Work, Energy and Power

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Board Coverage AQA Paper 1 | Edexcel CP1, CP2 | OCR (A) Paper 1 | CIE P2

Energy Skate Park: Basics

Explore the simulation above to develop intuition for this topic.

1. Work Done by a Force

Definition. The work done by a constant force $\mathbf{F}$ when its point of application moves through displacement $\mathbf{s}$ is:

$W = \mathbf{F} \cdot \mathbf{s} = Fs\cos\theta$

where $\theta$ is the angle between $\mathbf{F}$ and $\mathbf{s}$ .

Derivation. We decompose $\mathbf{F}$ into components parallel and perpendicular to $\mathbf{s}$ . Only the parallel component $F\cos\theta$ does work (the perpendicular component produces no displacement in its direction). Work is defined as force multiplied by the displacement in the direction of the force, hence $W = (F\cos\theta) \cdot s = Fs\cos\theta$ .

Units. Work has the unit joule (J), where $1 \mathrm{ J} = 1 \mathrm{ N m} = 1 \mathrm{ kg m}^2\mathrm{s}^{-2}$ .

Sign convention. Work is positive when the force has a component in the direction of motion, negative when opposing motion.

Work done by a variable force. For a force $F(x)$ that varies with position:

$W = \int_{x_1}^{x_2} F(x)\,dx$

2. Kinetic Energy

Definition. The kinetic energy of a body of mass $m$ moving with velocity $v$ is:

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

Derivation from the work-energy theorem. Consider a body of mass $m$ accelerated from rest to speed $v$ by a constant net force $F$ over distance $s$ . The work done is:

$W = Fs = mas$

Using $v^2 = u^2 + 2as$ with $u = 0$ : $s = v^2/(2a)$ . Therefore:

$W = ma \cdot \frac{v^2}{2a} = \frac{1}{2}mv^2$

We define this work done as the kinetic energy gained. $\square$

General proof (variable force). From Newton's second law: $F = m\frac{dv}{dt}$ . The work done is:

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

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

$\boxed{W_{\mathrm{net}} = \Delta E_k}$

3. Gravitational Potential Energy

Definition. The gravitational potential energy of a body of mass $m$ at height $h$ above a reference level is:

$\boxed{E_p = mgh}$

Derivation. The work done against gravity in lifting a mass $m$ through a vertical height $h$ at constant velocity (so $a = 0$ and applied force $= mg$ ):

$W = Fd = mg \times h = mgh$

This work is stored as gravitational potential energy. $\square$

Intuition. $E_p$ depends on the choice of reference level. Only changes in potential energy are physically meaningful. The convention is to take $E_p = 0$ at the lowest point of the system.

4. Elastic Potential Energy

Definition. The elastic potential energy stored in a spring (or any elastic body obeying Hooke's law) extended by $x$ from its natural length is:

$\boxed{E_e = \frac{1}{2}kx^2}$

where $k$ is the spring constant (force per unit extension).

Derivation. By Hooke's law, the force required to extend the spring by $x$ is $F = kx$ . Since the force varies with extension, we integrate:

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

This work is stored as elastic potential energy. $\square$

Graphical interpretation. On a force-extension graph, the elastic potential energy is the area under the line $F = kx$ , which is a triangle of base $x$ and height $kx$ , giving $\frac{1}{2}kx^2$ .

Springs in Series and Parallel

Parallel (same extension, forces add). Two springs with constants $k_1$ and $k_2$ share the same extension $x$ :

$E_e = \frac{1}{2}k_1 x^2 + \frac{1}{2}k_2 x^2 = \frac{1}{2}(k_1 + k_2)x^2 = \frac{1}{2}k_{\mathrm{eff}}x^2$

$\boxed{k_{\mathrm{parallel}} = k_1 + k_2}$

Proof from equilibrium. Both springs exert force on the mass: $F = k_1 x + k_2 x = (k_1 + k_2)x$ , so $k_{\mathrm{eff}} = k_1 + k_2$ . $\square$

Series (same force, extensions add). Two springs with constants $k_1$ and $k_2$ experience the same force $F$ :

$F = k_1 x_1 = k_2 x_2$ , so $x_1 = F/k_1$ and $x_2 = F/k_2$ . Total extension: $x = x_1 + x_2 = F(1/k_1 + 1/k_2)$ .

