Advanced Study Material for Class 9 Science: Chapter 10 Work and Energy
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WORK AND ENERGY
Work, energy and power are the core ideas of Class 9 Science Chapter 11. These notes explain work and its unit, kinetic energy and potential energy with their formulae, power and the commercial unit of energy, energy transformations and the law of conservation of energy, with every intext question, exercise question and practice numerical solved step by step.
WORK
Work (W) is said to be done when a force (F) acts on a body and the point of application of the force is displaced (s) in the direction of the force.
Work done = Force × Displacement, W = F × s
i) If the body is displaced in the same direction as the force, work is done by the force.
ii) If the displacement is against the force, work is done against the force.
iii) If the displacement is perpendicular to the direction of the force, the work done is zero.
UNIT OF WORK
The unit of work is the joule (J). One joule of work is said to be done when a force of 1 newton acting on a body displaces it by a distance of 1 m.
Larger units of work are:
i) kilojoule (1 kJ = 1000 J)
ii) megajoule (1 MJ = 10 lakh J = 1000000 J)
NUMERICALS ON WORK
Unless a question gives g, take g = 10 m/s2.
1. A force of 10 N acts on an object. The object is displaced through 12 m, in the direction of the force. If the force acts all through the displacement, find the work done by the force.
Answer:
1. W = F × s
2. F = 10 N, s = 12 m
3. W = 10 × 12 = 120 J
Teacher's Note:
a) Force and displacement are in the same direction, so the work is positive.
b) Check the unit: N × m = J.
2. A porter lifts a luggage of 15 kg from the ground and puts it on his head 1.5 m above the ground. Calculate the work done by him on the luggage.
Answer:
1. Force needed = weight = mg = 15 × 10 = 150 N
2. Displacement = 1.5 m
3. W = 150 × 1.5 = 225 J
Teacher's Note:
a) The porter lifts against gravity, so force = weight of the luggage.
b) With g = 9.8 m/s2 the answer would be 220.5 J.
3. A boy pushes a book by applying a force of 40 N. Find the work done by this force on the book if it is displaced through 25 cm along the path.
Answer:
1. s = 25 cm = 0.25 m
2. W = F × s = 40 × 0.25 = 10 J
Teacher's Note:
a) Convert centimetres to metres before multiplying.
b) The force acts along the path, so the work is positive.
4. A ball of mass 1 kg thrown upwards reaches a maximum height of 4 m. Calculate the work done by the force of gravity during the vertical displacement. (g = 10 m/s2)
Answer:
1. Force of gravity = mg = 1 × 10 = 10 N, acting downwards
2. Displacement = 4 m upwards, opposite to the force
3. W = -F × s = -10 × 4 = -40 J
Teacher's Note:
a) The work done by gravity is negative because the displacement is opposite to the force.
b) The same 40 J is stored as potential energy of the ball.
5. Find the amount of work done by a labourer who carries n bricks of m kg each to the roof of a house h metre high by climbing a ladder.
Answer:
1. Total mass carried = n × m kg
2. Force needed = weight = nmg
3. W = F × h = nmgh joule
Teacher's Note:
a) Work depends only on the vertical height h, not on the length of the ladder.
b) With g = 10 m/s2, W = 10nmh J.
6. An engine pulls a train 1 km over a level track. Calculate the work done by the engine given that the frictional resistance is 5 × 105 N.
Answer:
1. At constant speed the pulling force equals the friction, F = 5 × 105 N
2. s = 1 km = 1000 m
3. W = F × s = 5 × 105 × 1000 = 5 × 108 J
Teacher's Note:
a) The question says "work done by the train" but it is the engine that does this work against friction. The work done by friction itself is -5 × 108 J.
b) This assumes the train moves at a steady speed.
7. A man weighing 70 kg carries a weight of 10 kg on the top of a tower 100 m high. Calculate the work done by the man (g = 10 m/s2).
Answer:
1. Total mass lifted = 70 + 10 = 80 kg
2. Force = 80 × 10 = 800 N
3. W = 800 × 100 = 80000 J = 8 × 104 J
Teacher's Note:
a) The man lifts his own body as well as the load.
b) 1 kJ = 1000 J, so this is 80 kJ.
8. A boy of mass 55 kg runs up a flight of 40 stairs, each measuring 0.15 m. Calculate the work done by the boy.
Answer:
1. Height climbed = 40 × 0.15 = 6 m
2. W = mgh = 55 × 10 × 6 = 3300 J
Teacher's Note:
a) Only the vertical height of the stairs counts.
b) With g = 9.8 m/s2 the answer is 3234 J.
9. Calculate the work done in lifting 200 kg of water through a vertical height of 6 metres (g = 10 m/s2).
Answer:
1. W = mgh = 200 × 10 × 6 = 12000 J
Teacher's Note:
a) Mass is given in kg, so no conversion is needed.
b) 12000 J = 12 kJ.
10. A crane pulls up a car of mass 500 kg to a vertical height of 4 m. Calculate the work done by the crane.
Answer:
1. W = mgh = 500 × 10 × 4 = 20000 J
Teacher's Note:
a) g = 10 m/s2 is assumed as the question does not give it.
b) 20000 J = 20 kJ.
11. A force of 5 N acts on an object. The object is displaced through 8 m, in the direction of the force. If the force acts all through the displacement, find the work done by the force.
Answer:
1. W = F × s = 5 × 8 = 40 J
Teacher's Note:
a) The figure in the source shows 2 m, but the question text says 8 m. The text value 8 m is used. With 2 m the work would be 10 J.
b) Force and displacement are in the same direction.
12. A porter lifts a luggage of 15 kg from the ground and puts it on his head 1.5 m above the ground. Calculate the work done by him on the luggage.
Answer:
1. W = mgh = 15 × 10 × 1.5 = 225 J
Teacher's Note:
a) This is the same as Numerical 2, repeated in the source.
b) Weight = mg is the force needed to lift.
13. Calculate the work done by a student in lifting a 0.5 kg book from the ground and keeping it on a shelf 1.5 m high.
Answer:
1. W = mgh = 0.5 × 10 × 1.5 = 7.5 J
Teacher's Note:
a) With g = 9.8 m/s2 the work is 7.35 J.
b) The work is stored as potential energy of the book.
14. A coolie carries a load of 50 kg on his head and walks on a level road up to 100 m. What is the work done by him?
Answer: Zero.
