Work, Energy and Power — Easy Practice (Set 1)

Quick Recap — Work, Energy & Power Basics

• Work $W = Fd\cos\theta$ (unit joule, J); it is zero when the force is perpendicular to the displacement.
• Kinetic energy $KE = \tfrac12 mv^2$; gravitational PE $= mgh$; spring PE $= \tfrac12 kx^2$.
• Work-energy theorem: the net work done equals the change in kinetic energy, $W_{net} = \Delta KE$.
• Power $P = \dfrac{W}{t} = Fv$ (unit watt, W); 1 W = 1 J/s.
• Work and energy are scalars; the area under a force-displacement graph gives the work done.

Beyond-NCERT JEE Formulae

These go past the NCERT staples $W=Fd$, $KE=\dfrac{1}{2}mv^2$ and $PE=mgh$, and are the extras JEE Main leans on.

1. Work done by a variable force

When the force changes as the body moves, you cannot just multiply — you must integrate:

$$W=\int_{x_1}^{x_2}\vec F\cdot d\vec x$$

Numerically this is the area under the force-displacement ($F$ versus $x$) graph — split it into triangles and rectangles.

When to use: any position-dependent force, such as a spring ($F=kx$), a law like $F=ax^2$, or a force given only as a graph.

[JEE Tip] Area above the $x$-axis is positive work, area below is negative. For a spring taken from extension $x_1$ to $x_2$ the work you must supply is $\dfrac{1}{2}k\left(x_2^2-x_1^2\right)$, not $\dfrac{1}{2}kx_2^2$ — the spring already stored energy at $x_1$.

2. Force from a potential-energy curve

For a conservative force,

$$F=-\dfrac{dU}{dx}$$

so the force is minus the slope of $U(x)$ and always points "downhill" on the curve. Equilibrium is where $\dfrac{dU}{dx}=0$, and its type follows the curvature:

• Stable at a minimum of $U$, where $\dfrac{d^2U}{dx^2}>0$ (a valley — a small push gives a restoring force).
• Unstable at a maximum of $U$, where $\dfrac{d^2U}{dx^2}<0$ (a hilltop — a small push runs away).
• Neutral on a flat stretch, where $\dfrac{d^2U}{dx^2}=0$.

When to use: $U(x)$ is given as a formula or a graph and you need the force, the equilibrium points, or their nature.

[JEE Tip] Read the slope, not the height. The force is largest where the curve is steepest and vanishes at every peak and trough. The depth of a well measured up to $U=0$ is the least energy needed to just free the particle from it.

3. Vertical circular motion

For a mass $m$ on a string (or inside a circular track) of radius $r$ making a full vertical loop:

• Minimum speed at the top: $v_{top}=\sqrt{gr}$ (tension or normal reaction just falls to zero there).
• Minimum speed at the bottom: $v_{bot}=\sqrt{5gr}$.
• The two are linked by $v_{bot}^2=v_{top}^2+4gr$ (energy across the height $2r$).
• For any valid speed the tension difference is fixed:

$$T_{bot}-T_{top}=6mg$$

Condition to complete the loop (string, or the inside of a track): $v_{top}\ge\sqrt{gr}$, equivalently $v_{bot}\ge\sqrt{5gr}$.

When to use: a bob whirled on a string, a bead inside a loop, a "loop-the-loop" on the inside of a track.

[JEE Tip] The $6mg$ gap does not depend on speed — a classic one-line MCQ. It is $2mg$ (the two weight terms) plus $4mg$ from the extra centripetal demand $\dfrac{m}{r}\left(4gr\right)$ at the bottom. A rigid rod can push as well as pull, so with a rod the top condition relaxes to $v_{top}\ge 0$.

4. One-dimensional collisions

Coefficient of restitution:

$$e=\dfrac{\text{speed of separation}}{\text{speed of approach}}=\dfrac{v_2-v_1}{u_1-u_2}$$

with $e=1$ perfectly elastic, $0<e<1$ partially inelastic, and $e=0$ perfectly inelastic (the bodies move off together).

Elastic head-on onto a stationary target (mass $m_2$ at rest):

$$v_1=\dfrac{m_1-m_2}{m_1+m_2}\,u_1 \qquad v_2=\dfrac{2m_1}{m_1+m_2}\,u_1$$

The fraction of kinetic energy handed to the target is $\dfrac{4m_1m_2}{\left(m_1+m_2\right)^2}$, which reaches 100% only for equal masses.

