Kinetic energy ½ mv² and gravitational potential energy mgh

Work done by a constant force, including a force at an angle to the motion

Work done against friction and resistances, and against gravity on a slope

The work–energy principle

Conservation of mechanical energy when only gravity does work

Power as the rate of doing work, and P=Fv

Vehicles and cyclists working at constant power: driving force P/v, acceleration at a given speed and maximum speed, on level roads and on slopes

1. Energy methods account for what a particle had, what forces added or took away, and what is left, which is often quicker than finding an acceleration first.

2. KE=½ mv² and GPE=mgh, both in joules; GPE depends only on the vertical height above a level you choose.

3. Work done by a constant force = force × distance moved in the direction of the force, Fscos θ. Forces perpendicular to the motion (such as the normal reaction) do no work.

4. Work–energy principle: starting energy + work done by driving forces − work done against resistances = final energy (KE + PE).

5. If no work is done against resistances and no driving force acts, KE + PE stays constant.

6. Power is the rate of doing work: P=Fv, in watts. At constant power the driving force is P/v, so it falls as the vehicle speeds up.

7. Maximum speed is reached when the acceleration is zero: P/v= resistance (+ mgsin α uphill).

This topic gives you a second way into mechanics. Instead of finding an acceleration and then using the constant-acceleration formulae, you keep an account of energy. Two kinds of mechanical energy appear in M2.

Here m is the mass in kg, v the speed in m s⁻¹, g=9.8 m s⁻² and h the height in metres above a chosen zero level. Neither formula is in the formulae booklet: both must be known.

Energy is a scalar measured in joules (J), and 1 J = 1 N m. Kinetic energy depends on speed, not direction, so it is never negative. Only changes in GPE matter, so you may put the zero level anywhere; the lowest point of the motion is usually the best choice because every PE is then positive.

The h in mgh is always a vertical height. A particle that moves a distance d along a slope at angle α to the horizontal changes its height by dsin α.

A change of speed from u to v changes the KE by ½ m(v²-u²): square first, then subtract. It is not ½ m(v-u)².

WORKED EXAMPLE · QUESTION

(a) A car of mass 1200 kg travels at 54 km h⁻¹. Find its kinetic energy. (b) A walker of mass 70 kg follows a winding path 3.2 km long from a car park to a summit 240 m higher. Find the gain in her gravitational potential energy.

SOLUTION

(a) Convert the speed: 54×1000÷3600=15 m s⁻¹. Then KE=½×1200×15²= 135 000 J (135 kJ)

(b) GPE depends only on the vertical height gained, not on the length of the path: 70×9.8×240=164 640 J = 165 kJ (3 s.f.)

Here F is the force in N, s the distance moved in m, and θ the angle between the force and the direction of motion. This is not in the formulae booklet: learn the definition and the formula follows from it. Only the component of the force along the motion, Fcos θ, does work. A force perpendicular to the motion (θ=90°) does no work at all, because cos 90°=0.

Notice the last two rows of the table together. Pushing a particle up a slope against gravity costs the same as lifting it vertically through the same height: the component mgsin α is smaller than mg, but it acts over the longer distance d, and dsin α is just the height.

WORKED EXAMPLE · QUESTION

A crate of mass 40 kg is pulled 15 m along rough horizontal ground by a light rope inclined at 35° above the horizontal. The tension in the rope is 120 N and the coefficient of friction between the crate and the ground is 0.3. The crate starts from rest. Find (a) the work done by the tension, (b) the work done against friction, (c) the speed of the crate after 15 m.

SOLUTION

(a) Only the horizontal component of the tension does work: 120cos 35°×15=1474.5 J = 1470 J (3 s.f.)

(b) Resolve vertically to find R: R+120sin 35°=40×9.8, so R=392-68.83=323.17 N. Then F=0.3×323.17=96.95 N and the work against friction =96.95×15= 1450 J (3 s.f.)

(c) Weight and R do no work, being perpendicular to the motion. The difference is the KE gained: ½×40×v²=1474.47-1454.27=20.20 J, so v=1.0 m s⁻¹ (2 s.f.)

Keep unrounded values until the end: subtracting two numbers that are close together magnifies rounding errors. Using 1470-1450 would give 20 J and the same answer only by luck.

Written as an energy account, it reads

"Resistances" means friction, air resistance or any constant resistive force; "driving forces" means engine forces, pushes and pulls. This equation is not in the formulae booklet: build it from the definitions each time.

There is an equivalent form without PE: work done by the resultant force = change in KE. Use one form or the other, never both.

Gravity is either a PE term or a work term, not both, otherwise you count it twice.

A reliable method:

1. Choose the zero level for PE (usually the lowest point) and write the KE and PE at the start and the end.

2. Find the friction or resistance and the distance over which it acts; work against it = force × distance.

3. Write one energy equation and solve. No acceleration is needed, and the energy route also works when the acceleration is not constant.

WORKED EXAMPLE · QUESTION

A parcel of mass 3 kg is released from rest at the top A of a rough ramp inclined at 30° to the horizontal. It slides down a line of greatest slope to B, where AB = 5 m. The coefficient of friction between the parcel and the ramp is 0.2. Use the work–energy principle to find the speed of the parcel at B.

