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Solving problems with conservation of energy

Key idea: If no energy is dissipated, the kinetic energy gained by a falling object equals the gravitational potential energy it loses, so their total stays constant.

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When friction and air resistance are negligible, and no motor or push adds energy, a moving object’s kinetic and gravitational potential stores simply trade energy: whatever one loses, the other gains. That lets you find a speed from a height, or a height from a speed, without knowing the forces or the time taken.

Write an energy account

Energy is conserved. Under these conditions, the total of the kinetic energy (KE) and gravitational potential energy (GPE) stays constant. The GPE store belongs to the object and Earth together, so Earth is part of the system.

  1. Sketch the start and end positions, and note the speed at each.
  2. Choose the level where h = 0, usually the lowest point of the motion.
  3. Write the energy account in full, then cross out any term that is zero:
E_(p,i) + E_(k,i) = E_(p,f) + E_(k,f)
  1. Substitute Eₖ = (1/2)mv² and Eₚ = mgh in SI units, and solve for the unknown.

Writing the full account first stops you forgetting a store. An object that is thrown, rather than released from rest, starts with kinetic energy as well.

Worked example 1

Speed of a dropped ball

Problem

A ball is dropped from rest from 5.0 m above the ground. Take g = 10 N kg⁻¹ and ignore air resistance. Find its speed just before it hits the ground.
Show full solution

  1. Write the account and remove the zero terms

    Method

    Take the ground as h = 0. The ball starts at rest and ends at ground level.

    Reason

    So E_(k,i) = 0 and E_(p,f) = 0: all the GPE at the start becomes KE at the end.

    Working

    Answer

    mgh = (1/2)mv²

  2. Cancel the mass and rearrange

    Method

    Divide both sides by m and make v the subject.

    Reason

    Both stores are proportional to mass, so the mass cancels. With no air resistance, a heavy ball and a light ball dropped from the same height land at the same speed.

    Working

    Answer

    v = square root of 2gh

  3. Calculate the speed

    Working

    Answer

    v = square root of 2(10)(5.0) = 10 m s⁻¹

Check your understanding 1 How high does a thrown ball rise?

A ball is thrown vertically upwards at 12 m s⁻¹. Take g = 10 N kg⁻¹ and ignore air resistance. Find its maximum height above the release point.

At the highest point
Unit: m
Show answer

At the highest point, speed and KE are zero. Gravity still acts. Taking release as zero GPE gives

(1/2)m(12)² = mgh, h = 12²/2(10) = 7.2 m.

Compare your method: initial KE becomes GPE, and the height is measured above release.

On a slope, only the vertical drop matters

Gravitational potential energy depends only on vertical height. A slope makes the path longer, but the object still descends through the same vertical height, so the store decreases by the same amount. For an object released from rest on a smooth slope, with air resistance and additional stores such as rotational energy neglected, the final translational kinetic energy is the same as for a vertical fall through that height. The slope only spreads the drop over a longer distance and time.

Check your understanding 2 Predict before releasing the trolley

In the model, set Release height h to 3.0 m, Mass of the trolley to 2.0 kg and Friction to None. It uses g = 9.81 N/kg and neglects wheel rotation. Before releasing the trolley, predict its starting GPE and bottom speed. What happens to bottom speed if only mass doubles?

Show the steps Optional working
Unit: J
Unit: m/s
Bottom speed when mass doubles
Show answer

Starting GPE is 2.0 × 9.81 × 3.0 = 58.9 J. At the bottom it becomes KE, so v = square root of (2 × 9.81 × 3.0) = 7.67 m/s. With mass 4.0 kg, energy doubles to about 118 J but speed stays the same. Test these predictions using the model.

t = 0.00 s

A 2 kg trolley released from 3.0 metres with no friction. It is 3.00 metres up at 0.00 metres per second. GPE 58.9 J, KE 0.0 J, thermal 0.0 J.

Situation
m
kg
Friction
Height, h
3.00 m
Speed, v
0.00 m s−1
Kinetic energy
0.0 J
Gravitational potential energy
58.9 J
Internal energy of the surroundings
0.0 J
Energy stores against time

Try this

0 of 4 done

  1. With no friction, release the trolley from 3.0 m and find its speed at the bottom. (not done yet)

  2. With no friction, record the speed at the bottom for five release heights. (not done yet)

  3. Add friction and let the trolley come to rest. (not done yet)

  4. Lift the same crate with the new motor and the old motor at the same power. (not done yet)

Check your understanding 3 Account for the friction setting

Change Friction to Some and release the trolley again. How do the energy bars explain the lower bottom speed? Use the model or the energy account above.

Show answer

With friction, some energy enters internal stores of the trolley and surroundings, so less becomes KE at the bottom and the speed is lower. KE + GPE decreases, but including the dissipated energy preserves the total. At rest KE is zero; friction can stop the trolley above the lowest point, leaving some GPE.

Guided practice 2

Block on a frictionless slope

About 5 min

Problem

A 5.0 kg block slides from rest down a frictionless slope. Neglect air resistance. The distance along the slope is 30 m and the vertical height drop is 5.0 m. Take g = 10 N kg⁻¹. Find its kinetic energy at the bottom.