$\boxed{\frac◆LB◆1◆RB◆◆LB◆k_{\mathrm{series}}◆RB◆ = \frac{1}{k_1} + \frac{1}{k_2}}$

Intuition. Parallel springs are stiffer (each contributes to resisting displacement). Series springs are less stiff (each stretches under the same load, giving more total compliance). This is analogous to electrical circuits: parallel resistors decrease, series resistors increase.

5. Conservation of Energy

Principle of Conservation of Energy. Energy cannot be created or destroyed, only transferred from one form to another.

$\sum E_{\mathrm{initial}} = \sum E_{\mathrm{final}}$

For a conservative system (no dissipative forces like friction):

$E_k + E_p + E_e = \mathrm{constant}$

Proof for a mass falling under gravity. A mass $m$ falls from height $h_1$ (at rest) to height $h_2$ (speed $v$ ). By the work-energy theorem:

$W_{\mathrm{gravity}} = \Delta E_k$

But $W_{\mathrm{gravity}} = mg(h_1 - h_2) = -\Delta E_p$ . Therefore:

$-\Delta E_p = \Delta E_k \implies \Delta E_k + \Delta E_p = 0$

$E_k + E_p = \mathrm{constant} \quad \square$

With non-conservative forces:

$W_{\mathrm{nc}} = \Delta E_k + \Delta E_p$

where $W_{\mathrm{nc}}$ is the work done by non-conservative forces (e.g., friction does negative work, reducing total mechanical energy).

Energy Dissipation by Friction

When friction acts, mechanical energy is not conserved -- it decreases monotonically as energy is transferred to thermal energy. For a block sliding to rest on a rough horizontal surface:

$E_k(\mathrm{initial}) = E_k(\mathrm{final}) + W_f$

where $W_f = \mu mg d$ is the work done against friction over distance $d$ .

Setting $E_k(\mathrm{final}) = 0$ (block comes to rest):

$\frac{1}{2}mv^2 = \mu mgd$

$\boxed{d = \frac◆LB◆v^2◆RB◆◆LB◆2\mu g◆RB◆}$

Intuition. The stopping distance is proportional to $v^2$ (not $v$ ), which is why driving at twice the speed requires four times the braking distance. This quadratic dependence is a direct consequence of energy conservation: kinetic energy scales as $v^2$ , so removing that energy requires friction to act over a distance proportional to $v^2$ .

warning

warning never lost -- it is transferred from mechanical to thermal (internal) energy. The total energy of the closed system is always conserved. Only mechanical energy ( $E_k + E_p$ ) decreases.

Example: Braking distance comparison

A car travelling at 30 mph brakes to rest in 23 m. What is the braking distance at 60 mph?

Answer. Since $d \propto v^2$ : $d_2 = d_1 \times (v_2/v_1)^2 = 23 \times (60/30)^2 = 23 \times 4 = 92$ m.

Doubling the speed quadruples the braking distance -- a critical road safety result.

Enrichment: Noether's Theorem

Emmy Noether proved in 1915 that every conservation law corresponds to a symmetry of nature. Conservation of energy corresponds to time-translation symmetry — the laws of physics do not change over time. If you perform an experiment today or tomorrow, you expect the same result. This deep connection between symmetry and conservation is one of the most profound results in physics.

6. Power

Definition. Power is the rate at which work is done (or energy is transferred):

$\boxed{P = \frac{dW}{dt} = \frac◆LB◆\Delta E◆RB◆◆LB◆\Delta t◆RB◆}$

Units. 1 watt (W) $= 1$ J s $^{-1}$ .