Teacher's Note:
a) The force he applies on the load is upward, while the displacement is horizontal, so they are perpendicular.
b) Work done = F s cos 90° = 0.
15. A car weighing 1000 kg and travelling at 30 m/s stops at a distance of 50 m, decelerating uniformly. What is the force exerted on it by the brakes? What is the work done by the brakes?
Answer:
1. From v2 = u2 + 2as: 0 = 900 + 2 × a × 50, so a = -9 m/s2
2. Force = ma = 1000 × 9 = 9000 N, opposite to the motion
3. Work done by the brakes = -F × s = -9000 × 50 = -4.5 × 105 J
Teacher's Note:
a) Check using energy: initial KE = \( \frac{1}{2} \) mv2 = 0.5 × 1000 × 900 = 4.5 × 105 J, equal to the work done by the brakes.
b) The sign is negative because the force opposes the motion.
INTEXT QUESTIONS PAGE NO. 148
1. A force of 10 N acts on an object. The displacement is, say 8 m, in the direction of the force. Let us take it that the force acts on the object through the displacement. What is the work done in this case?
Answer:
1. Work done = Force × Displacement
2. F = 10 N, s = 8 m
3. W = 10 × 8 = 80 Nm = 80 J
Teacher's Note:
a) The source solution wrongly uses F = 7 N (giving 56 J), which is the figure from a different version of the question. The question states 10 N, so the answer is 80 J.
b) Force and displacement are in the same direction.
INTEXT QUESTIONS PAGE NO. 149
1. When do we say that work is done?
Answer: Work is done whenever two conditions are satisfied: (i) a force acts on the body, and (ii) there is a displacement of the body caused by the applied force, along the direction of the applied force.
Teacher's Note:
a) Both conditions must hold together.
b) If either force or displacement is zero, no work is done.
2. Write an expression for the work done when a force is acting on an object in the direction of its displacement.
Answer: When a force F displaces a body through a distance s in the direction of the force, the work done is W = F × s (Work done = Force × Displacement).
Teacher's Note:
a) Units: force in newton, displacement in metre, work in joule.
b) If the force is at an angle, only the component along the displacement is used.
3. Define 1 J of work.
Answer: 1 J is the amount of work done by a force of 1 N on an object that displaces it through a distance of 1 m in the direction of the applied force.
Teacher's Note:
a) 1 J = 1 N × 1 m = 1 Nm.
b) Remember the direction condition in the definition.
4. A pair of bullocks exerts a force of 140 N on a plough. The field being ploughed is 15 m long. How much work is done in ploughing the length of the field?
Answer:
1. W = F × d
2. F = 140 N, d = 15 m
3. W = 140 × 15 = 2100 J
Teacher's Note:
a) The force is along the direction of motion, so the work is positive.
b) Hence 2100 J of work is done.
ENERGY
The energy of a body is defined as its capacity to do work. Energy is measured in terms of work, so the unit of energy is also the joule. One joule of energy is required to do one joule of work.
DIFFERENT FORMS OF ENERGY
We live in a world where energy exists in many forms. Some important forms are mechanical energy, chemical energy, light energy, heat energy, electrical energy, nuclear energy and sound energy.
MECHANICAL ENERGY
The energy used to displace a body, to change the position of a body or to deform a body is known as mechanical energy. It is of two types: i) kinetic energy and ii) potential energy.
KINETIC ENERGY
The energy possessed by an object due to its motion is called kinetic energy. The kinetic energy of an object increases with its speed. The kinetic energy of an object moving with a velocity is equal to the work done on it to make it acquire that velocity.
Example 1: The kinetic energy of a hammer is used to drive a nail into the wall.
Example 2: A bullet fired from a gun can penetrate a target due to its kinetic energy.
EXPRESSION FOR KINETIC ENERGY
Let a body (ball) of mass m move with an initial velocity v. If it is brought to rest by applying a retarding (opposing) force F, it comes to rest after a displacement s. Let Ek = work done against the force used to stop it.
Ek = F × s ......(1)
The retarding force F = ma ......(2)
Initial velocity u = v, final velocity = 0. From the third equation of motion, 0 = v2 - 2as, so 2as = v2.
Displacement s = \( \frac{v2}{2a} \) ......(3)
Substituting (2) and (3) in (1): Ek = ma × \( \frac{v2}{2a} \)
Ek = \( \frac{1}{2} \) mv2
The kinetic energy of a moving object is half the product of the mass of the object and the square of its speed.
Work done (W) = Change in kinetic energy
W = \( \frac{1}{2} \) mv2 - \( \frac{1}{2} \) mu2 (when v > u)
or W = \( \frac{1}{2} \) mu2 - \( \frac{1}{2} \) mv2 (when u > v)
INTEXT QUESTIONS PAGE NO. 152
1. What is the kinetic energy of an object?
Answer: Kinetic energy is the energy possessed by a body by virtue of its motion. Every moving object possesses kinetic energy and uses it to do work. For example, the kinetic energy of a hammer drives a nail into a log of wood, and the kinetic energy of air runs wind mills.
Teacher's Note:
a) Kinetic energy depends on mass and speed.
b) It is always zero or positive.
2. Write an expression for the kinetic energy of an object.
Answer: If a body of mass m is moving with a velocity v, its kinetic energy is Ek = \( \frac{1}{2} \) mv2. Its SI unit is the joule (J).
Teacher's Note:
a) Velocity is squared, so doubling speed makes KE four times.
b) Mass is in kg and velocity in m/s.
3. The kinetic energy of an object of mass m moving with a velocity of 5 m/s is 25 J. What will be its kinetic energy when its velocity is doubled? What will be its kinetic energy when its velocity is increased three times?
Answer:
1. Ek is proportional to v2
2. (i) Velocity doubled to 10 m/s: KE becomes 22 = 4 times, so KE = 25 × 4 = 100 J
3. (ii) Velocity tripled to 15 m/s: KE becomes 32 = 9 times, so KE = 25 × 9 = 225 J
Teacher's Note:
a) Use the proportion, no need to find the mass (m = 2 kg).
b) KE changes with the square of the velocity ratio.
