Energy and Work

Work Done by a Force

Definition

Work is done when a force causes displacement in the direction of the force.

$W = Fs\cos\theta$

where $\theta$ is the angle between the force and the displacement. The SI unit of work is the joule (J), where $1 \mathrm{ J} = 1 \mathrm{ N\, m}$.

Condition	Work Done
$\theta = 0^\circ$	$W = Fs$ (maximum)
$\theta = 90^\circ$	$W = 0$ (no work)
$\theta \gt 90^\circ$	$W \lt 0$ (force opposes motion)

Work Done by a Variable Force

When force varies with displacement:

$W = \int_{s_1}^{s_2} F\, ds$

This equals the area under the force-displacement graph.

Work Done Against Gravity

Raising an object of mass $m$ through a vertical height $h$:

$W = mgh$

Work Done Stretching a Spring (Hooke's Law)

For a spring obeying $F = kx$:

$W = \frac{1}{2}kx^2$

where $k$ is the spring constant and $x$ is the extension from the natural length.

Worked Example 1

A spring of spring constant $500 \mathrm{ N/m}$ is stretched by $0.08 \mathrm{ m}$. Find the work done.

Solution

$W = \frac{1}{2}(500)(0.08)^2 = \frac{1}{2}(500)(0.0064) = 1.6 \mathrm{ J}$

Worked Example 2

A worker pushes a $40 \mathrm{ kg}$ crate across a floor by applying a force of $150 \mathrm{ N}$ at $25^\circ$ below the horizontal. The crate moves $8 \mathrm{ m}$. The coefficient of kinetic friction is $0.3$. Find the work done by the applied force, the work done by friction, and the net work done.

Solution

Work by applied force: $W_{\mathrm{app}} = Fs\cos\theta = 150 \times 8 \times \cos 25^\circ = 1200 \times 0.906 = 1088 \mathrm{ J}$

Normal reaction: $N = mg + F\sin\theta = 40 \times 9.81 + 150\sin 25^\circ = 392.4 + 63.4 = 455.8 \mathrm{ N}$

Friction force: $f_k = \mu_k N = 0.3 \times 455.8 = 136.7 \mathrm{ N}$

Work by friction: $W_f = -f_k s = -136.7 \times 8 = -1094 \mathrm{ J}$

$W_{\mathrm{net}} = 1088 + (-1094) = -6 \mathrm{ J}$

The small negative net work means the crate slightly decelerates.

Worked Example 3

A $5 \mathrm{ kg}$ box is lowered vertically by a rope with a constant downward acceleration of $2 \mathrm{ m/s}^2$. Find the tension in the rope and the work done by the tension as the box descends $4 \mathrm{ m}$.

Solution

Taking downward as positive: $mg - T = ma$

$T = m(g - a) = 5(9.81 - 2) = 5 \times 7.81 = 39.1 \mathrm{ N}$

The tension acts upward while displacement is downward, so $\theta = 180^\circ$:

$W = T s \cos 180^\circ = 39.1 \times 4 \times (-1) = -156 \mathrm{ J}$

---

Kinetic Energy

The kinetic energy of an object of mass $m$ moving at speed $v$:

$E_k = \frac{1}{2}mv^2$

Derivation from Newton's Second Law

Starting from $F = ma$ and using $v^2 = u^2 + 2as$ with constant force:

$W = Fs = mas = m\frac{v^2 - u^2}{2s}\cdot s = \frac{1}{2}m(v^2 - u^2) = \Delta E_k$

This establishes the work-energy theorem: the net work done on an object equals its change in kinetic energy.

---

Potential Energy

Gravitational Potential Energy

Near the Earth's surface:

$E_p = mgh$

where $h$ is the height above a chosen reference level.

Elastic Potential Energy

For a spring obeying Hooke's law with extension $x$:

$E_p = \frac{1}{2}kx^2$

---

Conservation of Energy

Principle

Energy cannot be created or destroyed, only transformed from one form to another. In a closed system with no non-conservative forces:

$E_{k1} + E_{p1} = E_{k2} + E_{p2}$

When friction or air resistance is present:

$E_{k1} + E_{p1} = E_{k2} + E_{p2} + W_{\mathrm{lost}}$

where $W_{\mathrm{lost}}$ is the energy dissipated as thermal energy.

Energy Skate Park

Observe the continuous interchange between kinetic and potential energy as the skater moves along the track.

Worked Example 3

A roller coaster car of mass $600 \mathrm{ kg}$ starts from rest at point A, $25 \mathrm{ m}$ above the ground. It descends to point B, $8 \mathrm{ m}$ above the ground. Find its speed at B, neglecting friction.

Solution

At A: $E_k = 0$, $E_p = 600 \times 9.81 \times 25 = 147150 \mathrm{ J}$

At B: $E_k = \frac{1}{2}(600)v^2$, $E_p = 600 \times 9.81 \times 8 = 47088 \mathrm{ J}$

$147150 = \frac{1}{2}(600)v^2 + 47088$

$\frac{1}{2}(600)v^2 = 100062$

$v^2 = \frac{200124}{600} = 333.54$

$v = 18.26 \mathrm{ m/s}$

Worked Example 4

A $2 \mathrm{ kg}$ block slides from rest down a rough curved ramp. The top of the ramp is $4 \mathrm{ m}$ above the ground. The block reaches the bottom with speed $7 \mathrm{ m/s}$. Find the energy lost to friction.