Power and Velocity

For a force $F$ moving a body at velocity $v$ :

$P = \frac{dW}{dt} = \frac{F \, ds}{dt} = Fv$

More generally, for a force at angle $\theta$ to the velocity:

$\boxed{P = Fv\cos\theta}$

Average Power for Constant Acceleration

When a constant force accelerates a body from $u$ to $v$ over distance $s$ :

$P_{\mathrm{avg}} = \frac{W}{t} = \frac{Fs}{t} = F \cdot \frac{s}{t} = F\bar{v}$

Since the average velocity is $\bar{v} = (u + v)/2$ :

$\boxed{P_{\mathrm{avg}} = \frac{F(u + v)}{2}}$

This is useful because it relates power to the initial and final velocities directly, without needing to know the distance or time separately.

tip

Exam Technique When asked for "the power developed by the engine" as a car accelerates, specify whether you mean instantaneous power ( $P = Fv$ ) or average power ( $P = Fs/t$ ). The instantaneous power at the end of the acceleration is $Fv_{\mathrm{final}}$ , while the average power over the whole acceleration is $F(u+v)/2$ .

Definition. The efficiency of an energy transfer is:

$\eta = \frac◆LB◆\mathrm{useful energy output}◆RB◆◆LB◆\mathrm{total energy input}◆RB◆ \times 100\%$

or equivalently for power:

$\eta = \frac◆LB◆P_{\mathrm{out}}◆RB◆◆LB◆P_{\mathrm{in}}◆RB◆ \times 100\%$

Efficiency is always between 0% and 100%. In practice, some energy is always dissipated (usually as thermal energy due to friction or resistance).

8. Energy as the Universal Currency

Energy is the most unifying concept in physics. It connects mechanics, thermodynamics, electromagnetism, waves, quantum physics, and relativity. The conservation of energy is the one principle that holds across all domains:

In mechanics, $E_k + E_p = \mathrm{const}$ .
In circuits, $E = VIt = I^2Rt$ is dissipated as heat.
In waves, intensity (energy per unit area per unit time) decreases with distance.
In nuclear physics, $E = \Delta mc^2$ converts mass to energy.
In thermodynamics, the first law $\Delta U = Q + W$ is a statement of energy conservation.

Problem Set

Details

Problem 1

A car of mass

1200

kg accelerates from rest to

25

m s

^{-1}

8.0

s. Calculate: (a) the kinetic energy gained, (b) the average power output, (c) the average force.

Answer. (a) $E_k = \frac{1}{2}(1200)(25)^2 = 375\,000$ J $= 375$ kJ.

(b) $P = \frac◆LB◆\Delta E_k◆RB◆◆LB◆\Delta t◆RB◆ = \frac{375\,000}{8.0} = 46\,900$ W $= 46.9$ kW.

(c) $P = Fv_{\mathrm{avg}}$ , $v_{\mathrm{avg}} = 12.5$ m s $^{-1}$ . $F = 46\,900/12.5 = 3750$ N. Alternatively, $F = ma = 1200 \times 25/8 = 3750$ N.

If you get this wrong, revise: Kinetic Energy and Power and Velocity

Details

Problem 2

A spring of spring constant

500

N m

^{-1}

is compressed by

0.08

m. A

0.50

kg ball is placed against it and released. Calculate the speed of the ball as it leaves the spring (assuming no energy losses).

Answer. Elastic PE converted to KE: $\frac{1}{2}kx^2 = \frac{1}{2}mv^2$ .

$v = \sqrt◆LB◆\frac{kx^2}{m}◆RB◆ = \sqrt◆LB◆\frac◆LB◆500 \times 0.0064◆RB◆◆LB◆0.50◆RB◆◆RB◆ = \sqrt{6.4} = 2.53$ m s $^{-1}$ .

If you get this wrong, revise: Conservation of Energy

Details

Problem 3

A roller coaster car of mass

500

kg starts from rest at point A, 20 m above the ground. It descends to point B at ground level, then rises to point C at 12 m above the ground. Assuming no friction, find the speed at B and C.