Choose the relevant height and energy balance

Show the steps Optional working
Unit: m
Unit: J
Unit: J

Hints

Hint 1

choose the height

Which distance tells you the change in vertical height?

Hint 2

use conservation

Friction and air resistance are both neglected, so all the GPE lost becomes KE.

Show solution step by step

  1. Calculate the GPE lost

    Method

    Use the vertical height drop.

    Reason

    Gravitational potential energy depends on vertical position, not path length.

    Working

    Answer

    Δ Eₚ = mgΔ h = (5.0)(10)(5.0) = 250 J

  2. Transfer GPE to KE

    Reason

    The block starts from rest and nothing dissipates energy in this model.

    Working

    Answer

    E_(k,bottom) = 250 J

A swinging pendulum

A pendulum bob trades energy between its two stores twice in every swing. Choose its lowest point, B, as h = 0. At the turning point A, the bob is momentarily at rest, so its KE is zero and its GPE is greatest.

Energy stores during one pendulum swingA pendulum moves from the left turning point A through the lowest point B towards the right. A and an ideal turning point D are at equal height. A real turning point is lower because some energy has been dissipated. Energy bars show all gravitational potential energy at A, all kinetic energy at B in the ideal model, and a mixture of gravitational potential and dissipated energy at the real turning point.Pendulum: choose the lowest point as zero GPEAt a turning point, speed and KE are momentarily zero.AreleaseBlowest pointDidealD′ realsame heightzero-GPE levelGPEKEdissipatedAGPEB idealKED′GPEdiss.
Energy stores during one pendulum swingA pendulum moves from the left turning point A through the lowest point B towards the right. A and an ideal turning point D are at equal height. A real turning point is lower because some energy has been dissipated. Energy bars show all gravitational potential energy at A, all kinetic energy at B in the ideal model, and a mixture of gravitational potential and dissipated energy at the real turning point.Pendulum: choose the lowest point as zero GPEAt a turning point, speed and KE are momentarily zero.AreleaseBlowest pointDidealD′ realsame heightzero-GPE levelGPEKEdissipatedAB idealD′GPEKEGPEdiss.
A and B show the loss-free swing, which reaches D at the same height as A. The separate real swing reaches the lower turning point D′ because energy is dissipated.

For a bob on a light string with a fixed pivot, neglect air resistance and pivot friction:

  • from A to B, the GPE decrease equals the KE increase, so E_(p,A) = E_(k,B);
  • from B to D, the KE decrease equals the GPE increase, so E_(k,B) = E_(p,D).

So the bob rises to D at the same height as A. The string’s tension does not change this: it pulls along the string, at right angles to the bob’s motion, so it does no work and transfers no energy.

In a real pendulum, air resistance and friction at the pivot transfer some energy to the internal stores of the bob, the air and the support. Each turning point is a little lower than the last, and the swing slowly dies away, although the total energy is still conserved.

Check your understanding 4 Speed of a pendulum bob

A pendulum bob starts from rest 0.45 m above its lowest point. Neglect air resistance and pivot friction; take g = 10 N/kg. Find its speed at the lowest point, then at 0.25 m above that point on the other side.

Unit: m/s
Show the steps Optional working
Unit: m
Unit: m/s
Show answer

At the lowest point, v = square root of 2(10)(0.45) = 3.0 m/s. At the later position, the drop is only 0.45-0.25 = 0.20 m, so v = square root of 2(10)(0.20) = 2.0 m/s.

The energy account still includes GPE at the later position. Compare the two height changes before calculating speed.

Check your understanding 5 Keep the initial kinetic energy

A 2.0 kg object is thrown vertically upwards at 4.0 m/s from 2.0 m above the ground. Ignore air resistance; use g = 10 N/kg. Find its greatest height above ground and its speed just before reaching the ground.

Show the steps Optional working
Unit: J
Unit: J
Unit: J
Unit: m
Show the steps Optional working
Unit: J
Unit: J
Unit: J
Unit: m/s
Show answer

Initially GPE is 2.0 × 10 × 2.0 = 40 J and KE is 1/2 × 2.0 × 4.0² = 16 J, giving 56 J in total.

At the highest point KE is zero, so h = 56/(2.0 × 10) = 2.8 m above ground. At ground level GPE is zero, so v = square root of (2 × 56/2.0) = 7.5 m/s, to two significant figures.

Compare your full account: including the initial KE, using ground as the common zero, and taking a square root to find speed.

Common mistakes

  • Using the distance along a slope instead of the vertical drop.
  • Leaving out the initial KE of an object that is thrown rather than released from rest.
  • Setting KE to zero at the top of a curved flight. Only at the top of a vertical throw, or at a pendulum’s turning point, is the speed zero; a ball thrown at an angle is still moving sideways at its highest point.
  • Using the whole starting height when the object ends part of the way down. Use the vertical height actually lost.
  • Assuming KE + GPE stays constant when friction or air resistance is significant.

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