NUMERICALS ON KINETIC ENERGY
21. How far should a man of mass 60 kg run so that his kinetic energy is 750 J?
Answer:
1. Ek = \( \frac{1}{2} \) mv2, so 750 = \( \frac{1}{2} \) × 60 × v2
2. v2 = 750 / 30 = 25, so v = 5 m/s
3. The kinetic energy depends on his speed, not on the distance run.
Teacher's Note:
a) The question is not well posed. It is the speed (5 m/s) that gives 750 J.
b) The distance cannot be found unless the time of running or the accelerating force is given.
22. Find the mass of the body which has 5 J of kinetic energy while moving at a speed of 2 m/s.
Answer:
1. m = 2Ek / v2 = (2 × 5) / 4 = 2.5 kg
Teacher's Note:
a) Rearrange Ek = \( \frac{1}{2} \) mv2 for m.
b) Speed is squared before dividing.
23. A player kicks a ball of mass 250 g at the centre of a field. The ball leaves his foot with a speed of 10 m/s. Find the work done by the player on the ball.
Answer:
1. m = 250 g = 0.25 kg
2. Work done = KE gained = \( \frac{1}{2} \) × 0.25 × 102 = 12.5 J
Teacher's Note:
a) Change grams to kilograms first.
b) The ball starts from rest, so work done equals its final KE.
24. A body of mass 5 kg, initially at rest, is subjected to a force of 20 N. What is the kinetic energy acquired by the body at the end of 10 s?
Answer:
1. a = F/m = 20/5 = 4 m/s2
2. v = u + at = 0 + 4 × 10 = 40 m/s
3. Ek = \( \frac{1}{2} \) × 5 × 402 = 4000 J
Teacher's Note:
a) First find the acceleration, then the velocity.
b) Check: distance = \( \frac{1}{2} \) × 4 × 100 = 200 m and F × s = 20 × 200 = 4000 J.
25. A bullet of mass 20 g moving with a velocity of 500 m/s strikes a tree and goes out from the other side with a velocity of 400 m/s. Calculate the work done by the bullet in joule in passing through the tree.
Answer:
1. m = 0.02 kg
2. Initial KE = \( \frac{1}{2} \) × 0.02 × 5002 = 2500 J
3. Final KE = \( \frac{1}{2} \) × 0.02 × 4002 = 1600 J
4. Work done by the bullet = loss of KE = 2500 - 1600 = 900 J
Teacher's Note:
a) The bullet loses 900 J of kinetic energy while pushing through the tree.
b) The tree does -900 J of work on the bullet.
26. An object of mass 15 kg is moving with a uniform velocity of 4 m/s. What is the kinetic energy possessed by the object?
Answer:
1. Ek = \( \frac{1}{2} \) × 15 × 42 = 120 J
Teacher's Note:
a) Square the velocity before multiplying.
b) Unit: kg m2/s2 = J.
27. What is the work done to increase the velocity of a car from 30 km/h to 60 km/h if the mass of the car is 1500 kg?
Answer:
1. u = 30 km/h = 25/3 m/s = 8.33 m/s, v = 60 km/h = 50/3 m/s = 16.67 m/s
2. W = \( \frac{1}{2} \) m (v2 - u2) = 750 × (277.78 - 69.44)
3. W = 750 × 208.33 = 156250 J = 1.56 × 105 J
Teacher's Note:
a) Convert km/h to m/s by multiplying by 5/18.
b) Work done is the change in kinetic energy, not the difference of the speeds.
28. A bullet of mass 0.03 kg moving with a velocity of 400 m/s penetrates 12 cm into a fixed target, which offers a constant resistive force of 1000 N to the motion of the bullet. Find (a) the initial kinetic energy of the bullet, (b) the distance through which the bullet has penetrated.
Answer:
1. (a) Ek = \( \frac{1}{2} \) × 0.03 × 4002 = 2400 J
2. (b) Work done against the resistance = KE lost, so F × d = 2400
3. d = 2400 / 1000 = 2.4 m
Teacher's Note:
a) The question as printed also says 12 cm, which does not agree with a 1000 N resistive force. With 12 cm penetration the resistive force would be 2400 / 0.12 = 20000 N.
b) Part (b) is answered from the 1000 N force as the question asks.
29. Two bodies of equal masses move with uniform velocities v and 3v respectively. Find the ratio of their kinetic energies.
Answer:
1. KE1 = \( \frac{1}{2} \) mv2
2. KE2 = \( \frac{1}{2} \) m (3v)2 = \( \frac{9}{2} \) mv2
3. KE1 : KE2 = 1 : 9
Teacher's Note:
a) For equal masses, KE is proportional to v2.
b) The faster body has 9 times the KE.
30. The mass of a ball A is double the mass of another ball B. The ball A moves at half the speed of the ball B. Calculate the ratio of the kinetic energy of A to the kinetic energy of B.
Answer:
1. Let mass of B = m and speed of B = v
2. KE of A = \( \frac{1}{2} \) × 2m × (v/2)2 = \( \frac{mv2}{4} \)
3. KE of B = \( \frac{1}{2} \) mv2
4. KEA : KEB = \( \frac{1}{4} \) : \( \frac{1}{2} \) = 1 : 2
Teacher's Note:
a) Doubling the mass doubles KE, halving the speed divides it by 4.
b) Net factor = 2 × \( \frac{1}{4} \) = \( \frac{1}{2} \).
31. A truck weighing 5000 kgf and a cart weighing 500 kgf are moving with the same speed. Compare their kinetic energies.
Answer:
1. Same speed, so KE is proportional to mass
2. KEtruck : KEcart = 5000 : 500 = 10 : 1
Teacher's Note:
a) Weights in kgf are proportional to the masses in kg.
b) The truck has 10 times the kinetic energy.
32. A bullet of mass 20 g is found to pass two points 30 m apart in 4 s. Assuming the speed to be constant, find its kinetic energy.
Answer:
1. v = distance / time = 30 / 4 = 7.5 m/s
2. m = 0.02 kg
3. Ek = \( \frac{1}{2} \) × 0.02 × 7.52 = 0.5625 J
Teacher's Note:
a) Speed is found from distance and time, since it is constant.
b) This is a very slow bullet, but the question gives these values.
POTENTIAL ENERGY
The energy possessed by a body by virtue of its position or due to a state of strain is called potential energy. The work done to lift a body above the ground level gives the potential energy of the body, for example in weight lifting.
Example: Water stored in a reservoir has a large amount of potential energy, due to which it can drive a water turbine when allowed to fall. This is the principle of production of hydroelectric energy.
