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Work, Energy and Power

Work, Energy and Power

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

Explore the simulation above to develop intuition for this topic.

Intuition

Energy is the ability to do work — and work is energy transferred by a force: Think of pushing a box across a floor. You do work on the box because you apply a force and the box moves. The energy you expend becomes kinetic energy of the box (if it speeds up) or heat (if there’s friction). Energy isn’t “used up” — it’s transformed from one form to another.

Why it matters: Work-energy principles explain why roller coasters work (potential → kinetic → potential), why brakes get hot (kinetic → heat), and why power stations exist (chemical/nuclear → electrical). Understanding energy transformations is the key to understanding all physical processes.

The key insight: Power is the rate of energy transfer, not the amount. A 100W bulb and a 1000W heater both convert electrical energy, but the heater does it 10 times faster. This is why a small fire can be more dangerous than a large one — it’s not the total energy but how quickly it’s delivered.

1. Work Done by a Force

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

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

Where θ\theta is the angle between F\mathbf{F} and s\mathbf{s}.

Derivation. We decompose F\mathbf{F} into components parallel and perpendicular to s\mathbf{s}. Only the parallel component FcosθF\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=(Fcosθ)s=FscosθW = (F\cos\theta) \cdot s = Fs\cos\theta.

Units. Work has the unit joule (J), where 1J=1Nm=1kgm2s21 \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)F(x) that varies with position:

W=x1x2F(x)dxW = \int_{x_1}^{x_2} F(x)\,dx

2. Kinetic Energy

Definition. The kinetic energy of a body of mass mm moving with velocity vv is:

Ek=12mv2\boxed{E_k = \frac{1}{2}mv^2}

Derivation from the work-energy theorem. Consider a body of mass mm accelerated from rest to Speed vv by a constant net force FF over distance ss. The work done is:

W=Fs=masW = Fs = mas

Using v2=u2+2asv^2 = u^2 + 2as with u=0u = 0: s=v2/(2a)s = v^2/(2a). Therefore:

W=mav22a=12mv2W = 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=mdvdtF = m\frac{dv}{dt}. The work done Is:

W=s1s2Fds=s1s2mdvdtds=v1v2mvdv=12mv2212mv12W = \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.

Wnet=ΔEk\boxed{W_{\mathrm{net}} = \Delta E_k}

3. Gravitational Potential Energy

Definition. The gravitational potential energy of a body of mass mm at height hh above a Reference level is:

Ep=mgh\boxed{E_p = mgh}

Derivation. The work done against gravity in lifting a mass mm through a vertical height hh at Constant velocity (so a=0a = 0 and applied force =mg= mg):

W=Fd=mg×h=mghW = Fd = mg \times h = mgh

This work is stored as gravitational potential energy. \square

Intuition. EpE_p depends on the choice of reference level. Only changes in potential energy Are physically meaningful. The convention is to take Ep=0E_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 xx from its natural length is:

Ee=12kx2\boxed{E_e = \frac{1}{2}kx^2}

Where kk is the spring constant (force per unit extension).

Derivation. By Hooke’s law, the force required to extend the spring by xx is F=kxF = kx. Since The force varies with extension, we integrate:

W=0xFdx=0xkxdx=[12kx2]0x=12kx2W = \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=kxF = kxWhich is a triangle of base xx and height kxkxGiving 12kx2\frac{1}{2}kx^2.

Springs in Series and Parallel

Parallel (same extension, forces add). Two springs with constants k1k_1 and k2k_2 share the same Extension xx:

Ee=12k1x2+12k2x2=12(k1+k2)x2=12keffx2E_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

kparallel=k1+k2\boxed{k_{\mathrm{parallel}} = k_1 + k_2}

Proof from equilibrium. Both springs exert force on the mass: F=k1x+k2x=(k1+k2)xF = k_1 x + k_2 x = (k_1 + k_2)xSo keff=k1+k2k_{\mathrm{eff}} = k_1 + k_2. \square

Series (same force, extensions add). Two springs with constants k1k_1 and k2k_2 experience the Same force FF:

F=k1x1=k2x2F = k_1 x_1 = k_2 x_2So x1=F/k1x_1 = F/k_1 and x2=F/k2x_2 = F/k_2. Total extension: x=x1+x2=F(1/k1+1/k2)x = x_1 + x_2 = F(1/k_1 + 1/k_2).