Solution

Initial energy: $E_i = mgh = 2 \times 9.81 \times 4 = 78.48 \mathrm{ J}$

Final kinetic energy: $E_{k,f} = \frac{1}{2}(2)(7^2) = 49.0 \mathrm{ J}$

$W_{\mathrm{lost}} = E_i - E_{k,f} = 78.48 - 49.0 = 29.5 \mathrm{ J}$

---

Power

Definition

Power is the rate of doing work:

$P = \frac{W}{t}$

For a force acting on a moving object:

$P = Fv\cos\theta$

When force and velocity are parallel ($\theta = 0^\circ$):

$P = Fv$

The SI unit of power is the watt (W), where $1 \mathrm{ W} = 1 \mathrm{ J/s}$.

Worked Example 5

A car of mass $1500 \mathrm{ kg}$ travels at a constant speed of $18 \mathrm{ m/s}$ up a slope of $\sin^{-1}(0.08)$. The total resistive force is $400 \mathrm{ N}$. Find the power output of the engine.

Solution

Component of weight along the slope: $mg\sin\theta = 1500 \times 9.81 \times 0.08 = 1177.2 \mathrm{ N}$

Total force: $F = 1177.2 + 400 = 1577.2 \mathrm{ N}$

$P = Fv = 1577.2 \times 18 = 28390 \mathrm{ W} = 28.4 \mathrm{ kW}$

Worked Example 6

A lift of mass $800 \mathrm{ kg}$ carries 5 passengers of average mass $70 \mathrm{ kg}$ each. The lift travels upward at a constant speed of $2 \mathrm{ m/s}$. The motor is $85\%$ efficient. Find the power input to the motor.

Solution

Total mass: $m = 800 + 5 \times 70 = 1150 \mathrm{ kg}$

Force to overcome gravity: $F = mg = 1150 \times 9.81 = 11282 \mathrm{ N}$

Useful power: $P_{\mathrm{out}} = Fv = 11282 \times 2 = 22563 \mathrm{ W}$

$P_{\mathrm{in}} = \frac{P_{\mathrm{out}}}{\eta} = \frac{22563}{0.85} = 26545 \mathrm{ W} = 26.5 \mathrm{ kW}$

---

Efficiency

Definition

$\mathrm{Efficiency} = \frac{\mathrm{Useful energy output}}{\mathrm{Total energy input}} \times 100\%$

Equivalently, using power:

$\mathrm{Efficiency} = \frac{P_{\mathrm{out}}}{P_{\mathrm{in}}} \times 100\%$

Efficiency is always less than 100% in practice because some energy is always dissipated as heat due to friction, air resistance, or electrical resistance.

Worked Example 7

A motor lifts a $200 \mathrm{ kg}$ load through $5 \mathrm{ m}$ in $10 \mathrm{ s}$. The motor is connected to a $240 \mathrm{ V}$ supply and draws a current of $5 \mathrm{ A}$. Find the efficiency of the motor.

Solution

Useful power output: $P_{\mathrm{out}} = \frac{mgh}{t} = \frac{200 \times 9.81 \times 5}{10} = 981 \mathrm{ W}$

Electrical power input: $P_{\mathrm{in}} = VI = 240 \times 5 = 1200 \mathrm{ W}$

$\mathrm{Efficiency} = \frac{981}{1200} \times 100\% = 81.8\%$

---

Energy in Simple Harmonic Motion

In SHM, kinetic and potential energy continuously interchange while the total energy remains constant.

$E_k = \frac{1}{2}mv^2 = \frac{1}{2}m\omega^2(A^2 - x^2)$

$E_p = \frac{1}{2}kx^2 = \frac{1}{2}m\omega^2 x^2$

$E_{\mathrm{total}} = \frac{1}{2}m\omega^2 A^2 = \frac{1}{2}kA^2$

At equilibrium ($x = 0$): all energy is kinetic. At maximum displacement ($x = A$): all energy is potential.

Worked Example 8

A mass-spring system has mass $0.3 \mathrm{ kg}$, spring constant $120 \mathrm{ N/m}$, and amplitude $0.04 \mathrm{ m}$. Find the total energy and the speed when $x = 0.02 \mathrm{ m}$.

Solution

$E_{\mathrm{total}} = \frac{1}{2}kA^2 = \frac{1}{2}(120)(0.04)^2 = \frac{1}{2}(120)(0.0016) = 0.096 \mathrm{ J}$

At $x = 0.02 \mathrm{ m}$:

$E_k = E_{\mathrm{total}} - \frac{1}{2}kx^2 = 0.096 - \frac{1}{2}(120)(0.0004) = 0.096 - 0.024 = 0.072 \mathrm{ J}$

$v = \sqrt{\frac{2E_k}{m}} = \sqrt{\frac{2 \times 0.072}{0.3}} = \sqrt{0.48} = 0.693 \mathrm{ m/s}$

Worked Example 9

A simple pendulum of length $1.5 \mathrm{ m}$ has a bob of mass $0.5 \mathrm{ kg}$. It is pulled aside until the string makes $30^\circ$ with the vertical and released from rest. Find the speed of the bob at the lowest point and the total energy, neglecting air resistance.

Solution

Height gain: $h = L(1 - \cos\theta) = 1.5(1 - \cos 30^\circ) = 1.5(1 - 0.866) = 0.201 \mathrm{ m}$

By conservation of energy: $mgh = \frac{1}{2}mv^2$

$v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 0.201} = \sqrt{3.94} = 1.99 \mathrm{ m/s}$

Total energy: $E_{\mathrm{total}} = mgh = 0.5 \times 9.81 \times 0.201 = 0.986 \mathrm{ J}$

---

Common Pitfalls

• Confusing work done on an object with the energy the object possesses. Work is a process; energy is a state.
• Forgetting that work is a scalar quantity. Even when a force acts at an angle, $W = Fs\cos\theta$ gives a signed scalar, not a vector.
• Applying $W = mgh$ when the height is large enough that $g$ varies significantly. For orbital problems, use $E_p = -GMm/r$ instead.
• Using $P = Fv$ when the force and velocity are not parallel. The correct form is $P = Fv\cos\theta$.
• Forgetting to include all forms of energy when applying conservation of energy. Missing a term (e.g., elastic potential energy or work done against friction) leads to incorrect results.