EXPRESSION FOR POTENTIAL ENERGY OF A BODY ABOVE THE GROUND LEVEL
The work done in raising an object from the ground to a certain height against gravity is stored in the body as potential energy. Consider an object of mass m raised through a height h from the ground. A force is needed to do this.
The downward force acting on the body due to gravity = mg.
The work has to be done to lift the body through a height h against the force of gravity.
Work done = force × displacement
w = F × h
w = mgh [since F = ma and a = g, therefore F = mg]
Work done is equal to the potential energy of the object.
Ep = mgh
Note: The potential energy of an object at a height depends on the ground level or zero level you choose. An object in a given position can have a certain potential energy with respect to one level and a different value with respect to another level.
NUMERICALS ON POTENTIAL ENERGY
Unless a question gives g, take g = 10 m/s2.
1. A body of mass 4 kg is taken from a height of 5 m to a height of 10 m. Find the increase in potential energy.
Answer:
1. Rise in height = 10 - 5 = 5 m
2. Increase in Ep = mgh = 4 × 10 × 5 = 200 J
Teacher's Note:
a) Only the change in height matters.
b) With g = 9.8 m/s2 the answer is 196 J.
2. An object of mass 1 kg is raised through a height h. Its potential energy increases by 1 J. Find the height h.
Answer:
1. Ep = mgh, so h = Ep / mg = 1 / (1 × 10) = 0.1 m = 10 cm
Teacher's Note:
a) Rearrange the formula for h.
b) With g = 9.8 m/s2, h = 0.102 m.
3. A 5 kg ball is thrown upwards with a speed of 10 m/s. (a) Find the potential energy when it reaches the highest point. (b) Calculate the maximum height attained by it.
Answer:
1. (a) At the highest point all the KE becomes PE: Ep = \( \frac{1}{2} \) × 5 × 102 = 250 J
2. (b) mgh = 250, so h = 250 / (5 × 10) = 5 m
Teacher's Note:
a) Use conservation of energy: KE at the start = PE at the top.
b) Check with v2 = 2gh: 100 = 2 × 10 × 5.
4. A 5 kg ball is dropped from a height of 10 m. (a) Find the initial potential energy of the ball. (b) Find the kinetic energy just before it reaches the ground. (c) Calculate the velocity before it reaches the ground.
Answer:
1. (a) Ep = mgh = 5 × 10 × 10 = 500 J
2. (b) By conservation of energy, KE just before landing = 500 J
3. (c) \( \frac{1}{2} \) × 5 × v2 = 500, so v2 = 200 and v = 14.1 m/s
Teacher's Note:
a) All the PE becomes KE just before the ball reaches the ground.
b) v = \( \sqrt{2gh} \) = \( \sqrt{200} \) = 14.1 m/s.
5. A body is thrown up with a kinetic energy of 10 J. If it attains a maximum height of 5 m, find the mass of the body.
Answer:
1. At the top, PE = initial KE, so mgh = 10
2. m = 10 / (10 × 5) = 0.2 kg
Teacher's Note:
a) Initial KE equals the PE at the highest point.
b) 0.2 kg = 200 g.
6. A rocket of mass 3 × 106 kg takes off from a launching pad and acquires a vertical velocity of 1 km/s and an altitude of 25 km. Calculate its (a) potential energy and (b) kinetic energy.
Answer:
1. m = 3 × 106 kg, h = 25 km = 25000 m, v = 1 km/s = 1000 m/s
2. (a) Ep = mgh = 3 × 106 × 10 × 25000 = 7.5 × 1011 J
3. (b) Ek = \( \frac{1}{2} \) × 3 × 106 × 10002 = 1.5 × 1012 J
Teacher's Note:
a) Convert km to m and km/s to m/s.
b) With g = 9.8 m/s2, Ep = 7.35 × 1011 J.
7. Find the energy possessed by an object of mass 10 kg when it is at a height of 6 m above the ground. Given g = 9.8 m/s2.
Answer:
1. Ep = mgh = 10 × 9.8 × 6 = 588 J
Teacher's Note:
a) The object at rest has only potential energy.
b) Use the g given in the question.
8. An object of mass 12 kg is at a certain height above the ground. If the potential energy of the object is 480 J, find the height at which the object is with respect to the ground. Given g = 10 m/s2.
Answer:
1. h = Ep / mg = 480 / (12 × 10) = 4 m
Teacher's Note:
a) Rearrange Ep = mgh.
b) Height is measured from the ground level.
9. Calculate the increase in potential energy as a block of 2 kg is lifted through 2 m.
Answer:
1. Increase in Ep = mgh = 2 × 10 × 2 = 40 J
Teacher's Note:
a) With g = 9.8 m/s2 the answer is 39.2 J.
b) Increase in PE equals the work done in lifting.
10. A ball of mass 1 kg is dropped from a height of 5 m. (a) Find the kinetic energy of the ball just before it reaches the ground. (b) What is the speed at this instant?
Answer:
1. (a) KE = loss of PE = mgh = 1 × 10 × 5 = 50 J
2. (b) \( \frac{1}{2} \) × 1 × v2 = 50, so v2 = 100 and v = 10 m/s
Teacher's Note:
a) Use conservation of mechanical energy.
b) Check: v2 = 2gh = 100.
11. A block of mass 30 kg is pulled up by a rope as shown in the figure with a constant speed by applying a force of 200 N parallel to the slope. A and B are the initial and final positions of the block. Calculate (a) the work done by the force in moving the block from A to B, (b) the potential energy gained by the block, (c) account for the difference between the work done by the force and the increase in potential energy of the block.
Answer:
1. From the figure, AB = 3 m along the slope and the vertical height BC = 1.5 m
2. (a) Work done by the force = 200 × 3 = 600 J
3. (b) PE gained = mgh = 30 × 10 × 1.5 = 450 J
4. (c) Difference = 600 - 450 = 150 J. This is the work done against friction between the block and the slope, which appears as heat
Teacher's Note:
a) The block moves at constant speed, so there is no gain in kinetic energy.
b) Efficiency of the slope = 450 / 600 = 75 %.
12. A body of mass 5 kg falls from a height of 5 m. How much energy does it possess at any instant?
Answer:
1. At the start the body has only PE = mgh = 5 × 10 × 5 = 250 J
2. During the fall PE is converted into KE, but the total mechanical energy stays constant
3. Total energy at any instant = 250 J
Teacher's Note:
a) This is the law of conservation of mechanical energy.
b) Air resistance is ignored.