1kseries=1k1+1k2\boxed{\frac{1}{k_{\mathrm{series}}} = \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.

Einitial=Efinal\sum E_{\mathrm{initial}} = \sum E_{\mathrm{final}}

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

Ek+Ep+Ee=constantE_k + E_p + E_e = \mathrm{constant}

Proof for a mass falling under gravity. A mass mm falls from height h1h_1 (at rest) to height h2h_2 (speed vv). By the work-energy theorem:

Wgravity=ΔEkW_{\mathrm{gravity}} = \Delta E_k

But Wgravity=mg(h1h2)=ΔEpW_{\mathrm{gravity}} = mg(h_1 - h_2) = -\Delta E_p. Therefore:

ΔEp=ΔEk    ΔEk+ΔEp=0-\Delta E_p = \Delta E_k \implies \Delta E_k + \Delta E_p = 0

Ek+Ep=constantE_k + E_p = \mathrm{constant} \quad \square

With non-conservative forces:

Wnc=ΔEk+ΔEpW_{\mathrm{nc}} = \Delta E_k + \Delta E_p

Where WncW_{\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:

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

Where Wf=μmgdW_f = \mu mg d is the work done against friction over distance dd.

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

12mv2=μmgd\frac{1}{2}mv^2 = \mu mgd

d=v22μg\boxed{d = \frac{v^2}{2\mu g}}

Intuition. The stopping distance is proportional to v2v^2 (not vv), 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 v2v^2So removing that energy requires Friction to act over a distance proportional to v2v^2.

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 dv2d \propto v^2: d2=d1×(v2/v1)2=23×(60/30)2=23×4=92d_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):

P=dWdt=ΔEΔt\boxed{P = \frac{dW}{dt} = \frac{\Delta E}{\Delta t}}

Units. 1 watt (W) =1= 1 J s1^{-1}.

Power and Velocity

For a force FF moving a body at velocity vv:

P=dWdt=Fdsdt=FvP = \frac{dW}{dt} = \frac{F \, ds}{dt} = Fv

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

P=Fvcosθ\boxed{P = Fv\cos\theta}

Average Power for Constant Acceleration

When a constant force accelerates a body from uu to vv over distance ss:

Pavg=Wt=Fst=Fst=FvˉP_{\mathrm{avg}} = \frac{W}{t} = \frac{Fs}{t} = F \cdot \frac{s}{t} = F\bar{v}

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

Pavg=F(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.

Definition. The efficiency of an energy transfer is:

η=usefulenergyoutputtotalenergyinput×100%\eta = \frac{\mathrm{useful energy output}}{\mathrm{total energy input}} \times 100\%

Or equivalently for power:

η=PoutPin×100%\eta = \frac{P_{\mathrm{out}}}{P_{\mathrm{in}}} \times 100\%

Efficiency is always between 0% and 100%. In practice, some energy is always dissipated ( 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, Ek+Ep=constE_k + E_p = \mathrm{const}.
  • In circuits, E=VIt=I2RtE = VIt = I^2Rt is dissipated as heat.
  • In waves, intensity (energy per unit area per unit time) decreases with distance.
  • In nuclear physics, E=Δmc2E = \Delta mc^2 converts mass to energy.
  • In thermodynamics, the first law ΔU=Q+W\Delta U = Q + W is a statement of energy conservation.

Problem Set

Problem 1A car of mass 12001200 kg accelerates from rest to 2525 m s1^{-1} in 8.08.0 s. Calculate: (a) the kinetic energy gained, (b) the average power output, (c) the average force.