---

Summary Table

Topic	Key Formula	Key Concept
Work	$W = Fs\cos\theta$	Energy transfer by a force
Work-energy theorem	$W_{\mathrm{net}} = \Delta E_k$	Net work = change in KE
Kinetic energy	$E_k = \frac{1}{2}mv^2$	Energy of motion
Gravitational PE	$E_p = mgh$	Energy due to position in a field
Elastic PE	$E_p = \frac{1}{2}kx^2$	Energy stored in a deformed spring
Conservation	$E_{k1} + E_{p1} = E_{k2} + E_{p2}$	No energy lost
Power	$P = W/t = Fv$	Rate of energy transfer
Efficiency	$\eta = E_{\mathrm{out}}/E_{\mathrm{in}}$	Always less than 100%

---

Problem Set

Problem 1. A crate of mass $50 \mathrm{ kg}$ is pushed $12 \mathrm{ m}$ up a rough ramp inclined at $25^\circ$ to the horizontal by a force of $350 \mathrm{ N}$ acting parallel to the ramp. The coefficient of kinetic friction is $0.2$. Find the work done by the applied force, the work done against gravity, the work done against friction, and the final speed if the crate starts from rest.

Solution

Work by applied force: $W_{\mathrm{app}} = 350 \times 12 = 4200 \mathrm{ J}$

Work against gravity: $W_g = mgh = 50 \times 9.81 \times 12\sin 25^\circ = 50 \times 9.81 \times 5.071 = 2487 \mathrm{ J}$

Work against friction: $W_f = \mu mg\cos\theta \times s = 0.2 \times 50 \times 9.81 \times \cos 25^\circ \times 12 = 0.2 \times 50 \times 9.81 \times 0.9063 \times 12 = 1066 \mathrm{ J}$

By work-energy theorem: $W_{\mathrm{net}} = W_{\mathrm{app}} - W_g - W_f = 4200 - 2487 - 1066 = 647 \mathrm{ J} = \frac{1}{2}mv^2$

$v = \sqrt{\frac{2 \times 647}{50}} = \sqrt{25.88} = 5.09 \mathrm{ m/s}$

If you get this wrong, revise: Work Done by a Force / Work-Energy Theorem

Problem 2. A ball of mass $0.1 \mathrm{ kg}$ is thrown vertically upward with speed $12 \mathrm{ m/s}$. Find the maximum height and the speed when it returns to its starting point, given that air resistance does $0.3 \mathrm{ J}$ of work on the ball during the ascent.

Solution

Going up: $E_{k1} = \frac{1}{2}(0.1)(144) = 7.2 \mathrm{ J}$

$7.2 = mgh + W_{\mathrm{air}} = 0.1 \times 9.81 \times h + 0.3$

$0.981h = 6.9 \implies h = 7.03 \mathrm{ m}$

Coming down: total energy lost to air resistance $= 2 \times 0.3 = 0.6 \mathrm{ J}$ (approximately, assuming similar dissipation on the way down).

$\frac{1}{2}(0.1)v^2 = 7.2 - 0.6 = 6.6 \mathrm{ J}$

$v = \sqrt{\frac{2 \times 6.6}{0.1}} = \sqrt{132} = 11.49 \mathrm{ m/s}$

If you get this wrong, revise: Conservation of Energy

Problem 3. A pump lifts $500 \mathrm{ kg}$ of water per minute from a well $15 \mathrm{ m}$ deep. If the pump is $75\%$ efficient, what is its power input?

Solution

Useful power: $P_{\mathrm{out}} = \frac{mgh}{t} = \frac{500 \times 9.81 \times 15}{60} = 1226.25 \mathrm{ W}$

$P_{\mathrm{in}} = \frac{P_{\mathrm{out}}}{\eta} = \frac{1226.25}{0.75} = 1635 \mathrm{ W} = 1.64 \mathrm{ kW}$

If you get this wrong, revise: Power and Efficiency

Problem 4. A spring with $k = 200 \mathrm{ N/m}$ is placed at the bottom of a ramp inclined at $30^\circ$. A block of mass $2 \mathrm{ kg}$ slides $0.5 \mathrm{ m}$ down the ramp (measured along the slope) before hitting the spring. The ramp is smooth. Find the maximum compression of the spring.

Solution

Height descended: $h = 0.5\sin 30^\circ = 0.25 \mathrm{ m}$

$mgh = \frac{1}{2}kx^2$

$2 \times 9.81 \times 0.25 = \frac{1}{2}(200)x^2$

$4.905 = 100x^2 \implies x^2 = 0.04905 \implies x = 0.222 \mathrm{ m}$

If you get this wrong, revise: Conservation of Energy / Elastic Potential Energy

Problem 5. A $1200 \mathrm{ kg}$ car accelerates from rest to $25 \mathrm{ m/s}$ in $8 \mathrm{ s}$ on a level road. The average resistive force is $400 \mathrm{ N}$. Find the average power output of the engine.

Solution

Final KE: $E_k = \frac{1}{2}(1200)(25^2) = 375000 \mathrm{ J}$

Work against resistance: $W_r = 400 \times d$, where $d = \frac{1}{2}(0 + 25) \times 8 = 100 \mathrm{ m}$

$W_r = 400 \times 100 = 40000 \mathrm{ J}$

Total work by engine: $W_{\mathrm{engine}} = 375000 + 40000 = 415000 \mathrm{ J}$

$P_{\mathrm{avg}} = \frac{W_{\mathrm{engine}}}{t} = \frac{415000}{8} = 51875 \mathrm{ W} = 51.9 \mathrm{ kW}$

If you get this wrong, revise: Power and Work Done by a Force

Problem 6. A pendulum bob of mass $0.2 \mathrm{ kg}$ is released from a height of $0.4 \mathrm{ m}$ above its lowest point. At the lowest point, $20\%$ of its energy is lost to air resistance during the swing. Find the speed at the lowest point and the maximum height on the other side.