SOLUTION

Step 1 Loss of PE, with the zero level at B: 3×9.8×5sin 30°=73.5 J.

Step 2 Friction. Perpendicular to the slope R=3×9.8cos 30°=25.461 N, so F=0.2R=5.0922 N and the work against friction is 5.0922×5=25.46 J.

Step 3 PE lost = KE gained + work done against friction: 73.5=½×3×v²+25.46, so ½×3×v²=48.04 and v=5.66 m s⁻¹ (3 s.f.)

Check: on a smooth ramp the speed would be √(2×9.8×2.5)=7.0 m s⁻¹, so friction has reduced it, as it should.

The energy method is particularly good at finding an unknown resistance. If you know the speeds at two points and the distance between them, the work done against the resistance is whatever energy is left over, and dividing by the distance gives the force.

WORKED EXAMPLE · QUESTION

A cyclist and her bicycle have a total mass of 80 kg. She freewheels (does not pedal) down a straight road inclined at an angle α to the horizontal, where sin α=(1/20). Her speed increases from 4 m s⁻¹ to 10 m s⁻¹ while she travels 120 m. The resistance to motion is modelled as a constant force. Find the magnitude of the resistance.

SOLUTION

Step 1 Vertical height lost: 120×(1/20)=6 m, so PE lost =80×9.8×6=4704 J.

Step 2 KE gained: ½×80×(10²-4²)=3360 J.

Step 3 The rest of the PE lost is used doing work against the resistance R over 120 m: 120R=4704-3360=1344, so R=11.2 N

This applies to a particle moving freely under gravity, such as a projectile with no air resistance, and to a particle on a smooth surface, where the normal reaction does no work. It does not depend on the shape of the path or the direction of projection: only the change in height matters. Once friction, resistance or a driving force does work, use the full work–energy principle instead.

WORKED EXAMPLE · QUESTION

(a) A ball of mass 0.5 kg is thrown from the top of a vertical cliff 30 m above the sea with speed 14 m s⁻¹. Air resistance is negligible. Find the speed with which the ball hits the sea. (b) A small block is projected with speed 8 m s⁻¹ up a line of greatest slope of a smooth plane inclined at 25° to the horizontal. Find the distance it travels up the plane before coming to instantaneous rest.

SOLUTION

(a) KE + PE is constant; take sea level as zero: ½×0.5×14²+0.5×9.8×30=½×0.5×v². The mass cancels: v²=196+588=784, so v=28 m s⁻¹

(b) KE lost = PE gained, with height dsin 25°: ½ m×8²=m×9.8×dsin 25°, so d=64/19.6sin 25°= 7.73 m (3 s.f.)

In (a) the angle of the throw is not given and is not needed — a strong hint that an energy method is expected.

Here P is the power in watts, F the driving force in N and v the speed in m s⁻¹. Neither form is in the formulae booklet: learn both. P=Fv follows from work = F× distance: divide both sides by time, and distance ÷ time is speed. 1 W = 1 J s⁻¹ and 1 kW = 1000 W.

A crane raising a 450 kg load at a steady 0.60 m s⁻¹ pulls with a force equal to the weight, so it works at 450×9.8×0.60=2646 W ≈ 2.65 kW.

The important consequence is for vehicles. If an engine works at a constant rate P, then the driving force at speed v is P/v: the faster the vehicle goes, the smaller the driving force. At the instant the speed is v, Newton's second law on a level road gives

As v rises the left-hand side shrinks, so the acceleration falls. The vehicle reaches its maximum speed when a=0, that is when P/v_(max)= resistance.

WORKED EXAMPLE · QUESTION

A car of mass 1000 kg has an engine working at a constant rate of 24 kW. The resistance to motion is constant, 800 N. (a) Find the maximum speed of the car on a straight horizontal road. (b) Find its acceleration on that road when its speed is 12 m s⁻¹. (c) The car now drives up a straight road inclined at α to the horizontal, where sin α=(1/20), with the same power and resistance. Find its maximum speed up the hill.

SOLUTION

(a) At maximum speed the driving force equals the resistance: 24 000/v=800, so v=30 m s⁻¹

(b) Driving force at 12 m s⁻¹: 24 000÷12=2000 N. Then 2000-800=1000a, so a=1.2 m s⁻²

(c) Uphill, the component of the weight mgsin α=1000×9.8×(1/20)=490 N also opposes the motion: 24 000/v=800+490=1290, so v=18.6 m s⁻¹ (3 s.f.)

Resolve along the slope. Up a slope at angle α, the forces opposing the motion are the resistance R and the weight component mgsin α:

Down a slope the weight component helps instead: P/v+mgsin α-R=ma.

Slopes are usually given as sin α=1/n, meaning a rise of 1 m for every n m travelled along the road. Use the fraction directly; there is no need to find α. The normal reaction and mgcos α are perpendicular to the motion and do not appear in the equation along the slope.