Answer. (a) Ek=12(1200)(25)2=375000E_k = \frac{1}{2}(1200)(25)^2 = 375\,000 J =375= 375 kJ.

(b) P=ΔEkΔt=3750008.0=46900P = \frac{\Delta E_k}{\Delta t} = \frac{375\,000}{8.0} = 46\,900 W =46.9= 46.9 kW.

(c) P=FvavgP = Fv_{\mathrm{avg}}, vavg=12.5v_{\mathrm{avg}} = 12.5 m s1^{-1}. F=46900/12.5=3750F = 46\,900/12.5 = 3750 N. Alternatively, F=ma=1200×25/8=3750F = ma = 1200 \times 25/8 = 3750 N.

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

Problem 2A spring of spring constant 500500 N m1^{-1} is compressed by 0.080.08 m. A 0.500.50 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: 12kx2=12mv2\frac{1}{2}kx^2 = \frac{1}{2}mv^2.

v=kx2m=500×0.00640.50=6.4=2.53v = \sqrt{\frac{kx^2}{m}} = \sqrt{\frac{500 \times 0.0064}{0.50}} = \sqrt{6.4} = 2.53 m s1^{-1}.

If you get this wrong, revise: Conservation of Energy

Problem 3A roller coaster car of mass 500500 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: 12(500)vB2=500×9.81×20\frac{1}{2}(500)v_B^2 = 500 \times 9.81 \times 20. vB=2×9.81×20=392.4=19.8v_B = \sqrt{2 \times 9.81 \times 20} = \sqrt{392.4} = 19.8 m s1^{-1}.

At C: Ek(C)=mg(hAhC)=500×9.81×8=39240E_k(C) = mg(h_A - h_C) = 500 \times 9.81 \times 8 = 39\,240 J. vC=2×39240/500=156.96=12.5v_C = \sqrt{2 \times 39\,240/500} = \sqrt{156.96} = 12.5 m s1^{-1}.

If you get this wrong, revise: Conservation of Energy

Problem 4A motor lifts a 200200 kg load through a height of 1515 m in 1010 s. If the motor is 80%80\% efficient and runs on a 240240 V supply, calculate the current it draws.

Answer. Useful power: Pout=mght=200×9.81×1510=2943P_{\mathrm{out}} = \frac{mgh}{t} = \frac{200 \times 9.81 \times 15}{10} = 2943 W.

Electrical power: Pin=Poutη=29430.80=3679P_{\mathrm{in}} = \frac{P_{\mathrm{out}}}{\eta} = \frac{2943}{0.80} = 3679 W.

Current: I=PinV=3679240=15.3I = \frac{P_{\mathrm{in}}}{V} = \frac{3679}{240} = 15.3 A.

If you get this wrong, revise: Efficiency

Problem 5A pendulum bob of mass 0.500.50 kg is pulled aside until it is 0.100.10 m above its lowest point and released. Find its speed at the lowest point.

Answer. mgh=12mv2    v=2gh=2×9.81×0.10=1.962=1.40mgh = \frac{1}{2}mv^2 \implies v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 0.10} = \sqrt{1.962} = 1.40 M s1^{-1}.

If you get this wrong, revise: Conservation of Energy

Problem 6A crate is pushed 8.0 m across a rough floor by a force of 150150 N at 3030^\circ below the horizontal. If μ=0.25\mu = 0.25 and the crate has mass 4040 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) WF=Fdcosθ=150×8.0×cos30°=150×8.0×0.866=1039W_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+Fsin30°=40×9.81+150×0.5=392.4+75=467.4R = mg + F\sin 30° = 40 \times 9.81 + 150 \times 0.5 = 392.4 + 75 = 467.4 N. Friction: Ff=0.25×467.4=116.9F_f = 0.25 \times 467.4 = 116.9 N. Wf=116.9×8.0=935W_f = 116.9 \times 8.0 = 935 J.