Solution

Initial PE: $E_p = mgh = 0.2 \times 9.81 \times 0.4 = 0.785 \mathrm{ J}$

KE at lowest point (80% of initial energy): $E_k = 0.80 \times 0.785 = 0.628 \mathrm{ J}$

$v = \sqrt{\frac{2E_k}{m}} = \sqrt{\frac{2 \times 0.628}{0.2}} = \sqrt{6.28} = 2.51 \mathrm{ m/s}$

If another $20\%$ is lost on the upswing: remaining energy $= 0.8^2 \times 0.785 = 0.502 \mathrm{ J}$

$h_{\mathrm{max}} = \frac{E_{\mathrm{remaining}}}{mg} = \frac{0.502}{0.2 \times 9.81} = 0.256 \mathrm{ m}$

If you get this wrong, revise: Conservation of Energy

Problem 7. A mass-spring system oscillates with amplitude $0.05 \mathrm{ m}$ and total energy $0.5 \mathrm{ J}$. The mass is $0.4 \mathrm{ kg}$. Find the spring constant and the speed when the displacement is $0.03 \mathrm{ m}$.

Solution

$E_{\mathrm{total}} = \frac{1}{2}kA^2 \implies k = \frac{2E_{\mathrm{total}}}{A^2} = \frac{2 \times 0.5}{0.05^2} = \frac{1}{0.0025} = 400 \mathrm{ N/m}$

At $x = 0.03 \mathrm{ m}$:

$E_k = E_{\mathrm{total}} - \frac{1}{2}kx^2 = 0.5 - \frac{1}{2}(400)(0.0009) = 0.5 - 0.18 = 0.32 \mathrm{ J}$

$v = \sqrt{\frac{2E_k}{m}} = \sqrt{\frac{2 \times 0.32}{0.4}} = \sqrt{1.6} = 1.26 \mathrm{ m/s}$

If you get this wrong, revise: Energy in Simple Harmonic Motion

Problem 8. An electric kettle rated at $2000 \mathrm{ W}$ takes $3$ minutes to boil $0.8 \mathrm{ kg}$ of water from $20^\circ\mathrm{C}$ to $100^\circ\mathrm{C}$. Find the efficiency of the kettle. (Specific heat capacity of water = 4200 \mathrm{ J/(kg\cdot}^\circ C)})

Solution

Useful energy: $Q = mc\Delta T = 0.8 \times 4200 \times (100 - 20) = 0.8 \times 4200 \times 80 = 268800 \mathrm{ J}$

Electrical energy supplied: $E_{\mathrm{in}} = Pt = 2000 \times 3 \times 60 = 360000 \mathrm{ J}$

$\mathrm{Efficiency} = \frac{268800}{360000} \times 100\% = 74.7\%$

If you get this wrong, revise: Efficiency

Problem 9. A force of $F = 3x^2 \mathrm{ N}$ (where $x$ is in metres) acts on an object moving along the x-axis from $x = 0$ to $x = 2 \mathrm{ m}$. Find the work done.

Solution

$W = \int_{0}^{2} 3x^2\, dx = \left[x^3\right]_0^2 = 8 - 0 = 8.0 \mathrm{ J}$

If you get this wrong, revise: Work Done by a Variable Force

Problem 10. A $60 \mathrm{ kg}$ student runs up a flight of stairs in $5 \mathrm{ s}$. The vertical height of the stairs is $6 \mathrm{ m}$. Find the average power developed by the student.

Solution

$P = \frac{mgh}{t} = \frac{60 \times 9.81 \times 6}{5} = \frac{3531.6}{5} = 706 \mathrm{ W}$

If you get this wrong, revise: Power

For the A-Level treatment of this topic, see Work, Energy and Power.

---

tip

tip Ready to test your understanding of Energy and Work? The diagnostic test contains the hardest questions within the DSE specification for this topic, each with a full worked solution.

Unit tests probe edge cases and common misconceptions. Integration tests combine Energy and Work with other physics topics to test synthesis under exam conditions.

See Diagnostic Guide for instructions on self-marking and building a personal test matrix.

---

Derivations

Derivation: Work-Energy Theorem

Starting from Newton's second law for constant mass:

$F = ma = m\frac{dv}{dt}$

The work done by a net force over a displacement from $s_1$ to $s_2$:

$W = \int_{s_1}^{s_2} F\, ds = \int_{s_1}^{s_2} m\frac{dv}{dt}\, ds = \int_{v_1}^{v_2} mv\, dv = \left[\frac{1}{2}mv^2\right]_{v_1}^{v_2} = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2$

Therefore:

$W_{\mathrm{net}} = \Delta E_k = E_{k,f} - E_{k,i}$

This is the work-energy theorem: the net work done on an object equals its change in kinetic energy.

Derivation: Elastic Potential Energy of a Spring

For a spring obeying Hooke's law, $F = kx$, the force varies linearly with extension. The work done in stretching the spring from $x = 0$ to $x = x$ is:

$W = \int_0^x F\, dx' = \int_0^x kx'\, dx' = \left[\frac{1}{2}kx'^2\right]_0^x = \frac{1}{2}kx^2$

This work is stored as elastic potential energy: $E_p = \frac{1}{2}kx^2$.