13. An 800 g ball is pulled up a slope as shown in the diagram. Calculate the potential energy it gains.
Answer:
1. m = 800 g = 0.8 kg
2. Vertical height of the slope = 20 cm = 0.2 m (the 50 cm is the length along the slope)
3. Ep = mgh = 0.8 × 10 × 0.2 = 1.6 J
Teacher's Note:
a) Only the vertical height counts for potential energy.
b) The slope length of 50 cm is not needed.
14. A spring is compressed by a toy cart of mass 150 g. On releasing, the cart moves with a speed of 0.2 m/s. Calculate the elastic potential energy of the spring.
Answer:
1. m = 150 g = 0.15 kg
2. Elastic PE stored in the spring = KE given to the cart
3. E = \( \frac{1}{2} \) × 0.15 × 0.22 = 0.003 J
Teacher's Note:
a) The stored spring energy is completely converted to kinetic energy of the cart.
b) 0.003 J = 3 mJ.
15. An object of mass 40 kg is raised to a height of 5 m above the ground. What is its potential energy? If the object is allowed to fall, find its kinetic energy when it is half-way down.
Answer:
1. Ep = mgh = 40 × 10 × 5 = 2000 J
2. Half-way down, the PE left is 1000 J, so KE gained = 2000 - 1000 = 1000 J
Teacher's Note:
a) At the half-way point PE and KE are equal.
b) With g = 9.8 m/s2 the values are 1960 J and 980 J.
16. A box has a mass of 5.8 kg. The box is lifted from the garage floor and placed on a shelf. If the box gains 145 J of potential energy, how high is the shelf?
Answer:
1. h = Ep / mg = 145 / (5.8 × 10) = 2.5 m
Teacher's Note:
a) Rearrange Ep = mgh for h.
b) Height is measured from the garage floor.
17. A man climbs on to a wall that is 3.6 m high and gains 2268 J of potential energy. What is the mass of the man?
Answer:
1. m = Ep / gh = 2268 / (10 × 3.6) = 63 kg
Teacher's Note:
a) With g = 10 m/s2 the mass is a round 63 kg.
b) Rearrange Ep = mgh for m.
18. The figure shows a ski-jump. A skier of mass 60 kg stands at A at the top of the ski-jump. He moves from A to B and takes off for his jump at B. (a) Calculate the change in the gravitational potential energy of the skier between A and B. (b) If 75% of the energy in part (a) becomes the kinetic energy at B, calculate the speed at which the skier arrives at B.
Answer:
1. From the figure, A is at 75 m and B at 15 m above the ground, so the drop = 75 - 15 = 60 m
2. (a) Change in PE = mgh = 60 × 10 × 60 = 36000 J (a decrease)
3. (b) KE at B = 75% of 36000 = 27000 J
4. 27000 = \( \frac{1}{2} \) × 60 × v2, so v2 = 900 and v = 30 m/s
Teacher's Note:
a) The remaining 25% (9000 J) is lost as heat due to friction and air resistance.
b) The speed is 30 m/s = 108 km/h.
19. Consider the case of a freely falling body in the figure. At A, kinetic energy = 0 and potential energy = mgh. At B, kinetic energy = mgx. At C, kinetic energy = mgh and potential energy = 0. (a) Find the potential energy of the body at B. (b) Find the total energy at A, B and C. (c) Is there any variation in total energy? What do you infer from the result?
Answer:
1. (a) At B the body is at a height (h - x) above the ground, so PE = mg(h - x)
2. (b) Total at A = 0 + mgh = mgh. Total at B = mgx + mg(h - x) = mgh. Total at C = mgh + 0 = mgh
3. (c) No, the total energy is the same at A, B and C. Mechanical energy is conserved during free fall: PE is converted into KE without any loss
Teacher's Note:
a) The body has fallen a distance x when at B, so KE = mgx.
b) This is the law of conservation of energy.
20. A bag of wheat weighs 200 kg. To what height should it be raised so that its potential energy may be 9800 joules? (g = 9.8 m/s2)
Answer:
1. h = Ep / mg = 9800 / (200 × 9.8) = 5 m
Teacher's Note:
a) Use g = 9.8 m/s2 as given.
b) Rearrange Ep = mgh for h.
POWER
Power is defined as the rate of doing work or the rate of transfer of energy. If an agent does work W in time t, then power is given by:
Power = Work / Time = Energy / Time, P = W / t
The unit of power is the watt (symbol W). 1 watt is the power of an agent which does work at the rate of 1 joule per second. Power is 1 W when the rate of consumption of energy is 1 J/s.
1 watt = 1 joule/second or 1 W = 1 J/s
LARGER UNITS OF POWER
1 kilowatt = 1000 watts, 1 kW = 1000 W = 1000 J/s
A unit used exclusively in engineering is the horse power (hp): 1 hp = 746 W
The commercial unit of energy is the kilowatt hour (kWh).
1 kWh = 1 kW × 1 h = 1000 W × 3600 s = 3600000 J = 3.6 × 106 J = 3.6 MJ (megajoule)
One kilowatt hour is the amount of energy consumed by an agent in one hour working at a constant rate of one kilowatt. It is also called the unit of electrical energy.
INTEXT QUESTIONS PAGE NO. 156
1. What is power?
Answer: Power is the rate of doing work or the rate of transfer of energy. If W is the work done in time t, then Power = Work / Time = Energy / Time, so P = W / t. It is expressed in watt (W).
Teacher's Note:
a) Power tells how fast work is done.
b) Two agents doing the same work differ in power if they take different times.
2. Define 1 watt of power.
Answer: 1 watt is the power of an agent which does work at the rate of 1 joule per second. Power is 1 W when the rate of consumption of energy is 1 J/s, that is, 1 W = 1 J/s.
Teacher's Note:
a) 1 kW = 1000 W.
b) 1 hp = 746 W.
3. A lamp consumes 1000 J of electrical energy in 10 s. What is its power?
Answer:
1. W = 1000 J, t = 10 s
2. P = W / t = 1000 / 10 = 100 W
Teacher's Note:
a) Power = energy / time.
b) The unit is the watt.
4. Define average power.
Answer: A body can do different amounts of work in different time intervals, so it is better to define average power. Average power = total energy consumed / total time taken, that is, the total work done divided by the total time taken to do it.