(c) Net work = 1039935=1041039 - 935 = 104 J. Wnet=ΔEk=12mv2W_{\mathrm{net}} = \Delta E_k = \frac{1}{2}mv^2. v=2×104/40=5.2=2.28v = \sqrt{2 \times 104/40} = \sqrt{5.2} = 2.28 m s1^{-1}.

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

Problem 7A car of mass 800800 kg travels at constant speed 2020 m s1^{-1} up a hill inclined at 55^\circ to the horizontal. The engine has an efficiency of 25%25\%. Calculate the rate of fuel energy consumption (in watts).

Answer. Component of weight down the slope: mgsin5°=800×9.81×0.0872=684mg\sin 5° = 800 \times 9.81 \times 0.0872 = 684 N.

At constant speed, engine force = resistance = 684684 N (ignoring other losses).

Power output: P=Fv=684×20=13680P = Fv = 684 \times 20 = 13\,680 W.

Fuel power: Pfuel=P/η=13680/0.25=54700P_{\mathrm{fuel}} = P/\eta = 13\,680/0.25 = 54\,700 W =54.7= 54.7 kW.

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

Problem 8Two springs are connected in parallel (both attached to the same mass). Spring A has kA=200k_A = 200 N m1^{-1}Spring B has kB=300k_B = 300 N m1^{-1}. The mass is displaced 0.050.05 m. Find the total elastic PE stored.

Answer. For parallel springs, the effective spring constant is keff=kA+kB=500k_{\mathrm{eff}} = k_A + k_B = 500 N m1^{-1}.

Ee=12keffx2=12(500)(0.0025)=0.625E_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

Problem 9A ball is thrown vertically upward with speed 1515 m s1^{-1} from a height of 2.02.0 m. Use energy conservation to find the maximum height above the ground.

Answer. At launch: E=Ek+Ep=12m(15)2+mg(2)=112.5m+19.62m=132.1mE = E_k + E_p = \frac{1}{2}m(15)^2 + mg(2) = 112.5m + 19.62m = 132.1m.

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

If you get this wrong, revise: Conservation of Energy

Problem 10A satellite of mass 500500 kg is in a circular orbit at altitude 300300 km above Earth’s surface. Given g=9.81g = 9.81 m s2^{-2} at the surface and RE=6370R_E = 6370 km, calculate the total energy of the satellite. (The total energy of a circular orbit is E=GMm2r=12mghE = -\frac{GMm}{2r} = -\frac{1}{2}mgh is not valid here — use E=GMm/(2r)E = -GMm/(2r) with GM=gRE2GM = gR_E^2.)

Answer. r=RE+h=6670000r = R_E + h = 6\,670\,000 m. GM=gRE2=9.81×(6.37×106)2=3.976×1014GM = gR_E^2 = 9.81 \times (6.37 \times 10^6)^2 = 3.976 \times 10^{14}.

E=GMm2r=3.976×1014×5002×6.67×106=1.988×10171.334×107=1.49×1010E = -\frac{GMm}{2r} = -\frac{3.976 \times 10^{14} \times 500}{2 \times 6.67 \times 10^6} = -\frac{1.988 \times 10^{17}}{1.334 \times 10^7} = -1.49 \times 10^{10} J =14.9= -14.9 GJ.

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

Problem 11

A spring of constant 200200 N m1^{-1} is connected in series with a spring of constant 300300 N M1^{-1}. A 2.02.0 kg mass is attached and displaced 0.100.10 m. Find: (a) the total extension, (b) the Total elastic potential energy stored.

Answer. (a) For series: 1/keff=1/200+1/300=5/6001/k_{\mathrm{eff}} = 1/200 + 1/300 = 5/600. keff=120k_{\mathrm{eff}} = 120 N M1^{-1}.

