JEE Physics Work Energy and Power Notes Set 05

Work Done by a Force

1. Work is the scalar product of force and displacement: \( W = \vec F\cdot\vec s = Fs\cos\theta \). It is positive for \( \theta < 90^\circ \), zero at \( 90^\circ \) and negative for \( \theta > 90^\circ \). The unit is the joule (1 J = 1 N m).
2. Work depends on the frame of reference, because the displacement of the point of application depends on the observer. A force that does no work in one frame can do work in another.
3. Forces perpendicular to the motion do no work. Examples: the normal force on a block sliding on a fixed surface, the centripetal force in uniform circular motion, the magnetic force on a charge.
4. Static friction can do work on a body when the surface itself moves. A box carried on an accelerating truck without slipping gains kinetic energy only because static friction does positive work on it.
5. A force that acts at a point which does not move does zero work, even though it can cause acceleration of the whole system. Internal forces can change the kinetic energy of a system because their points of application move relative to each other (a person pushing off a wall gains kinetic energy from internal work).

Work Done by a Variable Force

1. For a force that changes along a path: \( W = \int \vec F\cdot d\vec r = \int (F_x\,dx + F_y\,dy + F_z\,dz) \).
2. In one dimension \( W = \int_{x_1}^{x_2} F(x)\,dx \), which equals the area under the F-x graph. Area above the x-axis counts as positive work and area below it as negative work.
3. If the force is given as a function of time, first find the velocity or displacement as a function of time, then use \( dW = F\,v\,dt \).
4. When \( F_x\,dx + F_y\,dy \) is a perfect differential, the work depends only on the end points. For \( \vec F = y\hat i + x\hat j \), \( dW = d(xy) \), so \( W = (xy)_f - (xy)_i \). For \( \vec F = y\hat i \) the work depends on the path chosen between the same two points.
5. Work done by a constant force \( \vec F \) between points A and B is \( \vec F\cdot(\vec r_B - \vec r_A) \) and does not depend on the path.

Conservative and Non-Conservative Forces

1. A force is conservative if the work it does between two points is independent of the path, or equivalently if the work in any closed path is zero.
2. Conservative forces: gravity, the spring force, electrostatic force, and any central force that depends only on distance from a fixed centre. Non-conservative forces: friction, air resistance, viscous drag, and in general any velocity-dependent force.
3. The work done by a conservative force is fully recoverable. The work done against friction is dissipated as heat and depends on the length of the path.
4. A potential energy can be defined only for a conservative force. The work done by the force equals the decrease in potential energy: \( W_c = -\Delta U \).
5. In one dimension \( F = -\frac{dU}{dx} \). In three dimensions \( \vec F = -\left(\frac{\partial U}{\partial x}\hat i + \frac{\partial U}{\partial y}\hat j + \frac{\partial U}{\partial z}\hat k\right) \).

Kinetic Energy and Potential Energy

1. Kinetic energy \( K = \frac12 mv^2 = \frac{p^2}{2m} \). It is never negative and does not depend on direction. For a given momentum, a lighter body has more kinetic energy. For a given kinetic energy, a heavier body has more momentum.
2. Gravitational potential energy near the Earth is \( U = mgh \), measured from a chosen reference level. A body below the reference level has negative U. Only the change in U has physical meaning, and it does not depend on the frame.
3. Elastic potential energy of a spring is \( U = \frac12kx^2 \), where x is the extension or compression from the natural length. It is never negative. The work done by an external agent in stretching the spring from \( x_1 \) to \( x_2 \) is \( \frac12k(x_2^2 - x_1^2) \).
4. Springs in series: \( \frac1{k} = \frac1{k_1}+\frac1{k_2} \). Springs in parallel: \( k = k_1 + k_2 \). A spring cut to a fraction \( 1/n \) of its length has constant \( nk \).
5. For a spring stretched by a load, the energy stored is \( \frac12kx^2 = \frac{(mg)^2}{2k} \) when the load is hung gently and brought to equilibrium. The loss of gravitational potential energy is \( mgx = 2\times \) this energy. The remainder is lost to oscillation, which is eventually damped.

Work-Energy Theorem

1. The net work done by all forces on a particle equals its change in kinetic energy: \( W_{net} = \Delta K = \frac12mv^2 - \frac12mu^2 \). It holds for conservative as well as non-conservative forces, and in every inertial frame.
2. Split the net work as \( W_c + W_{nc} + W_{ext} = \Delta K \). With \( W_c = -\Delta U \) this gives \( W_{nc} + W_{ext} = \Delta K + \Delta U = \Delta E_{mech} \).
3. For a constant retarding force F stopping a body of speed v, the stopping distance is \( \frac{mv^2}{2F} \). For the same retarding force the stopping distance is proportional to \( v^2 \) and to the mass, and for the same stopping time, to \( v \).
4. Work done by gravity depends only on the vertical drop, \( W_g = mg\Delta h \). A body sliding down a smooth surface from height h reaches speed \( \sqrt{2gh} \) whatever the shape of the surface.
5. When a particle moves in a circle at constant speed the net work is zero, since \( \Delta K = 0 \) even though the force is non-zero.

Worked Examples

1. A force \( \vec F = (4x\hat i + 3y\hat j) \) N moves a particle along the x-axis from the origin to x = 5 m. Only the x-component acts along the displacement, so \( W = \int_0^5 4x\,dx = 2\times 25 = 50 \) J.
2. A spring of constant 200 N/m is stretched from 10 cm to 20 cm. Work needed \( = \frac12(200)(0.2^2 - 0.1^2) = 3 \) J.
3. A 2 kg block at rest is acted on by \( F = 2t^2 \) N. Then \( a = t^2 \), \( v = \frac{t^3}{3} \) and at t = 2 s the speed is 8/3 m/s. By the work-energy theorem \( W = \frac12(2)(8/3)^2 = \frac{64}{9} \) J.
4. A force acts on a body so that the F-x graph is a rectangle of height 5 N from x = 0 to 4 m, followed by a triangle falling linearly to zero at x = 6 m. Work \( = 5\times4 + \frac12\times 2\times 5 = 25 \) J.
5. A 0.5 kg ball is thrown vertically up at 20 m/s. At the top its kinetic energy is zero, so by the work-energy theorem \( -mgh = -\frac12mu^2 \) gives \( h = \frac{u^2}{2g} = 20 \) m (g = 10 m/s²).
6. A body of mass 4 kg moving at 10 m/s is stopped by friction in 20 m. The retarding force is \( F = \frac{mv^2}{2s} = \frac{4\times100}{40} = 10 \) N.