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 $1 \mathrm{ J} = 1 \mathrm{ N\, m}$.

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.

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 $4 \mathrm{ m}$.

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.

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.

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 $x = 0.02 \mathrm{ m}$.

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

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.

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?

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.

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 $0.03 \mathrm{ m}$.

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.

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

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 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 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.

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.

(a)

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.

(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.

(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.

Details

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.

Worked Examples

Example 1: Work-energy theorem

Problem. A $4 \mathrm{ kg}$ block is pushed $5 \mathrm{ m}$ across a rough floor ($\mu_k = 0.2$) by a horizontal force of $30 \mathrm{ N}$, starting from rest. Find its final speed.

Solution. Friction: $f = 0.2 \times 4 \times 9.81 = 7.85 \mathrm{ N}$. Net work: $W_{\text{net}} = (30 - 7.85) \times 5 = 110.75 \mathrm{ J}$. $v = \sqrt{\frac{2 \times 110.75}{4}} = \sqrt{55.38} = 7.44 \mathrm{ m/s}$.

$\blacksquare$

Example 2: Conservation of energy

Problem. A $500 \mathrm{ kg}$ roller coaster starts from rest $20 \mathrm{ m}$ above ground and descends to $5 \mathrm{ m}$ above ground. Find the speed at the lower point (ignore friction).