Extension: x=F/keff=keff×0.10/keff=0.10x = F/k_{\mathrm{eff}} = k_{\mathrm{eff}} \times 0.10 / k_{\mathrm{eff}} = 0.10 m. (Each spring shares the same force: F=k1x1=k2x2F = k_1 x_1 = k_2 x_2. x1=F/k1=0.10×120/200=0.060x_1 = F/k_1 = 0.10 \times 120/200 = 0.060 m, x2=0.10×120/300=0.040x_2 = 0.10 \times 120/300 = 0.040 m. Total =0.100= 0.100 m, confirming consistency.)

(b) Ee=12keffx2=12(120)(0.010)=0.60E_e = \frac{1}{2}k_{\mathrm{eff}}x^2 = \frac{1}{2}(120)(0.010) = 0.60 J.

Alternatively: Ee=12k1x12+12k2x22=12(200)(0.0036)+12(300)(0.0016)=0.36+0.24=0.60E_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 3.03.0 kg is projected up a rough incline at 3030^\circ with speed 8.08.0 m s1^{-1}. The coefficient of friction is 0.200.20. Use energy methods to find how far up the incline it travels Before stopping.

Answer. Initial KE: 12(3.0)(64)=96\frac{1}{2}(3.0)(64) = 96 J. Gain in PE: mgΔssin30°=3.0×9.81×0.5×Δs=14.7Δsmg\Delta s \sin 30° = 3.0 \times 9.81 \times 0.5 \times \Delta s = 14.7\,\Delta s.

Work against friction: μmgcos30°×Δs=0.20×3.0×9.81×0.866×Δs=5.10Δs\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: Ek=ΔEp+WfE_k = \Delta E_p + W_f. 96=14.7Δs+5.10Δs=19.8Δs96 = 14.7\,\Delta s + 5.10\,\Delta s = 19.8\,\Delta s.

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

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

Problem 13

A 7070 kg student runs up a flight of stairs, gaining 5.05.0 m of height in 4.04.0 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=70×9.81×5.0=3434W = mgh = 70 \times 9.81 \times 5.0 = 3434 J.

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

(c) Starting from rest with constant acceleration, h=12at2h = \frac{1}{2}at^2 and v=atv = at.

a=2h/t2=2×5.0/16=0.625a = 2h/t^2 = 2 \times 5.0/16 = 0.625 m s2^{-2}. vfinal=0.625×4=2.5v_{\mathrm{final}} = 0.625 \times 4 = 2.5 m S1^{-1}.

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

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

If you get this wrong, revise: Power and Velocity

Problem 14

A mass of 0.500.50 kg is attached to a spring of constant 200200 N m1^{-1} on a smooth horizontal Surface. The mass is pulled 0.150.15 m from equilibrium and released. Find: (a) the maximum speed, (b) The speed when the mass is 0.050.05 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: Ee+Ek=EtotalE_e + E_k = E_{\mathrm{total}}.

Etotal=12(200)(0.15)2=2.25E_{\mathrm{total}} = \frac{1}{2}(200)(0.15)^2 = 2.25 J. Maximum speed at equilibrium (x=0x = 0): Ek=EtotalE_k = E_{\mathrm{total}}. vmax=2Etotal/m=2×2.25/0.50=9=3.0v_{\mathrm{max}} = \sqrt{2E_{\mathrm{total}}/m} = \sqrt{2 \times 2.25/0.50} = \sqrt{9} = 3.0 m S1^{-1}.

(b) At x=0.05x = 0.05 m: Ee=12(200)(0.0025)=0.25E_e = \frac{1}{2}(200)(0.0025) = 0.25 J. Ek=2.250.25=2.0E_k = 2.25 - 0.25 = 2.0 J. v=2×2.0/0.50=8=2.83v = \sqrt{2 \times 2.0/0.50} = \sqrt{8} = 2.83 m s1^{-1}.

(c) Etotal=2.25E_{\mathrm{total}} = 2.25 J (constant, since no friction).

If you get this wrong, revise: Conservation of Energy

Problem 15

A car engine delivers a constant driving force of 30003000 N. The total resistive force (drag + Rolling resistance) is Fres=300+2v2F_{\mathrm{res}} = 300 + 2v^2 N (where vv is in m s1^{-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+2v23000 = 300 + 2v^2.