This is the area of a triangle under the force-extension graph (base $= x$, height $= kx$):

$E_p = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times x \times kx = \frac{1}{2}kx^2$

Derivation: Escape Velocity

The escape velocity is the minimum speed needed for an object to escape a gravitational field (i.e., reach infinity with zero kinetic energy). By conservation of energy:

$\frac{1}{2}mv_e^2 - \frac{GMm}{R} = 0 + 0$

$\frac{1}{2}mv_e^2 = \frac{GMm}{R}$

$v_e = \sqrt{\frac{2GM}{R}}$

For Earth: $v_e = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{6.37 \times 10^6}} = \sqrt{1.25 \times 10^8} = 11200 \mathrm{ m/s} \approx 11.2 \mathrm{ km/s}$

Derivation: Power-Velocity Relation

$P = \frac{dW}{dt} = \frac{d(Fs)}{dt} = F\frac{ds}{dt} = Fv$

More generally, $P = \vec{F} \cdot \vec{v} = Fv\cos\theta$, where $\theta$ is the angle between the force and velocity vectors.

---

Experimental Methods

Determining the Spring Constant Using Energy Conservation

Apparatus: A spring, a set of known masses, a metre rule, and a motion sensor (or video analysis).

Procedure:

1. Hang the spring vertically and attach a mass $m$.
2. Pull the mass down a known distance $x_0$ from the equilibrium position and release.
3. Measure the maximum speed $v_{\max}$ at the equilibrium position using a motion sensor.
4. By energy conservation: $\frac{1}{2}kx_0^2 = \frac{1}{2}mv_{\max}^2$
5. Calculate: $k = m\frac{v_{\max}^2}{x_0^2}$
6. Repeat for different masses and extensions, plot $v_{\max}^2$ versus $x_0^2$, and find the gradient $= k/m$.

Comparison with static method: The static method (measuring extension under different loads) assumes Hooke's law is obeyed. The dynamic method verifies this independently through energy conservation.

Verifying Conservation of Energy on an Inclined Plane

Apparatus: An inclined plane, a trolley, light gates, a metre rule, and a mass balance.

Procedure:

1. Measure the mass $m$ of the trolley.
2. Set the inclined plane at angle $\theta$ and measure the height $h$ from the top to the bottom.
3. Release the trolley from rest at the top and use light gates to measure the speed $v$ at the bottom.
4. Calculate: $\Delta E_p = mgh$ and $E_k = \frac{1}{2}mv^2$.
5. Compare $\Delta E_p$ with $E_k$. The difference is the work done against friction.
6. Vary $\theta$ and plot $E_k/E_p$ versus $\theta$ to see how the fraction of energy conserved changes.

Investigating Power Output of a Motor

Apparatus: A small electric motor, a string, a set of masses, a metre rule, a stopwatch, an ammeter, and a voltmeter.

Procedure:

1. Attach a mass $m$ to the motor via a string over a pulley.
2. Measure the time $t$ for the motor to lift the mass through a height $h$.
3. Record the voltage $V$ and current $I$.
4. Calculate useful power: $P_{\mathrm{out}} = mgh/t$.
5. Calculate electrical power input: $P_{\mathrm{in}} = VI$.
6. Calculate efficiency: $\eta = P_{\mathrm{out}} / P_{\mathrm{in}}$.
7. Repeat for different masses and plot efficiency versus load.

Expected result: Efficiency is low for very light loads (most energy lost to overcoming internal friction) and for very heavy loads (motor draws high current, high copper losses). Maximum efficiency occurs at intermediate loads.

---

Data Analysis and Uncertainty

Uncertainty in Energy Calculations

When calculating $E_k = \frac{1}{2}mv^2$, the percentage uncertainty is:

$\frac{\Delta E_k}{E_k} = \sqrt{\left(\frac{\Delta m}{m}\right)^2 + \left(2\frac{\Delta v}{v}\right)^2}$

Note the factor of 2 on the velocity uncertainty because $E_k \propto v^2$.

Example: Mass $(0.200 \pm 0.001) \mathrm{ kg}$, velocity $(3.00 \pm 0.05) \mathrm{ m/s}$:

$E_k = \frac{1}{2}(0.200)(3.00)^2 = 0.900 \mathrm{ J}$

$\frac{\Delta E_k}{E_k} = \sqrt{\left(\frac{0.001}{0.200}\right)^2 + \left(2 \times \frac{0.05}{3.00}\right)^2} = \sqrt{0.000025 + 0.001111} = \sqrt{0.001136} = 0.0337 = 3.4\%$

$\Delta E_k = 0.034 \times 0.900 = 0.031 \mathrm{ J}$

$E_k = (0.90 \pm 0.03) \mathrm{ J}$

Linearising Energy Data

To verify $E_k \propto v^2$: plot $E_k$ (y-axis) versus $v^2$ (x-axis). A straight line through the origin confirms the relationship, and the gradient equals $m/2$.

To verify $E_p \propto h$: plot $E_p$ (y-axis) versus $h$ (x-axis). A straight line through the origin with gradient $mg$ confirms the relationship.

---

Additional Worked Examples

Worked Example 10

A $0.5 \mathrm{ kg}$ ball is thrown vertically upward with speed $15 \mathrm{ m/s}$ from the top of a building $20 \mathrm{ m}$ tall. Air resistance is negligible. Find: (a) the maximum height above the ground reached by the ball, (b) the speed of the ball just before it hits the ground.