Perfectly inelastic onto a stationary target: the fraction of kinetic energy lost is

$$\dfrac{\Delta KE}{KE_i}=\dfrac{m_2}{m_1+m_2}$$

When to use: carts or balls colliding along a line; bounce problems, where $e=\sqrt{\dfrac{h'}{h}}$ for a drop from $h$ rebounding to $h'$.

[JEE Tip] Against a fixed wall $e=\dfrac{\text{rebound speed}}{\text{incoming speed}}$, and the fraction of kinetic energy surviving one bounce is $e^2$. A light body hitting a heavy one bounces almost straight back; a heavy body ploughs on almost unchanged.

5. Power: instantaneous versus average

$$P_{inst}=\vec F\cdot\vec v=Fv\cos\theta \qquad P_{avg}=\dfrac{W}{t}=\dfrac{\Delta E}{t}$$

When to use: engines and pumps ($P=Fv$ at steady speed), a variable driving force, or "the power at the instant the speed is …".

[JEE Tip] For a body starting from rest under a constant force, the instantaneous power at the end of an interval is exactly TWICE the average, since $P=Fv$ climbs linearly from zero. At constant engine power the drive force is $F=\dfrac{P}{v}$, so the top speed (where $a=0$) is $v_{max}=\dfrac{P}{F_{resist}}$.

Solved Examples — Beyond-NCERT Formulae

Example 1 — Work by a variable force (integration). A single force $F=\left(3x^2-2x\right)$ N acts along the motion of a $1$ kg particle as it moves from $x=1$ m to $x=3$ m on a smooth floor, starting from rest. Find the work done and the final speed.

Formula: the force varies with position, so $W=\displaystyle\int_{x_1}^{x_2}F\,dx$, and then $W=\dfrac{1}{2}mv^2$.

Working: $W=\displaystyle\int_{1}^{3}\left(3x^2-2x\right)dx=\left[x^3-x^2\right]_{1}^{3}=\left(27-9\right)-\left(1-1\right)=18$ J. Then $\dfrac{1}{2}\left(1\right)v^2=18$ gives $v^2=36$, so $v=6$ m/s.

Answer: $W=18$ J, and $v=6$ m/s.

Example 2 — Work as the area under an $F$-$x$ graph. A horizontal force on a $5$ kg block (started from rest) rises linearly from $0$ at $x=0$ to $40$ N at $x=2$ m, then stays constant at $40$ N up to $x=5$ m. Find the block's speed at $x=5$ m.

Formula: the work is the area under the $F$ versus $x$ graph, and then $W=\dfrac{1}{2}mv^2$.

Working: the triangle from $x=0$ to $x=2$ m has area $\dfrac{1}{2}\left(2\right)\left(40\right)=40$ J; the rectangle from $x=2$ m to $x=5$ m has area $40\times 3=120$ J. Total work $W=40+120=160$ J. Then $\dfrac{1}{2}\left(5\right)v^2=160$ gives $v^2=64$, so $v=8$ m/s.

Answer: $v=8$ m/s.

Example 3 — Vertical circle: minimum speeds and tensions. A $0.4$ kg ball on a light string of radius $0.9$ m is whirled in a vertical circle with just enough speed to complete it. Take $g=10$ m/s squared. Find the speeds at the top and bottom and the difference in string tension.

Formula: at the top, "just completes" means zero tension, so $v_{top}=\sqrt{gr}$; energy over the height $2r$ gives $v_{bot}=\sqrt{5gr}$; the tension difference is $T_{bot}-T_{top}=6mg$.

Working: $v_{top}=\sqrt{\left(10\right)\left(0.9\right)}=\sqrt{9}=3$ m/s, and $v_{bot}=\sqrt{5\left(10\right)\left(0.9\right)}=\sqrt{45}\approx 6.71$ m/s. Since it barely completes the loop, $T_{top}=0$ and $T_{bot}=6mg=6\left(0.4\right)\left(10\right)=24$ N, so the difference is $24$ N.

Answer: $v_{top}=3$ m/s, $v_{bot}\approx 6.71$ m/s, and $T_{bot}-T_{top}=24$ N.

Example 4 — Elastic head-on collision: velocities and energy transfer. On a smooth line a $2$ kg ball moving at $12$ m/s strikes a stationary $6$ kg ball elastically. Find both final velocities and the fraction of kinetic energy transferred.

Formula: for an elastic hit on a target at rest, $v_1=\dfrac{m_1-m_2}{m_1+m_2}u_1$ and $v_2=\dfrac{2m_1}{m_1+m_2}u_1$, while the transferred fraction is $\dfrac{4m_1m_2}{\left(m_1+m_2\right)^2}$.