Teacher's Note:
a) Average power takes the whole interval into account.
b) If power is constant, average power equals the power at any instant.
NUMERICALS ON POWER
Unless a question gives g, take g = 10 m/s2.
1. Two girls each of weight 400 N climb up a rope through a height of 8 m. We name one of the girls A and the other B. Girl A takes 20 s while B takes 50 s to accomplish this task. What is the power expended by each girl?
Answer:
1. Work done by each girl = weight × height = 400 × 8 = 3200 J
2. Power of A = 3200 / 20 = 160 W
3. Power of B = 3200 / 50 = 64 W
Teacher's Note:
a) Both do the same work, but A does it faster, so A has more power.
b) Here the weight is given in newton, so g is not needed.
2. A boy of mass 50 kg runs up a staircase of 45 steps in 9 s. If the height of each step is 15 cm, find his power. Take g = 10 m/s2.
Answer:
1. Height = 45 × 0.15 = 6.75 m
2. Work = mgh = 50 × 10 × 6.75 = 3375 J
3. Power = 3375 / 9 = 375 W
Teacher's Note:
a) Change 15 cm to 0.15 m.
b) 375 W is about half a horse power.
3. An electric bulb of 60 W is used for 6 hours per day. Calculate the units of energy consumed in one day by the bulb.
Answer:
1. P = 60 W = 0.06 kW, t = 6 h
2. Energy = P × t = 0.06 × 6 = 0.36 kWh
3. Units consumed = 0.36 unit
Teacher's Note:
a) 1 unit = 1 kWh.
b) Convert watts to kilowatts before multiplying by hours.
4. A 60 kg person climbs stairs of total height 20 m in 2 min. Calculate the power delivered.
Answer:
1. Work = mgh = 60 × 10 × 20 = 12000 J
2. t = 2 min = 120 s
3. Power = 12000 / 120 = 100 W
Teacher's Note:
a) Convert minutes to seconds.
b) With g = 9.8 m/s2 the power is 98 W.
5. The work done by the heart is 1 J per beat. Calculate the power of the heart if it beats 72 times/min.
Answer:
1. Work in 1 minute = 72 × 1 = 72 J
2. t = 60 s
3. Power = 72 / 60 = 1.2 W
Teacher's Note:
a) Total work over a minute divided by 60 s.
b) The heart works at about 1.2 W.
6. A man exerts a force of 200 N in pulling a cart at a constant speed of 16 m/s. Calculate the power spent by the man.
Answer:
1. P = W / t = F × s / t = F × v
2. P = 200 × 16 = 3200 W = 3.2 kW
Teacher's Note:
a) Power = force × velocity when the force is along the motion.
b) 16 m/s is very fast for pulling a cart, but this is the value given.
7. Calculate the power of an engine required to lift 105 kg of coal per hour from a mine 360 m deep.
Answer:
1. Work per hour = mgh = 105 × 10 × 360 = 3.6 × 108 J
2. t = 1 h = 3600 s
3. Power = 3.6 × 108 / 3600 = 105 W = 100 kW
Teacher's Note:
a) The mass is 105 kg, read from the printed page as the text layer loses the superscript.
b) Power = work / time with time in seconds.
8. A man does 200 J of work in 10 s and a boy does 100 J of work in 4 s. (a) Who is delivering more power? (b) Find the ratio of the power delivered by the man to that delivered by the boy.
Answer:
1. Power of man = 200 / 10 = 20 W
2. Power of boy = 100 / 4 = 25 W
3. (a) The boy delivers more power
4. (b) Ratio = 20 : 25 = 4 : 5
Teacher's Note:
a) Compare power, not work done.
b) The man does more work but takes longer.
9. A boy of mass 40 kg runs up a flight of 50 steps, each of 10 cm high, in 5 s. Find the power developed by the boy.
Answer:
1. Height = 50 × 0.1 = 5 m
2. Work = mgh = 40 × 10 × 5 = 2000 J
3. Power = 2000 / 5 = 400 W
Teacher's Note:
a) Convert cm to m.
b) Power = work / time.
10. A car of mass 2000 kg is lifted up a distance of 30 m by a crane in 1 min. A second crane does the same job in 2 min. What is the power applied by each crane? Do the cranes consume the same or different amounts of fuel? Neglect power dissipation against friction.
Answer:
1. Work = mgh = 2000 × 10 × 30 = 6 × 105 J
2. Power of crane 1 = 6 × 105 / 60 = 10000 W = 10 kW
3. Power of crane 2 = 6 × 105 / 120 = 5000 W = 5 kW
4. Both do the same work, so with no losses they consume the same amount of fuel
Teacher's Note:
a) Fuel used depends on the energy supplied, which equals the work done.
b) The faster crane has the higher power, not a higher fuel use.
11. What should be the power of an engine required to lift 90 metric tonnes of coal per hour from a mine whose depth is 200 m?
Answer:
1. m = 90 tonnes = 90000 kg, t = 3600 s
2. Work = mgh = 90000 × 10 × 200 = 1.8 × 108 J
3. Power = 1.8 × 108 / 3600 = 50000 W = 50 kW
Teacher's Note:
a) 1 metric tonne = 1000 kg.
b) Power = work / time.
12. How much time does it take to perform 500 J of work at a rate of 10 W?
Answer:
1. t = W / P = 500 / 10 = 50 s
Teacher's Note:
a) Rearrange P = W / t.
b) Rate of work is the power.
13. Calculate the units of energy consumed by a 100 W electric bulb in 5 hours.
Answer:
1. P = 100 W = 0.1 kW
2. Energy = 0.1 × 5 = 0.5 kWh = 0.5 unit
Teacher's Note:
a) 1 unit = 1 kWh.
b) 0.5 kWh = 1.8 × 106 J.
14. A lift is designed to carry a load of 4000 kg through 10 floors of a building, averaging 6 m per floor, in 10 s. Calculate the power of the lift.
Answer:
1. Total height = 10 × 6 = 60 m
2. Work = mgh = 4000 × 10 × 60 = 2.4 × 106 J
3. Power = 2.4 × 106 / 10 = 2.4 × 105 W = 240 kW
Teacher's Note:
a) Find the total height first.
b) This ignores friction and the weight of the lift cage.