Solution

(a) At maximum height above the launch point, $v = 0$:

$h_{\max} = \frac{v_0^2}{2g} = \frac{15^2}{2 \times 9.81} = \frac{225}{19.62} = 11.47 \mathrm{ m}$

Maximum height above ground: $H = 20 + 11.47 = 31.5 \mathrm{ m}$

(b) Taking ground as reference. Total energy at launch: $E = E_k + E_p = \frac{1}{2}(0.5)(225) + 0.5 \times 9.81 \times 20 = 56.25 + 98.1 = 154.35 \mathrm{ J}$

At ground level, all energy is kinetic:

$\frac{1}{2}(0.5)v^2 = 154.35$

$v = \sqrt{\frac{2 \times 154.35}{0.5}} = \sqrt{617.4} = 24.8 \mathrm{ m/s}$

Worked Example 11

A spring of spring constant $200 \mathrm{ N/m}$ is compressed by $0.05 \mathrm{ m}$ and used to launch a $0.1 \mathrm{ kg}$ ball horizontally from a table of height $1.5 \mathrm{ m}$. The spring transfers $80\%$ of its energy to the ball. Find the horizontal distance the ball travels before hitting the ground.

Solution

Energy stored in spring: $E_p = \frac{1}{2}(200)(0.05)^2 = 0.25 \mathrm{ J}$

Kinetic energy of ball: $E_k = 0.80 \times 0.25 = 0.20 \mathrm{ J}$

$v = \sqrt{\frac{2E_k}{m}} = \sqrt{\frac{2 \times 0.20}{0.1}} = \sqrt{4.0} = 2.0 \mathrm{ m/s}$

Time to fall $1.5 \mathrm{ m}$: $h = \frac{1}{2}gt^2 \implies t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 1.5}{9.81}} = \sqrt{0.306} = 0.553 \mathrm{ s}$

Horizontal distance: $d = vt = 2.0 \times 0.553 = 1.11 \mathrm{ m}$

Worked Example 12

A car of mass $1200 \mathrm{ kg}$ accelerates uniformly from $10 \mathrm{ m/s}$ to $25 \mathrm{ m/s}$ over a distance of $200 \mathrm{ m}$ against a constant resistive force of $600 \mathrm{ N}$. Find the average force developed by the engine and the average power.

Solution

Change in kinetic energy: $\Delta E_k = \frac{1}{2}(1200)(25^2 - 10^2) = 600(625 - 100) = 600 \times 525 = 315000 \mathrm{ J}$

Work against resistance: $W_r = 600 \times 200 = 120000 \mathrm{ J}$

Total work by engine: $W_{\mathrm{engine}} = \Delta E_k + W_r = 315000 + 120000 = 435000 \mathrm{ J}$

$F_{\mathrm{engine}} = \frac{W_{\mathrm{engine}}}{d} = \frac{435000}{200} = 2175 \mathrm{ N}$

Time taken: $d = \frac{1}{2}(u + v)t \implies t = \frac{2d}{u + v} = \frac{400}{35} = 11.43 \mathrm{ s}$

$P_{\mathrm{avg}} = \frac{W_{\mathrm{engine}}}{t} = \frac{435000}{11.43} = 38060 \mathrm{ W} = 38.1 \mathrm{ kW}$

---

Exam-Style Questions

Question 1 (DSE Structured)

A student investigates how the stopping distance of a car depends on its speed. She measures the stopping distance $d$ from speed $v$ for several trials on a level road.

Speed $v$ (m/s)	Stopping distance $d$ (m)
5.0	4.2
10.0	16.8
15.0	37.5
20.0	66.0
25.0	103.0

(a) Plot a graph of $d$ against $v^2$. What relationship does this suggest?

(b) The student suggests that the work done by friction equals the initial kinetic energy: $\mu mgd = \frac{1}{2}mv^2$. Use the graph to find the coefficient of friction $\mu$.

(c) State two assumptions made in this model.

(d) Explain why the actual stopping distance is typically longer than the value predicted by this model.

Solution

(a)

$v^2$ (m$^2$/s$^2$)	$d$ (m)
25	4.2
100	16.8
225	37.5
400	66.0
625	103.0

The graph of $d$ versus $v^2$ is approximately a straight line through the origin, confirming $d \propto v^2$.

(b) Gradient of the line of best fit:

$\mathrm{Gradient} = \frac{\Delta d}{\Delta v^2} \approx \frac{103.0 - 4.2}{625 - 25} = \frac{98.8}{600} = 0.165 \mathrm{ s}^2$

From $\mu mgd = \frac{1}{2}mv^2$: $d = \frac{v^2}{2\mu g}$, so gradient $= \frac{1}{2\mu g}$.

$\mu = \frac{1}{2g \times \mathrm{gradient}} = \frac{1}{2 \times 9.81 \times 0.165} = \frac{1}{3.237} = 0.309$

(c) Assumptions:

1. The braking force (friction) is constant throughout the stopping distance.
2. The road is level (no component of weight assists or opposes braking).
3. All the initial kinetic energy is converted to work against friction (no other energy losses or gains).

(d) In practice, the braking force is not constant: it builds up as the brakes engage, and may decrease if the brakes overheat. Additionally, the driver's reaction time adds to the total stopping distance (thinking distance + braking distance), and road conditions (wet, icy) may reduce the friction coefficient.

Question 2 (DSE Structured)

A roller coaster car of mass $500 \mathrm{ kg}$ starts from rest at point A, height $30 \mathrm{ m}$ above the ground. It descends to point B at height $5 \mathrm{ m}$, then rises to point C at height $20 \mathrm{ m}$. The total energy lost to friction between A and C is $5000 \mathrm{ J}$.

(a) Calculate the speed of the car at point B, neglecting friction.

(b) Calculate the speed of the car at point C, including friction.

(c) If the average frictional force over the track from A to C is $200 \mathrm{ N}$, estimate the total track length from A to C.

(d) The car then descends from C to D at ground level. If the same average frictional force acts, find the speed at D.

