weight (dieter) lifts a 10 kg mass 0.5 m 1000 times. Assume that the potential energy lost each time she lowers the mass is dissipated?
(a) How much work does she do against the gravitational force?
(b) Fat supplies \( 3.8 \times 10^7 \) J of energy per kilogram, which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up?
Answer:
(a) Given values: Mass \( m = 10 \) kg Height per lift \( h = 0.5 \) m Number of lifts \( n = 1000 \) The total height lifted, \( H = n \times h = 1000 \times 0.5 \text{ m} = 500 \text{ m} \). The work done against the gravitational force is: \( \text{Work done} = mgh = 10 \times 9.8 \times 500 = 49000 \) J.
(b) Energy supplied by fat per kilogram \( = 3.8 \times 10^7 \) J/kg. The efficiency of converting fat energy to mechanical energy is \( 20\% \). From part (a), the total mechanical energy output (work done) \( = 49000 \) J. We know that: \( \text{Efficiency} = \frac{\text{Mechanical Energy Output}}{\text{Energy Input (from fat)}} \times 100\% \) So, \( 20\% = \frac{49000 \text{ J}}{\text{Energy Input}} \times 100\% \) Rearranging to find Energy Input: \( \text{Energy Input} = \frac{49000 \times 100}{20} = 49000 \times 5 = 245000 \) J. Now, to find the mass of fat used, we divide the total energy input by the energy supplied per kilogram of fat: \( \text{Mass of fat used} = \frac{245000 \text{ J}}{3.8 \times 10^7 \text{ J/kg}} \) \( = \frac{2.45 \times 10^5}{3.8 \times 10^7} \text{ kg} = 0.006447 \text{ kg} \) \( = 6.447 \times 10^{-3} \text{ kg} \). Rounding off, the mass of fat used \( = 6.45 \times 10^{-3} \text{ kg} \).In simple words: First, we calculate the total work done by the person in lifting the mass. This is the useful energy output. Then, using the efficiency, we find the total energy the body needed from fat. Finally, we convert this energy into the amount of fat consumed using the given energy content of fat.

Exam Tip: Pay close attention to efficiency calculations. Remember that efficiency is (useful output energy / total input energy) * 100%, and units must be consistent throughout the calculations.

Question 23. A large family uses 8kW of power.
(a) Direct solar energy is incident on the horizontal surface at an average rate of 200W per square metre. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8kW?
(b) Compare this area to that of the roof of a typical house.
Answer:
Total power required by the family \( = 8 \text{ kW} = 8000 \text{ W} \).
(a) The average rate of incidence of direct solar energy on the horizontal surface \( = 200 \text{ W/m}^2 \). The energy converted into useful electrical energy per square metre is \( 20\% \) of the incident energy: \( \text{Useful electrical energy} = 20\% \text{ of } 200 \text{ W/m}^2 = \frac{20}{100} \times 200 \text{ W/m}^2 = 40 \text{ W/m}^2 \). The total power required by the family is \( 8000 \text{ W} \). Therefore, the area required to produce this amount of electrical energy is: \( \text{Area} = \frac{\text{Total Power Required}}{\text{Useful Electrical Energy per } \text{m}^2} = \frac{8000 \text{ W}}{40 \text{ W/m}^2} = 200 \text{ m}^2 \).
(b) To compare this area to that of a typical house roof, let \( 'a' \) be the side length of a square roof. The area of a square roof \( = a \times a = a^2 \). We found that \( a^2 = 200 \text{ m}^2 \). So, \( a = \sqrt{200} \text{ m} \approx 14.14 \text{ m} \). Thus, the area required is \( 14.14 \times 14.14 \text{ m}^2 \). This \( 200 \text{ m}^2 \) area is comparable to the roof of a large house, which might have dimensions of approximately \( 14 \text{ m} \times 14 \text{ m} \).In simple words: First, we calculate how much useful electrical power can be made from solar energy per square meter. Then, we use the family's total power need to find out what size of solar panel area is required. Finally, we compare this area to the size of a typical large house roof.

Exam Tip: When dealing with energy conversion and area calculations, ensure all units are consistent (e.g., kW to W). Pay attention to percentages to correctly calculate useful output from total input.