Answer. At B: $\frac{1}{2}(500)v_B^2 = 500 \times 9.81 \times 20$ . $v_B = \sqrt◆LB◆2 \times 9.81 \times 20◆RB◆ = \sqrt{392.4} = 19.8$ m s $^{-1}$ .

At C: $E_k(C) = mg(h_A - h_C) = 500 \times 9.81 \times 8 = 39\,240$ J. $v_C = \sqrt◆LB◆2 \times 39\,240/500◆RB◆ = \sqrt{156.96} = 12.5$ m s $^{-1}$ .

If you get this wrong, revise: Conservation of Energy

Details

Problem 4

A motor lifts a

200

kg load through a height of

15

m in

10

s. If the motor is

80\%

efficient and runs on a

240

V supply, calculate the current it draws.

Answer. Useful power: $P_{\mathrm{out}} = \frac{mgh}{t} = \frac◆LB◆200 \times 9.81 \times 15◆RB◆◆LB◆10◆RB◆ = 2943$ W.

Electrical power: $P_{\mathrm{in}} = \frac◆LB◆P_{\mathrm{out}}◆RB◆◆LB◆\eta◆RB◆ = \frac{2943}{0.80} = 3679$ W.

Current: $I = \frac◆LB◆P_{\mathrm{in}}◆RB◆◆LB◆V◆RB◆ = \frac{3679}{240} = 15.3$ A.

If you get this wrong, revise: Efficiency

Details

Problem 5

A pendulum bob of mass

0.50

kg is pulled aside until it is

0.10

m above its lowest point and released. Find its speed at the lowest point.

Answer. $mgh = \frac{1}{2}mv^2 \implies v = \sqrt{2gh} = \sqrt◆LB◆2 \times 9.81 \times 0.10◆RB◆ = \sqrt{1.962} = 1.40$ m s $^{-1}$ .

If you get this wrong, revise: Conservation of Energy

Details

Problem 6

A crate is pushed 8.0 m across a rough floor by a force of

150

N at

30^\circ

below the horizontal. If

\mu = 0.25

and the crate has mass

40

kg, find: (a) the work done by the applied force, (b) the work done against friction, (c) the final speed (starting from rest).

Answer. (a) $W_F = Fd\cos\theta = 150 \times 8.0 \times \cos 30° = 150 \times 8.0 \times 0.866 = 1039$ J.

(b) Normal reaction: $R = mg + F\sin 30° = 40 \times 9.81 + 150 \times 0.5 = 392.4 + 75 = 467.4$ N. Friction: $F_f = 0.25 \times 467.4 = 116.9$ N. $W_f = 116.9 \times 8.0 = 935$ J.

(c) Net work = $1039 - 935 = 104$ J. $W_{\mathrm{net}} = \Delta E_k = \frac{1}{2}mv^2$ . $v = \sqrt◆LB◆2 \times 104/40◆RB◆ = \sqrt{5.2} = 2.28$ m s $^{-1}$ .

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

Details

Problem 7

A car of mass

800

kg travels at constant speed

20

m s

^{-1}

up a hill inclined at

5^\circ

to the horizontal. The engine has an efficiency of

25\%

. Calculate the rate of fuel energy consumption (in watts).

Answer. Component of weight down the slope: $mg\sin 5° = 800 \times 9.81 \times 0.0872 = 684$ N.

At constant speed, engine force = resistance = $684$ N (ignoring other losses).

Power output: $P = Fv = 684 \times 20 = 13\,680$ W.

Fuel power: $P_{\mathrm{fuel}} = P/\eta = 13\,680/0.25 = 54\,700$ W $= 54.7$ kW.

If you get this wrong, revise: Power and Velocity and Efficiency

Details

Problem 8

Two springs are connected in parallel (both attached to the same mass). Spring A has

k_A = 200

N m

^{-1}

, spring B has

k_B = 300

N m

^{-1}

. The mass is displaced

0.05

m. Find the total elastic PE stored.