2v^2 = 2700$$v^2 = 1350$$v = 36.7 m s1132^{-1} \approx 132 km h1^{-1}.

(b) P=Fv=3000×36.7=110100P = Fv = 3000 \times 36.7 = 110\,100 W =110= 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 7575 kg jumps from a platform. The bungee cord has an unstretched length of 2525 m and spring constant 5050 N m1^{-1}. Taking the reference level as the jump platform, find The speed of the jumper when the cord is stretched 1515 m beyond its natural length.

Answer. Height below platform: h=25+15=40h = 25 + 15 = 40 m. Loss of PE: mgh=75×9.81×40=29430mgh = 75 \times 9.81 \times 40 = 29\,430 J.

Elastic PE in cord: Ee=12(50)(15)2=5625E_e = \frac{1}{2}(50)(15)^2 = 5625 J.

By conservation: mgh=Ek+Eemgh = E_k + E_e. 29430=12(75)v2+562529\,430 = \frac{1}{2}(75)v^2 + 5625.

12(75)v2=23805\frac{1}{2}(75)v^2 = 23\,805. v2=634.8v^2 = 634.8. v=25.2v = 25.2 m s1^{-1}.

If you get this wrong, revise: Conservation of Energy

Problem 17

An electric motor is 85% efficient. It lifts a 200200 kg crate vertically at a steady speed of 0.800.80 M s1^{-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 1212 m.

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

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

(c) Energy wasted per second = PinPout=18471570=277P_{\mathrm{in}} - P_{\mathrm{out}} = 1847 - 1570 = 277 J s1^{-1}.

(d) Time to lift 12 m: t=12/0.80=15t = 12/0.80 = 15 s. Total electrical energy = Pin×t=1847×15=27705P_{\mathrm{in}} \times t = 1847 \times 15 = 27\,705 J =27.7= 27.7 kJ.

If you get this wrong, revise: Efficiency

Problem 18

Two identical springs, each of constant k=400k = 400 N m1^{-1}Are connected in parallel and support A mass of 5.05.0 kg. The mass is displaced 0.050.05 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: keff=2k=800k_{\mathrm{eff}} = 2k = 800 N m1^{-1}.

Angular frequency: ω=keff/m=800/5.0=160=12.6\omega = \sqrt{k_{\mathrm{eff}}/m} = \sqrt{800/5.0} = \sqrt{160} = 12.6 rad S1^{-1}.

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

(b) Maximum KE = total energy =12keffA2=12(800)(0.05)2=1.0= \frac{1}{2}k_{\mathrm{eff}}A^2 = \frac{1}{2}(800)(0.05)^2 = 1.0 J.

(c) Total elastic PE = 1.0 J (when KE = 0). Each spring stores half: Ee,each=0.50E_{e,\mathrm{each}} = 0.50 J.

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


Common Pitfalls

  1. Misidentifying the system boundary when applying conservation laws. Define what is included before writing equations.

  2. Using the wrong equation from the data sheet. Take time to read the full equation, including conditions and variable definitions.

  3. Forgetting to include units in final answers, especially when working with derived units like Nkg1m2\text{N}\,\text{kg}^{-1}\,\text{m}^2.

  4. Confusing scalar and vector quantities. Always check whether direction matters for the quantity in question.

Summary

The key principles covered in this topic are linked in the sub-pages above. Focus on understanding the definitions, applying the formulas or frameworks, and evaluating strengths and limitations of each approach.

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

Cross-References

  • Dynamics — The work-energy theorem is derived from Newton’s second law, connecting force, displacement, and kinetic energy.
  • Momentum — Impulse relates force applied over time to change in momentum, complementary to the work-energy approach.
  • Oscillations — Energy conservation in SHM uses elastic potential energy from springs, which obeys Hooke’s law.
  • Gravitational Fields — Gravitational potential energy near the surface is the small-height approximation of the full gravitational potential.