Solution

(a) Conservation of energy from A to B (no friction):

$\frac{1}{2}mv_B^2 = mg(h_A - h_B) = 500 \times 9.81 \times (30 - 5) = 500 \times 9.81 \times 25 = 122625 \mathrm{ J}$

$v_B = \sqrt{\frac{2 \times 122625}{500}} = \sqrt{490.5} = 22.1 \mathrm{ m/s}$

(b) With friction from A to C:

$E_{\mathrm{initial}} = mgh_A = 500 \times 9.81 \times 30 = 147150 \mathrm{ J}$

$E_{\mathrm{final}} = mgh_C + \frac{1}{2}mv_C^2 = 500 \times 9.81 \times 20 + \frac{1}{2}(500)v_C^2 = 98100 + 250v_C^2$

$147150 = 98100 + 250v_C^2 + 5000$

$250v_C^2 = 147150 - 98100 - 5000 = 44050$

$v_C = \sqrt{\frac{44050}{250}} = \sqrt{176.2} = 13.3 \mathrm{ m/s}$

(c) Energy lost to friction $= 5000 \mathrm{ J} = f \times L = 200 \times L$:

$L = \frac{5000}{200} = 25 \mathrm{ m}$

(d) From C to D: $\Delta h = 20 \mathrm{ m}$.

Energy available: $mgh_C - W_{\mathrm{friction}}$

$\frac{1}{2}mv_D^2 = \frac{1}{2}mv_C^2 + mg(h_C - h_D) - f \times L_{CD}$

We need $L_{CD}$. From the height difference: the track length is at least $20 \mathrm{ m}$ (if straight down), but the actual length depends on the track shape. Assuming similar track geometry to the A-to-C section, we need more information. If we assume the track from C to D is $20 \mathrm{ m}$ (a minimum estimate):

$\frac{1}{2}(500)v_D^2 = 44050 + 500 \times 9.81 \times 20 - 200 \times 20 = 44050 + 98100 - 4000 = 138150$

$v_D = \sqrt{\frac{2 \times 138150}{500}} = \sqrt{552.6} = 23.5 \mathrm{ m/s}$

Question 3 (DSE Structured)

(a) Define power and state its SI unit.

(b) A lift motor has a power output of $15 \mathrm{ kW}$. It lifts a total mass of $1200 \mathrm{ kg}$ at constant speed. Calculate the speed of the lift.

(c) The lift is only $85\%$ efficient. Calculate the electrical power input.

(d) The motor has a label that says "15 kW, 240 V". Calculate the current it draws and the cost of running it for 8 hours at USD 1.20 per kWh.

Solution

(a) Power is the rate of doing work (or rate of energy transfer): $P = W/t$. SI unit: watt (W), where $1 \mathrm{ W} = 1 \mathrm{ J/s}$.

(b) At constant speed, the force equals the weight: $F = mg = 1200 \times 9.81 = 11772 \mathrm{ N}$

$v = \frac{P}{F} = \frac{15000}{11772} = 1.27 \mathrm{ m/s}$

(c) $P_{\mathrm{in}} = \frac{P_{\mathrm{out}}}{\eta} = \frac{15000}{0.85} = 17647 \mathrm{ W} = 17.6 \mathrm{ kW}$

(d) Current at rated power: $I = \frac{P}{V} = \frac{15000}{240} = 62.5 \mathrm{ A}$

Energy consumed: $E = P_{\mathrm{in}} \times t = 17.647 \times 8 = 141.2 \mathrm{ kWh}$

Cost: 141.2 \times 1.20 = \169.44$

Question 4 (DSE Structured)

Two trolleys A and B are on a smooth horizontal track. Trolley A has mass $2.0 \mathrm{ kg}$ and trolley B has mass $1.0 \mathrm{ kg}$. Trolley A moves towards B at $4.0 \mathrm{ m/s}$ and trolley B is stationary. They collide and stick together.

(a) Calculate the velocity of the combined trolleys after the collision.

(b) Calculate the kinetic energy before and after the collision, and the energy lost.

(c) Explain why kinetic energy is not conserved in this collision, even though momentum is.

(d) If the collision were elastic instead, calculate the velocities of both trolleys after the collision.

Solution

(a) By conservation of momentum:

$m_A u_A = (m_A + m_B)v$

$2.0 \times 4.0 = (2.0 + 1.0)v$

$v = \frac{8.0}{3.0} = 2.67 \mathrm{ m/s}$

(b) Before: $E_{k,i} = \frac{1}{2}(2.0)(4.0)^2 = 16.0 \mathrm{ J}$

After: $E_{k,f} = \frac{1}{2}(3.0)(2.67)^2 = \frac{1}{2}(3.0)(7.13) = 10.7 \mathrm{ J}$

Energy lost: $\Delta E_k = 16.0 - 10.7 = 5.3 \mathrm{ J}$ (converted to thermal energy, sound, and deformation)

(c) Momentum is always conserved in a closed system because there is no external force. Kinetic energy is only conserved in perfectly elastic collisions. In this perfectly inelastic collision, some kinetic energy is converted to other forms (heat, sound, permanent deformation) because the objects stick together and deform. The work done in deforming the objects accounts for the "missing" kinetic energy.

(d) For an elastic collision, both momentum and kinetic energy are conserved.