Answer. For parallel springs, the effective spring constant is $k_{\mathrm{eff}} = k_A + k_B = 500$ N m $^{-1}$ .

$E_e = \frac{1}{2}k_{\mathrm{eff}}x^2 = \frac{1}{2}(500)(0.0025) = 0.625$ J.

If you get this wrong, revise: Elastic Potential Energy

Details

Problem 9

A ball is thrown vertically upward with speed

15

m s

^{-1}

from a height of

2.0

m. Use energy conservation to find the maximum height above the ground.

Answer. At launch: $E = E_k + E_p = \frac{1}{2}m(15)^2 + mg(2) = 112.5m + 19.62m = 132.1m$ .

At max height: $E = mgh_{\max}$ . So $h_{\max} = 132.1/9.81 = 13.5$ m.

If you get this wrong, revise: Conservation of Energy

Details

Problem 10

A satellite of mass

500

kg is in a circular orbit at altitude

300

km above Earth's surface. Given

g = 9.81

m s

^{-2}

at the surface and

R_E = 6370

km, calculate the total energy of the satellite. (The total energy of a circular orbit is

E = -\frac{GMm}{2r} = -\frac{1}{2}mgh

is not valid here — use

E = -GMm/(2r)

with

GM = gR_E^2

Answer. $r = R_E + h = 6\,670\,000$ m. $GM = gR_E^2 = 9.81 \times (6.37 \times 10^6)^2 = 3.976 \times 10^{14}$ .

$E = -\frac{GMm}{2r} = -\frac◆LB◆3.976 \times 10^{14} \times 500◆RB◆◆LB◆2 \times 6.67 \times 10^6◆RB◆ = -\frac◆LB◆1.988 \times 10^{17}◆RB◆◆LB◆1.334 \times 10^7◆RB◆ = -1.49 \times 10^{10}$ J $= -14.9$ GJ.

If you get this wrong, revise: Conservation of Energy and 08-gravitational-fields

Problem 11

A spring of constant $200$ N m $^{-1}$ is connected in series with a spring of constant $300$ N m $^{-1}$ . A $2.0$ kg mass is attached and displaced $0.10$ m. Find: (a) the total extension, (b) the total elastic potential energy stored.

Answer. (a) For series: $1/k_{\mathrm{eff}} = 1/200 + 1/300 = 5/600$ . $k_{\mathrm{eff}} = 120$ N m $^{-1}$ .

Extension: $x = F/k_{\mathrm{eff}} = k_{\mathrm{eff}} \times 0.10 / k_{\mathrm{eff}} = 0.10$ m. (Each spring shares the same force: $F = k_1 x_1 = k_2 x_2$ . $x_1 = F/k_1 = 0.10 \times 120/200 = 0.060$ m, $x_2 = 0.10 \times 120/300 = 0.040$ m. Total $= 0.100$ m, confirming consistency.)

(b) $E_e = \frac{1}{2}k_{\mathrm{eff}}x^2 = \frac{1}{2}(120)(0.010) = 0.60$ J.

Alternatively: $E_e = \frac{1}{2}k_1 x_1^2 + \frac{1}{2}k_2 x_2^2 = \frac{1}{2}(200)(0.0036) + \frac{1}{2}(300)(0.0016) = 0.36 + 0.24 = 0.60$ J.

If you get this wrong, revise: Springs in Series and Parallel

Problem 12

A block of mass $3.0$ kg is projected up a rough incline at $30^\circ$ with speed $8.0$ m s $^{-1}$ . The coefficient of friction is $0.20$ . Use energy methods to find how far up the incline it travels before stopping.

Answer. Initial KE: $\frac{1}{2}(3.0)(64) = 96$ J. Gain in PE: $mg\Delta s \sin 30° = 3.0 \times 9.81 \times 0.5 \times \Delta s = 14.7\,\Delta s$ .