ENERGY TRANSFORMATION
Water from a dam: potential energy into kinetic energy
Microphone: sound energy into electrical energy
TV camera: light energy into electrical energy
Solar cell: light energy into electrical energy
Iron box: electrical energy into heat energy
Loudspeaker: electrical energy into sound energy
Fan: electrical energy into mechanical energy
Light: electrical energy into light energy
LAW OF CONSERVATION OF ENERGY
Energy in a system may take on various forms (for example kinetic, potential, heat, light). The law of conservation of energy states that energy may neither be created nor destroyed. Therefore the sum of all the energies in the system is a constant.
The most commonly used example is the pendulum.
The formula to calculate the potential energy is PE = mgh.
The mass of the ball = 10 kg, the height h = 0.2 m, g = 9.8 m/s2.
PE = 10 × 9.8 × 0.2 = 19.6 J
At the extreme position A (blue ball), the potential energy is 19.6 J and the kinetic energy is 0. As the ball approaches the mean position P, PE decreases and KE increases. At exactly halfway between A and P, PE = KE.
At the mean position P (purple ball), KE is maximum while PE = 0. Theoretically, all the PE has changed into KE, so KE = 19.6 J and PE = 0.
At the other extreme position B (pink ball), PE is again maximum and KE = 0.
The sum of PE and KE is the total mechanical energy: Total mechanical energy = PE + KE = constant. The source writes this as PE + KE = 0 and PE = - KE, which is meant to say that any decrease in PE equals the increase in KE (the change in PE + KE is zero).
COMMERCIAL UNIT OF ENERGY
The joule is too small a unit to express large quantities of energy, so a bigger unit called the kilowatt hour (kWh) is used. The commercial unit of electrical energy is the kilowatt hour, commonly known as the "unit". 1 kWh is the amount of energy consumed by an electrical gadget in one hour at the rate of 1000 J/s or 1 kW.
1 kWh = 1 kW × 1 h = 1000 W × 3600 s = 3600000 J = 3.6 × 106 J
Energy used in households, industries and commercial establishments is usually expressed in kilowatt hour. For example, the electrical energy used during a month is expressed in "units", where 1 unit means 1 kilowatt hour.
EXERCISE QUESTIONS PAGE NO. 158 AND 159
1. Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term work. (a) Suma is swimming in a pond. (b) A donkey is carrying a load on its back. (c) A wind-mill is lifting water from a well. (d) A green plant is carrying out photosynthesis. (e) An engine is pulling a train. (f) Food grains are getting dried in the sun. (g) A sailboat is moving due to wind energy.
Answer:
1. Work is done when (i) a force acts on the body and (ii) there is a displacement caused by the force, along or opposite to its direction
2. (a) Suma pushes the water backwards and the forward reaction of water moves her forward. Force causes displacement, so work is done by Suma
3. (b) The donkey applies an upward force on the load but the displacement is forward, perpendicular to the force, so work done is zero
4. (c) The wind-mill works against gravity to lift water, so work is done
5. (d) There is no displacement of any body, so the work done is zero
6. (e) The engine applies a force and the train moves in the same direction, so work is done
7. (f) The grains do not move, so the work done is zero
8. (g) Wind applies a force on the sailboat and the boat moves in that direction, so work is done by the wind
Teacher's Note:
a) The source solution calls Suma "Seema" in part (a); the name in the question is Suma.
b) Force and displacement must both be present, and the displacement must not be perpendicular to the force.
2. An object thrown at a certain angle to the ground moves in a curved path and falls back to the ground. The initial and the final points of the path of the object lie on the same horizontal line. What is the work done by the force of gravity on the object?
Answer:
1. Work done by gravity depends only on the vertical displacement
2. Vertical displacement = final height - initial height = 0
3. W = mgh = mg × 0 = 0 J
Teacher's Note:
a) The path taken does not matter, only the net vertical change.
b) The work done by gravity is zero during the whole flight.
3. A battery lights a bulb. Describe the energy changes involved in the process.
Answer:
1. The chemical energy of the battery is changed into electrical energy
2. The bulb converts the electrical energy into light energy and heat energy
3. Chemical energy → Electrical energy → Light energy + Heat energy
Teacher's Note:
a) The source writes "Chemistry Energy" in the chain; it should be Chemical Energy.
b) Total energy is conserved at every step.
4. A certain force acting on a 20 kg mass changes its velocity from 5 m/s to 2 m/s. Calculate the work done by the force.
Answer:
1. KE at 5 m/s = \( \frac{1}{2} \) × 20 × 52 = 250 J
2. KE at 2 m/s = \( \frac{1}{2} \) × 20 × 22 = 40 J
3. Work done = change in KE = 40 - 250 = -210 J
Teacher's Note:
a) The negative sign shows the force acts opposite to the motion.
b) Work done equals change in kinetic energy.
5. A mass of 10 kg is at a point A on a table. It is moved to a point B. If the line joining A and B is horizontal, what is the work done on the object by the gravitational force? Explain your answer.
Answer:
1. Work done by gravity depends only on the vertical displacement, not on the path
2. Vertical displacement h = 0
3. W = mgh = mg × 0 = 0
Teacher's Note:
a) Gravity acts downwards while the motion is horizontal, so they are perpendicular.
b) Hence the work done by gravity is zero.
6. The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?
Answer:
1. No, it does not violate the law of conservation of energy
2. As the body falls, its potential energy changes into kinetic energy progressively
3. The decrease in PE equals the increase in KE, so the total mechanical energy stays constant
Teacher's Note:
a) Energy is changing form, not disappearing.
b) Total energy = PE + KE is the same at every height.
7. What are the various energy transformations that occur when you are riding a bicycle?
Answer:
1. The muscular (chemical) energy of the rider changes into heat energy and kinetic energy of the bicycle
2. Heat energy warms the rider and the parts of the bicycle, and kinetic energy gives velocity to the bicycle
3. Muscular energy → Kinetic energy + Heat energy, with the total energy conserved
Teacher's Note:
a) Some energy is always lost as heat due to friction.
b) Total energy remains the same.
8. Does the transfer of energy take place when you push a huge rock with all your might and fail to move it? Where is the energy you spend going?
Answer:
1. There is no displacement of the rock, so no work is done on it and no energy is transferred to it
2. The muscular energy we spend is converted into heat energy, which makes our body hot
Teacher's Note:
a) The source solution says "there is no loss of energy because muscular energy is transferred into heat". This is reworded: no energy goes to the rock, and the energy spent is changed into heat in the body.
b) No displacement means zero work.
