JEE Physics Work Energy and Power Notes Set 06

Conservation of Mechanical Energy

1. If only conservative forces do work, the total mechanical energy is constant: \( K + U = \text{constant} \), that is \( \Delta K = -\Delta U \).
2. If non-conservative forces also act, \( W_{nc} = \Delta(K+U) \). Friction does negative work, so the mechanical energy decreases by the heat produced, \( f\times \) (path length).
3. A body released from rest on any smooth curve of height h has speed \( \sqrt{2gh} \) at the bottom. The time taken depends on the shape, but the speed does not.
4. A simple pendulum released from angle \( \theta \) with length L has speed at the lowest point \( v = \sqrt{2gL(1-\cos\theta)} \). The tension there is \( T = mg + \frac{mv^2}{L} = mg(3 - 2\cos\theta) \).
5. For a block on a spring dropped or hung on a vertical spring, include the elastic energy as well. The block's maximum speed occurs where the net force is zero, at extension \( x = \frac{mg}{k} \).
6. A block released on a spring from the natural length falls a maximum distance \( \frac{2mg}{k} \), twice the equilibrium extension.

Potential Energy Curves and Equilibrium

1. The force at any point is minus the slope of the U-x curve: \( F = -\frac{dU}{dx} \). A negative slope means the force pushes towards +x.
2. Equilibrium occurs where \( \frac{dU}{dx} = 0 \). Stable equilibrium is a minimum of U (\( \frac{d^2U}{dx^2} > 0 \)). Unstable equilibrium is a maximum (\( \frac{d^2U}{dx^2} < 0 \)). Neutral equilibrium has U constant over a region.
3. A particle of total energy E can move only where \( U \le E \). The points where \( U = E \) are turning points, where \( K = 0 \).
4. Near a stable equilibrium the particle oscillates with angular frequency \( \omega = \sqrt{\frac{U''(x_0)}{m}} \).
5. Spring-mass system: \( U = \frac12kx^2 \), minimum at x = 0, a parabola. Gravitational potential energy of two masses \( U = -\frac{GMm}{r} \), which rises towards zero at infinity.
6. For a force like \( F = -kx + \) constant, the equilibrium is shifted but the oscillation is still about the new equilibrium with the same \( \omega = \sqrt{k/m} \).

Work-Energy in a Vertical Circle

1. A particle on a string of length R moves in a vertical circle. Energy conservation from the bottom (speed \( u \)) to a point at height h above the bottom gives \( v^2 = u^2 - 2gh \). The tension at that point is \( T = \frac{mv^2}{R} + mg\cos\phi \), where \( \phi \) is the angle of the string from the downward vertical.
2. To complete the circle on a string the tension at the top must be non-negative. Then \( v_{top} \ge \sqrt{gR} \) and \( u_{bottom} \ge \sqrt{5gR} \). The tension at the bottom is then at least 6 mg, and the tension difference between bottom and top is 6 mg whatever the speed.
3. For a rod or a bead on a ring, the support can push as well as pull, so the particle only needs to reach the top with speed above zero: \( u_{bottom} > \sqrt{4gR} \).
4. If \( \sqrt{2gR} < u < \sqrt{5gR} \) on a string, the string goes slack somewhere above the horizontal level and the particle leaves the circle to follow a projectile path. If \( u \le \sqrt{2gR} \), the particle oscillates and never rises above the horizontal level of the centre.
5. A particle sliding on the outside of a smooth sphere of radius R from the top leaves the surface when \( \cos\theta = \frac23 \), where \( \theta \) is measured from the vertical through the centre. This is at a depth \( \frac R3 \) below the top, and the speed there is \( \sqrt{\frac{2gR}{3}} \).
6. Loop-the-loop on a smooth track needs a release height of at least \( \frac{5R}{2} \) above the bottom of the loop.

Power

1. Average power \( P_{av} = \frac{W}{t} \). Instantaneous power \( P = \frac{dW}{dt} = \vec F\cdot\vec v = Fv\cos\theta \). The unit is the watt (1 W = 1 J/s). 1 horsepower = 746 W.
2. Work done in a time interval is the area under the P-t graph: \( W = \int P\,dt \). The kilowatt-hour is a unit of energy: 1 kWh = \( 3.6\times10^6 \) J.
3. A machine of constant power P moving a body against a constant resistance F has a top speed \( v_{max} = \frac PF \). The acceleration at speed v is \( a = \frac{1}{m}\left(\frac Pv - F\right) \).
4. A body accelerated from rest by constant power P has \( \frac12mv^2 = Pt \), so \( v \propto \sqrt t \), and the distance covered is \( s \propto t^{3/2} \).
5. Under a constant force the power grows linearly with time, \( P = F\,at \).
6. A pump raising water of mass flow rate \( \mu \) (kg/s) through height h and giving it speed v needs power \( P = \mu\left(gh + \frac{v^2}{2}\right) \).
7. Efficiency \( \eta = \frac{\text{useful output}}{\text{input}} \), always less than 1 for a real machine.

Springs and Energy

1. A block of mass m moving at speed v hits a spring of constant k. Maximum compression \( x = v\sqrt{\frac mk} \) on a smooth surface. With friction \( \mu \) the compression satisfies \( \frac12mv^2 = \frac12kx^2 + \mu mgx \).
2. A spring compressed by x and released on a rough surface moves the block a distance d until it stops, with \( \frac12kx^2 = \mu mg d \) if the block is not attached and the spring is at its natural length at release.
3. A block is just able to lift another block off the ground through a spring when the spring is stretched by \( \frac{Mg}{k} \).
4. In a system acted on by gravity and a spring only, \( K + U_g + U_s \) is constant. At the lowest point of a vertical oscillation the speed is zero and the spring energy is largest.

Worked Examples

1. A body is released from rest at the top of a smooth incline of height 5 m. At the bottom \( v = \sqrt{2gh} = \sqrt{2\times10\times5} = 10 \) m/s.
2. A pendulum of length 1 m is released from 60° to the vertical. Speed at the bottom \( = \sqrt{2(10)(1)(1-\cos60^\circ)} = \sqrt{10} \approx 3.16 \) m/s. Tension \( = mg(3-2\cos60^\circ) = 2mg \).
3. A particle has \( U = 3x^2 - 6x \) joule. Then \( F = -(6x - 6) \) N. Equilibrium at x = 1 m, and \( U'' = 6 > 0 \), so it is stable.
4. A car engine delivers 40 kW against a constant resistance of 800 N. Top speed \( = \frac{40000}{800} = 50 \) m/s. At 20 m/s the pull is 2000 N, so with m = 1000 kg the acceleration is \( \frac{2000 - 800}{1000} = 1.2 \) m/s².
5. A pump lifts 100 kg of water through 10 m in 20 s. Power \( = \frac{100\times10\times10}{20} = 500 \) W (g = 10 m/s²).
6. A bob on a string of length 0.4 m must have bottom speed at least \( \sqrt{5gR} = \sqrt{5\times10\times0.4} = \sqrt{20} \approx 4.47 \) m/s to complete the vertical circle.
7. A 2 kg block moving at 4 m/s hits a spring of constant 200 N/m on a smooth surface. Maximum compression \( x = 4\sqrt{2/200} = 0.4 \) m.