Working: $v_1=\dfrac{2-6}{2+6}\left(12\right)=-6$ m/s (the light ball rebounds) and $v_2=\dfrac{2\left(2\right)}{8}\left(12\right)=6$ m/s. The transferred fraction is $\dfrac{4\left(2\right)\left(6\right)}{8^2}=\dfrac{48}{64}=0.75$. Check: momentum $2\left(12\right)=24=2\left(-6\right)+6\left(6\right)$, and the kinetic energy is $144$ J both before and after.

Answer: $v_1=-6$ m/s and $v_2=6$ m/s, with $75\%$ of the kinetic energy transferred.

Example 5 — Perfectly inelastic collision: kinetic energy lost. A $6$ kg block moving at $8$ m/s strikes and sticks to a stationary $2$ kg block on a smooth floor. Find the common velocity and the fraction and amount of kinetic energy lost.

Formula: momentum is conserved, $v=\dfrac{m_1u_1}{m_1+m_2}$, and the fraction of kinetic energy lost is $\dfrac{m_2}{m_1+m_2}$.

Working: $v=\dfrac{6\left(8\right)}{6+2}=6$ m/s. The fraction lost is $\dfrac{2}{6+2}=\dfrac{1}{4}=25\%$. The initial kinetic energy is $\dfrac{1}{2}\left(6\right)\left(8^2\right)=192$ J, so the loss is $0.25\times 192=48$ J, leaving $144$ J — which checks against $\dfrac{1}{2}\left(8\right)\left(6^2\right)=144$ J.

Answer: $v=6$ m/s, with $25\%$ (i.e. $48$ J) lost.

Example 6 — Force from a potential curve and equilibrium. A particle on the $x$-axis has potential energy $U\left(x\right)=x^4-2x^2$ J (with $x$ in m). Locate the equilibrium positions, classify each, and find the force at $x=2$ m.

Formula: $F=-\dfrac{dU}{dx}$; equilibrium where $\dfrac{dU}{dx}=0$; stability from the sign of $\dfrac{d^2U}{dx^2}$.

Working: $\dfrac{dU}{dx}=4x^3-4x=4x\left(x^2-1\right)$, which is zero at $x=0$ and $x=\pm 1$ m. The curvature $\dfrac{d^2U}{dx^2}=12x^2-4$ equals $-4$ at $x=0$ (a maximum, hence unstable) and $+8$ at $x=\pm 1$ m (minima, hence stable). The force is $F=-\left(4x^3-4x\right)$, so at $x=2$ m it is $F=-\left(32-8\right)=-24$ N, pointing back toward the wells.

Answer: unstable equilibrium at $x=0$; stable equilibria at $x=\pm 1$ m; and $F=-24$ N at $x=2$ m.

Example 7 — Power: instantaneous versus average. A $1000$ kg car accelerates uniformly from rest to $20$ m/s in $8$ s on a level road with negligible resistance. Find the average power over the interval and the instantaneous power at the end.

Formula: $P_{avg}=\dfrac{\Delta KE}{t}$, while $P_{inst}=Fv$ with $F=ma$.

Working: $a=\dfrac{20}{8}=2.5$ m/s squared, so $F=1000\times 2.5=2500$ N. The kinetic energy gained is $\dfrac{1}{2}\left(1000\right)\left(20^2\right)=200000$ J, giving $P_{avg}=\dfrac{200000}{8}=25000$ W. At the end $v=20$ m/s, so $P_{inst}=2500\times 20=50000$ W — exactly twice the average, as expected for a start-from-rest constant force.

Answer: $P_{avg}=25000$ W $=25$ kW, and $P_{inst}=50000$ W $=50$ kW.

Example 8 — Coefficient of restitution from bounce heights. A ball dropped from $5$ m onto a floor rebounds to $1.8$ m. Find the coefficient of restitution and the percentage of kinetic energy lost in the bounce.

Formula: for a drop from height $h$ rebounding to $h'$, $e=\sqrt{\dfrac{h'}{h}}$, and the fraction of kinetic energy surviving is $e^2$.

Working: $e=\sqrt{\dfrac{1.8}{5}}=\sqrt{0.36}=0.6$. The retained fraction is $e^2=0.36$, so the lost fraction is $1-0.36=0.64$.

Answer: $e=0.6$, and $64\%$ of the kinetic energy is lost.