Momentum: $2.0 \times 4.0 = 2.0v_A + 1.0v_B \implies 8.0 = 2v_A + v_B \quad (1)$

KE: $\frac{1}{2}(2.0)(4.0)^2 = \frac{1}{2}(2.0)v_A^2 + \frac{1}{2}(1.0)v_B^2 \implies 16 = 2v_A^2 + \frac{1}{2}v_B^2 \quad (2)$

From (1): $v_B = 8 - 2v_A$. Substituting into (2):

$16 = 2v_A^2 + \frac{1}{2}(8 - 2v_A)^2 = 2v_A^2 + \frac{1}{2}(64 - 32v_A + 4v_A^2) = 2v_A^2 + 32 - 16v_A + 2v_A^2$

$4v_A^2 - 16v_A + 32 = 16$

$4v_A^2 - 16v_A + 16 = 0$

$v_A^2 - 4v_A + 4 = 0$

$(v_A - 2)^2 = 0 \implies v_A = 2.0 \mathrm{ m/s}$

$v_B = 8 - 2(2) = 4.0 \mathrm{ m/s}$

After the elastic collision: A moves at $2.0 \mathrm{ m/s}$ and B moves at $4.0 \mathrm{ m/s}$. (A transfers all its "excess" speed to B.)

Question 5 (DSE Structured)

A $3.0 \mathrm{ kg}$ block is attached to a spring of spring constant $150 \mathrm{ N/m}$ on a smooth horizontal surface. The block is displaced $0.10 \mathrm{ m}$ from the equilibrium position and released from rest.

(a) Calculate the total energy of the system.

(b) Calculate the maximum speed of the block.

(c) Calculate the speed of the block when it is $0.05 \mathrm{ m}$ from the equilibrium position.

(d) Calculate the acceleration of the block when it is $0.05 \mathrm{ m}$ from the equilibrium position.

(e) Sketch a graph showing how the kinetic energy and potential energy vary with displacement over one complete oscillation.

Solution

(a) $E_{\mathrm{total}} = \frac{1}{2}kA^2 = \frac{1}{2}(150)(0.10)^2 = \frac{1}{2}(150)(0.01) = 0.75 \mathrm{ J}$

(b) Maximum speed occurs at equilibrium ($x = 0$), where all energy is kinetic:

$\frac{1}{2}mv_{\max}^2 = 0.75 \implies v_{\max} = \sqrt{\frac{2 \times 0.75}{3.0}} = \sqrt{0.50} = 0.707 \mathrm{ m/s}$

(c) At $x = 0.05 \mathrm{ m}$:

$E_k = E_{\mathrm{total}} - \frac{1}{2}kx^2 = 0.75 - \frac{1}{2}(150)(0.0025) = 0.75 - 0.1875 = 0.5625 \mathrm{ J}$

$v = \sqrt{\frac{2E_k}{m}} = \sqrt{\frac{2 \times 0.5625}{3.0}} = \sqrt{0.375} = 0.612 \mathrm{ m/s}$

(d) $F = -kx = -150 \times 0.05 = -7.5 \mathrm{ N}$

$a = \frac{F}{m} = \frac{-7.5}{3.0} = -2.5 \mathrm{ m/s}^2$

(The negative sign indicates the acceleration is directed towards the equilibrium position.)

(e) The KE is maximum at $x = 0$ (parabolic decrease with $x$): $E_k = \frac{1}{2}k(A^2 - x^2)$. The PE is maximum at $x = \pm A$ (parabolic increase with $x$): $E_p = \frac{1}{2}kx^2$. The total energy $E_k + E_p = 0.75 \mathrm{ J}$ is constant (a horizontal line). The KE and PE curves are inverted parabolas that sum to the constant total.

Extended Derivation: Power Dissipated by a Falling Object

An object of mass $m$ falls from height $h$. The power dissipated by air resistance at any instant is:

$P_{\mathrm{air}} = F_{\mathrm{air}} \times v = (mg - ma) \times v$

At terminal velocity, $a = 0$, so $F_{\mathrm{air}} = mg$ and:

$P_{\mathrm{terminal}} = mgv_{\mathrm{terminal}}$

The gravitational power input ($mgv$) exactly equals the power dissipated by air resistance.

Extended Worked Example: Terminal Velocity

A raindrop of mass $5.0 \times 10^{-7} \mathrm{ kg}$ falls through air. The air resistance force is given by $F_{\mathrm{air}} = kv^2$, where $k = 2.0 \times 10^{-5} \mathrm{ kg/m}$. Calculate the terminal velocity.

Solution

At terminal velocity: $mg = kv_{\mathrm{terminal}}^2$

$v_{\mathrm{terminal}} = \sqrt{\frac{mg}{k}} = \sqrt{\frac{5.0 \times 10^{-7} \times 9.81}{2.0 \times 10^{-5}}} = \sqrt{\frac{4.905 \times 10^{-6}}{2.0 \times 10^{-5}}} = \sqrt{0.245} = 0.495 \mathrm{ m/s}$

Extended Worked Example: Efficiency of a Machine

A machine lifts a load of $800 \mathrm{ kg}$ through $5.0 \mathrm{ m}$ in $30 \mathrm{ s}$. The machine is powered by an electric motor connected to a $240 \mathrm{ V}$ supply drawing $12 \mathrm{ A}$.

(a) Calculate the useful power output. (b) Calculate the electrical power input. (c) Calculate the efficiency. (d) If the motor runs for 8 hours per day, calculate the daily energy cost at USD 1.50 per kWh.

Solution

(a) $P_{\mathrm{out}} = \frac{mgh}{t} = \frac{800 \times 9.81 \times 5.0}{30} = \frac{39240}{30} = 1308 \mathrm{ W}$

(b) $P_{\mathrm{in}} = VI = 240 \times 12 = 2880 \mathrm{ W}$

(c) $\eta = \frac{1308}{2880} \times 100\% = 45.4\%$

(d) Daily energy consumption: $E = P_{\mathrm{in}} \times t = 2880 \times 8 = 23040 \mathrm{ Wh} = 23.04 \mathrm{ kWh}$

Daily cost: 23.04 \times 1.50 = \34.56$