Work against friction: $\mu mg\cos 30° \times \Delta s = 0.20 \times 3.0 \times 9.81 \times 0.866 \times \Delta s = 5.10\,\Delta s$ .

Energy equation: $E_k = \Delta E_p + W_f$ . $96 = 14.7\,\Delta s + 5.10\,\Delta s = 19.8\,\Delta s$ .

$\Delta s = 96/19.8 = 4.85$ m.

If you get this wrong, revise: Energy Dissipation by Friction

Problem 13

A $70$ kg student runs up a flight of stairs, gaining $5.0$ m of height in $4.0$ s. Calculate: (a) the work done against gravity, (b) the average power output, (c) the minimum instantaneous power if the student starts from rest and accelerates uniformly.

Answer. (a) $W = mgh = 70 \times 9.81 \times 5.0 = 3434$ J.

(b) $P_{\mathrm{avg}} = W/t = 3434/4.0 = 859$ W $\approx 860$ W. (About 1.15 horsepower.)

$a = 2h/t^2 = 2 \times 5.0/16 = 0.625$ m s $^{-2}$ . $v_{\mathrm{final}} = 0.625 \times 4 = 2.5$ m s $^{-1}$ .

Vertical force = $mg + ma = 70(9.81 + 0.625) = 731$ N.

$P_{\mathrm{max}} = Fv = 731 \times 2.5 = 1828$ W. The minimum instantaneous power is the power at the start (when $v = 0$ ): $P_{\mathrm{min}} = 0$ . Power increases linearly with speed.

If you get this wrong, revise: Power and Velocity

Problem 14

A mass of $0.50$ kg is attached to a spring of constant $200$ N m $^{-1}$ on a smooth horizontal surface. The mass is pulled $0.15$ m from equilibrium and released. Find: (a) the maximum speed, (b) the speed when the mass is $0.05$ m from equilibrium, (c) the total energy of the system.

Answer. (a) At equilibrium: all energy is elastic. At maximum displacement: all energy is elastic. At any other point: $E_e + E_k = E_{\mathrm{total}}$ .

$E_{\mathrm{total}} = \frac{1}{2}(200)(0.15)^2 = 2.25$ J. Maximum speed at equilibrium ( $x = 0$ ): $E_k = E_{\mathrm{total}}$ . $v_{\mathrm{max}} = \sqrt◆LB◆2E_{\mathrm{total}}/m◆RB◆ = \sqrt◆LB◆2 \times 2.25/0.50◆RB◆ = \sqrt{9} = 3.0$ m s $^{-1}$ .

(b) At $x = 0.05$ m: $E_e = \frac{1}{2}(200)(0.0025) = 0.25$ J. $E_k = 2.25 - 0.25 = 2.0$ J. $v = \sqrt◆LB◆2 \times 2.0/0.50◆RB◆ = \sqrt{8} = 2.83$ m s $^{-1}$ .

If you get this wrong, revise: Conservation of Energy

Problem 15

A car engine delivers a constant driving force of $3000$ N. The total resistive force (drag + rolling resistance) is $F_{\mathrm{res}} = 300 + 2v^2$ N (where $v$ is in m s $^{-1}$ ). Find: (a) the maximum speed of the car (terminal velocity), (b) the power at this speed.

Answer. (a) At maximum speed, driving force equals resistive force: $3000 = 300 + 2v^2$ .

$2v^2 = 2700$ , $v^2 = 1350$ , $v = 36.7$ m s $^{-1} \approx 132$ km h $^{-1}$ .

(b) $P = Fv = 3000 \times 36.7 = 110\,100$ W $= 110$ kW.

Note: at maximum speed, all the engine's power goes into overcoming resistance. If the car tried to go faster, resistance would exceed the driving force and it would decelerate.