9. A certain household has consumed 250 units of energy during a month. How much energy is this in joules?
Answer:
1. 1 unit = 1 kWh = 3.6 × 106 J
2. 250 units = 250 × 3.6 × 106 = 9 × 108 J
Teacher's Note:
a) 1 unit of electrical energy is 1 kWh.
b) 9 × 108 J = 900 MJ.
10. An object of mass 40 kg is raised to a height of 5 m above the ground. What is its potential energy? If the object is allowed to fall, find its kinetic energy when it is half-way down.
Answer:
1. Ep = mgh = 40 × 9.8 × 5 = 1960 J
2. Half-way down, PE = 1960 / 2 = 980 J
3. Total energy is conserved, so KE = 1960 - 980 = 980 J
Teacher's Note:
a) Here g = 9.8 m/s2 as in the book.
b) At half-way, PE and KE are equal.
11. What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer.
Answer:
1. The force of gravity on the satellite acts towards the centre of the earth
2. The displacement of the satellite is along its path, which is perpendicular to this force
3. Hence the work done by gravity on the satellite is zero
Teacher's Note:
a) For a circular orbit the angle between force and displacement is 90°.
b) So the speed and KE of the satellite stay constant.
12. Can there be displacement of an object in the absence of any force acting on it? Think. Discuss this question with your friends and teacher.
Answer:
1. Yes, for an object moving with constant velocity
2. The net force on it is zero, yet it is displaced along its path
Teacher's Note:
a) By Newton's first law the object continues in uniform motion without a force.
b) Work done is zero because there is no force.
13. A person holds a bundle of hay over his head for 30 minutes and gets tired. Has he done some work or not? Justify your answer.
Answer:
1. There is no displacement of the bundle of hay
2. Although he applies a force, with no displacement the work done on the bundle is zero
3. He gets tired because of muscular effort inside the body, not because of work done on the hay
Teacher's Note:
a) The source solution says no force is applied by the person, which is wrong: he applies an upward force equal to the weight of the hay.
b) Work = force × displacement = F × 0 = 0.
14. An electric heater is rated 1500 W. How much energy does it use in 10 hours?
Answer:
1. P = 1500 W = 1.5 kW, t = 10 h
2. Energy = P × t = 1.5 × 10 = 15 kWh
Teacher's Note:
a) Energy consumed equals work done.
b) 15 kWh = 15 units.
15. Illustrate the law of conservation of energy by discussing the energy changes which occur when we draw a pendulum bob to one side and allow it to oscillate. Why does the bob eventually come to rest? What happens to its energy eventually? Is it a violation of the law of conservation of energy?
Answer:
1. The law states that energy can be neither created nor destroyed; it only changes from one form to another
2. When the bob is at an extreme position A or B it is raised by a height h above the mean position P and has only PE; its KE is zero
3. As it moves towards P, PE decreases and KE increases; at P the PE is zero and the bob has only KE. This repeats as it oscillates
4. The bob finally stops because of air resistance and friction at the support; its mechanical energy turns into heat, which passes to the surroundings
5. This is not a violation: the energy lost by the pendulum is gained by the surroundings, so the total energy is conserved
Teacher's Note:
a) Mechanical energy of the bob alone is not conserved, but total energy is.
b) Draw the pendulum figure with A, P and B in the answer.
16. An object of mass m is moving with a constant velocity v. How much work should be done on the object in order to bring the object to rest?
Answer:
1. KE of the object = \( \frac{1}{2} \) mv2
2. To bring it to rest, \( \frac{1}{2} \) mv2 of work has to be done on it (negative work, by a retarding force)
Teacher's Note:
a) Work done = change in KE = 0 - \( \frac{1}{2} \) mv2.
b) The work done is negative.
17. Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km/h.
Answer:
1. v = 60 km/h = 60 × 5/18 = 16.67 m/s
2. Ek = \( \frac{1}{2} \) × 1500 × (16.67)2 = 2.08 × 105 J
3. So 2.08 × 105 J (20.8 × 104 J) of work is required to stop the car
Teacher's Note:
a) Convert km/h to m/s first.
b) The work done by the brakes is negative: -2.08 × 105 J.
18. In each of the following a force F is acting on an object of mass m. The direction of displacement is from west to east shown by the longer arrow. Observe the diagrams carefully and state whether the work done by the force is negative, positive or zero.
Answer:
1. Case I: the force is perpendicular to the displacement, so the work done is zero
2. Case II: the force is in the direction of the displacement, so the work done is positive
3. Case III: the force is opposite to the displacement, so the work done is negative
Teacher's Note:
a) Work is positive, zero or negative depending on the angle between force and displacement.
b) Angles 0°, 90° and 180° give these three cases.
19. Soni says that the acceleration in an object could be zero even when several forces are acting on it. Do you agree with her? Why?
Answer:
1. Yes, I agree with Soni
2. When the forces cancel each other, the net force on the object is zero
3. With zero net force the acceleration is zero, as for an object moving with uniform velocity
Teacher's Note:
a) Acceleration depends on the net force, not on each force.
b) Example: a book resting on a table.
20. Find the energy in kWh consumed in 10 hours by four devices of power 500 W each.
Answer:
1. Power of one device = 500 W = 0.5 kW
2. Energy of one device in 10 h = 0.5 × 10 = 5 kWh
3. Energy of four devices = 4 × 5 = 20 kWh = 20 units
Teacher's Note:
a) Total power = 4 × 0.5 = 2 kW, and 2 × 10 = 20 kWh.
b) 1 kWh = 1 unit.
21. A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy?
Answer:
1. While falling, PE decreases and KE increases; just before touching the ground all the PE has become KE
2. When the object hits the ground its KE is changed into heat energy and sound energy, and may deform the ground
Teacher's Note:
a) The energy is not destroyed, it is changed into other forms.
b) The amount of deformation depends on the ground and on the KE.
i) kilojoule (1000 joule)
ii) megajoule (10 lakh joule)
v = Velocity of the object = 5 m/s
Given that kinetic energy, Ek = 25 J
(i) If the velocity of an object is doubled, then v = 5 × 2 = 10 m/s.
Therefore, its kinetic energy becomes 4 times its original value, because it is proportional to the square of the velocity. Hence, kinetic energy = 25 × 4 = 100 J.
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