WORKED EXAMPLE · QUESTION

A lorry of mass 12 000 kg moves along a straight road inclined at α to the horizontal, where sin α=(1/40). The resistance to motion from non-gravitational forces is constant, 4200 N. (a) The lorry moves up the road with its engine working at 90 kW. Find its acceleration when its speed is 10 m s⁻¹, and its maximum speed. (b) The lorry then moves down the same road at a constant speed of 20 m s⁻¹. Find the rate at which its engine is now working.

SOLUTION

Step 1 Weight component along the road: 12 000×9.8×(1/40)=2940 N.

Step 2 (a) Driving force 90 000÷10=9000 N. Newton's second law up the slope: 9000-4200-2940=12 000a, so a=0.155 m s⁻²

Step 3 Maximum speed when a=0: 90 000/v=4200+2940=7140, so v=12.6 m s⁻¹ (3 s.f.)

Step 4 (b) Constant speed downhill: driving force + weight component = resistance, so D=4200-2940=1260 N and P=1260×20=25 200 W = 25.2 kW

Longer questions mix the two halves of this topic, and the first job is to decide which tool each part needs.

At an instant (acceleration at a given speed, maximum speed): use P=Fv and Newton's second law.

Over a distance (speed changes between two points, a time taken at constant power): use the work–energy principle, with work done by the engine = P× time.

An engine working at constant power does not give constant acceleration, so the constant-acceleration formulae cannot be used for that stage.

WORKED EXAMPLE · QUESTION

A car of mass 1100 kg drives up a straight road inclined at α to the horizontal, where sin α=(1/25). The resistance to motion from non-gravitational forces is constant, 650 N. (a) The engine works at a constant rate of 30 kW. Find the acceleration of the car when its speed is 12 m s⁻¹. (b) Find the maximum speed of the car up the road at this rate of working. (c) On another occasion the car passes a point A on the road at 20 m s⁻¹ and a point B, 400 m further up the road, at 14 m s⁻¹. It takes 25 s to go from A to B. Find the work done by the engine between A and B, and the average rate at which it works.

SOLUTION

(a) Weight component mgsin α=1100×9.8÷25=431.2 N; driving force =30 000÷12=2500 N. Newton's second law up the slope: 2500-650-431.2=1100a, so a=1.2898…= 1.29 m s⁻² (3 s.f.)

(b) Maximum speed when the resultant force is zero: 30 000/v=650+431.2=1081.2, so v=27.747…= 27.7 m s⁻¹ (3 s.f.)

(c) Height gained =400÷25=16 m. KE lost =½×1100×(20²-14²)=112 200 J; PE gained =1100×9.8×16=172 480 J; work against resistance =650×400=260 000 J.

KE lost + work by engine = PE gained + work against resistance, so work by engine =172 480+260 000-112 200=320 280 J ≈ 3.20×10⁵ J

Average power = work ÷ time =320 280÷25=12 811.2 W ≈ 12.8 kW (3 s.f.)

Sense check: the average power is well below 30 kW, which fits a car that is slowing down on the hill.

M1 Dynamics

Friction F=μR, resolving on inclined planes and Newton's second law are all reused here; the energy method is an alternative to F=ma plus the constant-acceleration formulae whenever forces are constant.

M2.5 Collisions and M2 projectiles

Loss of kinetic energy in an impact is calculated with ½ mv², and the speed of a projectile at a given height comes straight from conservation of energy, without the angle of projection.

Pure and IAL Physics

Rearranging P/v-R=ma for v and solving for unknown resistances is straightforward algebra; work done by a variable force (an integral) is a natural extension met in later mechanics. In IAL Physics Unit 1 the same ideas of work, kinetic and gravitational potential energy and power appear, with efficiency added.

1. Find the kinetic energy of a ball of mass 0.06 kg moving at 25 m s⁻¹.

2. A box of mass 15 kg is lifted at constant speed onto a shelf 1.8 m higher. Find the work done against gravity.

3. A car travels at a constant 25 m s⁻¹ along a level road against a constant resistance of 720 N. Find the rate at which its engine is working.

4. Explain why the normal reaction does no work on a particle sliding down a slope.

5. A particle slides from rest down a smooth curved surface and ends 2.5 m lower than it started. Find its speed.

6. A cyclist works at a constant 150 W on a level road against a constant resistance of 20 N. Find her maximum speed, and her acceleration at 5 m s⁻¹ if she and her bicycle have mass 75 kg.

1. ½×0.06×25²=18.75 J (18.8 J to 3 s.f.).

2. mgh=15×9.8×1.8=264.6 J ≈ 265 J.

3. Constant speed, so the driving force = 720 N. P=Fv=720×25=18 000 W = 18 kW.

4. The normal reaction is perpendicular to the slope, and so perpendicular to the direction of motion. Its component along the motion is zero (cos 90°=0), so the work it does is zero.

5. Only the weight does work, so mechanical energy is conserved: ½ v²=9.8×2.5, v=7 m s⁻¹. The shape of the surface does not matter.

6. Maximum speed when 150/v=20: v=7.5 m s⁻¹. At 5 m s⁻¹ the driving force is 150÷5=30 N, so 30-20=75a and a=0.133 m s⁻² (3 s.f.).