If you get this wrong, revise: Power and Velocity

Problem 16

A bungee jumper of mass $75$ kg jumps from a platform. The bungee cord has an unstretched length of $25$ m and spring constant $50$ N m $^{-1}$ . Taking the reference level as the jump platform, find the speed of the jumper when the cord is stretched $15$ m beyond its natural length.

Answer. Height below platform: $h = 25 + 15 = 40$ m. Loss of PE: $mgh = 75 \times 9.81 \times 40 = 29\,430$ J.

Elastic PE in cord: $E_e = \frac{1}{2}(50)(15)^2 = 5625$ J.

By conservation: $mgh = E_k + E_e$ . $29\,430 = \frac{1}{2}(75)v^2 + 5625$ .

$\frac{1}{2}(75)v^2 = 23\,805$ . $v^2 = 634.8$ . $v = 25.2$ m s $^{-1}$ .

If you get this wrong, revise: Conservation of Energy

Problem 17

An electric motor is 85% efficient. It lifts a $200$ kg crate vertically at a steady speed of $0.80$ m s $^{-1}$ . Calculate: (a) the useful power output, (b) the total electrical power input, (c) the energy wasted per second, (d) the total electrical energy consumed to lift the crate $12$ m.

Answer. (a) $P_{\mathrm{out}} = Fv = mgv = 200 \times 9.81 \times 0.80 = 1570$ W.

(b) $P_{\mathrm{in}} = P_{\mathrm{out}}/\eta = 1570/0.85 = 1847$ W.

(d) Time to lift 12 m: $t = 12/0.80 = 15$ s. Total electrical energy = $P_{\mathrm{in}} \times t = 1847 \times 15 = 27\,705$ J $= 27.7$ kJ.

If you get this wrong, revise: Efficiency

Problem 18

Two identical springs, each of constant $k = 400$ N m $^{-1}$ , are connected in parallel and support a mass of $5.0$ kg. The mass is displaced $0.05$ m downward and released. Calculate: (a) the period of the resulting oscillation (you may assume SHM), (b) the maximum kinetic energy, (c) the maximum elastic potential energy in each spring.

Answer. (a) For parallel springs: $k_{\mathrm{eff}} = 2k = 800$ N m $^{-1}$ .

Angular frequency: $\omega = \sqrt◆LB◆k_{\mathrm{eff}}/m◆RB◆ = \sqrt{800/5.0} = \sqrt{160} = 12.6$ rad s $^{-1}$ .

Period: $T = 2\pi/\omega = 2\pi/12.6 = 0.499$ s.

(b) Maximum KE = total energy $= \frac{1}{2}k_{\mathrm{eff}}A^2 = \frac{1}{2}(800)(0.05)^2 = 1.0$ J.

If you get this wrong, revise: Springs in Series and Parallel

tip

Diagnostic Test Ready to test your understanding of Work, Energy and Power? The diagnostic test contains the hardest questions within the A-Level specification for this topic, each with a full worked solution.

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

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

danger

Common Pitfalls

Confusing work done on an object with work done by gravity: Work done by gravity when an object falls is POSITIVE (force and displacement are in the same direction). Work done AGAINST gravity when lifting is NEGATIVE. When using energy conservation, gravitational PE LOSS equals kinetic energy GAIN -- be careful with signs.
Applying conservation of energy when non-conservative forces act: The work-energy principle (W_net = delta(E_k)) always applies, but conservation of mechanical energy (E_k + E_p = constant) only holds when no non-conservative forces (friction, air resistance) do work. If friction is present, mechanical energy is lost to heat.
Forgetting the angle in W = Fd cos(theta): Work is maximum when the force is parallel to displacement (theta = 0, cos = 1) and zero when perpendicular (theta = 90, cos = 0). A force applied at right angles to motion (e.g., centripetal force in circular motion) does NO work and does not change kinetic energy.
Confusing power and energy: Power is the RATE of doing work (P = W/t = Fv), measured in watts. Energy is the total work done or transferred, measured in joules. A machine that is twice as powerful delivers the same energy in half the time, not twice the energy.