K326 / K327 / 2027
Energy overview

Topic 3 of 4

Work and conservation of energy

Work is a mechanical energy transfer. Conservation lets you account for that energy across all the stores and transfers that matter.

Kinetic energy is 1/2 mv2; a gravitational change is mgh. These describe stores. The work done by a force describes a transfer.

Work done = force x distance moved in the direction of the force
W = FdW = work in J; F = force in N; d = distance in m along the force. Apply this relationship to a constant force and the movement along its direction.

Work done may be written W or E, since it is energy transferred. Here W means work in joules. In W = mg, W instead means weight in newtons; after a power value, such as 60 W, W is the unit watt. Read the quantity and unit together.

Worked example

A 12 N pull through 3.0 m

A constant force of 12 N acts on a body while it moves 3.0 m along the force's direction.

Work done by the force = 12 x 3.0 = 36 J.

This tells us the mechanical transfer by that force. To say that the kinetic store increases by 36 J, we must also establish that no other transfer or store change takes part. For example, friction could transfer some of the input to internal stores.

A force can act without doing work on the object

  • Pushing a stationary wall: the wall does not move, so the work done on it by the push is zero.
  • Supporting a load during purely horizontal motion: the upward support is perpendicular to the movement. There is no distance moved in that force's direction, so its work on the load is zero.

These statements concern the specified force and object. They do not mean that a person's body transfers no energy internally while pushing or supporting.

For the 2.0 kg load lifted at constant speed through 1.5 m, take g = 10 N/kg and neglect losses. The lifting force balances its 20 N weight, so the lift does 20 x 1.5 = 30 J of work. That equals the 30 J gravitational store increase; the speed is unchanged, so kinetic energy is unchanged.

Conservation is an account of the whole change

The principle of conservation of energy states that energy cannot be created or destroyed. For an isolated system, the total energy remains constant. More generally:

Initial energy + energy transferred in
= final energy + energy transferred outChoose the objects included in the system and the same start and end points for every term. Do not count a transfer twice.

If no energy enters or leaves the system and only kinetic and gravitational stores change, their total remains constant. These are model assumptions, not a rule that air resistance or friction must be absent from every real situation.

Account for the initial 9.0 J throughout the fall

A 0.50 kg object is released from rest at height 1.8 m; g = 10 N/kg. First, take the object and Earth as the system, with negligible air resistance and no other store changes.

Ep: gravitational energy of the object and Earth

Ek: kinetic energy of the object

Released from rest: h = 1.8 m

Released from rest: h = 1.8 m: 9.0 J gravitational energy and 0.0 J kinetic energyA stacked bar has a total length representing 9.0 J. The gravitational part is 9.0 J and the kinetic part is 0.0 J. All bars for this fall use the same energy scale. Kinetic energy is zero because the object is released from rest.Ep9.0 JEk = 0.0 J; total = 9.0 J

During the fall: h = 0.80 m

During the fall: h = 0.80 m: 4.0 J gravitational energy and 5.0 J kinetic energyA stacked bar has a total length representing 9.0 J. The gravitational part is 4.0 J and the kinetic part is 5.0 J. All bars for this fall use the same energy scale. At 0.80 metres, gravitational energy is 0.50 times 10 times 0.80, or 4.0 J. The remaining 5.0 J is kinetic energy.Ep4.0 JEk5.0 J4.0 J + 5.0 J = 9.0 J

At the chosen zero height: h = 0 m

At the chosen zero height: h = 0 m: 0.0 J gravitational energy and 9.0 J kinetic energyA stacked bar has a total length representing 9.0 J. The gravitational part is 0.0 J and the kinetic part is 9.0 J. All bars for this fall use the same energy scale. Gravitational energy is zero at the chosen height reference. This is the instant the object reaches that level, before any impact.Ek9.0 JEp = 0.0 J; total = 9.0 J

The zero-height level sets the gravitational reference. The last bar is before any impact.

Separate case: resistance increases internal stores

Start with the same 9.0 J. Now include the object, Earth and surrounding air in the account. At zero height, internal stores have increased by 2.0 J.

At zero height with resistance: 7.0 J kinetic energy and 2.0 J increase in internal energyThis is a different fall with the same initial gravitational energy of 9.0 J. At the chosen zero-height level, before impact, gravitational energy is zero. Kinetic energy is 7.0 J and the internal energy of the object and surroundings has increased by 2.0 J. The account still totals 9.0 J. The kinetic and internal-increase parts have widths in the exact ratio seven to two, at the same scale as the earlier bars. The 2.0 J is an increase, not the total internal energy already present.Ek 7.0 J2.0 J7.0 J + 2.0 J = 9.0 J

2.0 J is the increase in internal energy of the object and surroundings.

A 0.50 kg body is released from rest at a height of 1.8 m, with g = 10 N/kg. The first three bars show a fall with negligible air resistance: kinetic and gravitational stores total 9.0 J. The separate resistance case accounts for 7.0 J kinetic and a 2.0 J internal increase. All heights use the same chosen zero level.

Worked example

Find the speed during a fall

A 0.50 kg body is released from rest 1.8 m above the chosen zero-height level. Use g = 10 N/kg. Neglect air resistance and all other store changes, and consider the body just before it reaches that level, before any impact.

  1. Initial account: Ek = 0 because it starts from rest. Ep = 0.50 x 10 x 1.8 = 9.0 J.
  2. At zero height: Ep = 0, so conservation gives Ek = 9.0 J.
  3. Use the kinetic-energy equation: 9.0 = 0.5 x 0.50 x v2, so v2 = 36 and v = 6.0 m/s.

At a height of 0.80 m, the remaining gravitational energy is 0.50 x 10 x 0.80 = 4.0 J. The kinetic energy is 9.0 - 4.0 = 5.0 J. Then v2 = (2 x 5.0) / 0.50 = 20, giving v = about 4.5 m/s.

Include a transfer to internal stores

Now keep the same initial conditions, but suppose 2.0 J increases internal stores of the body and surrounding air by the time the body reaches zero height. Its final kinetic energy is 9.0 - 2.0 = 7.0 J.

The speed follows from v2 = (2 x 7.0) / 0.50 = 28, giving about 5.3 m/s. It is smaller than 6.0 m/s because less energy remains in the kinetic store.

The final account is 7.0 J kinetic + 2.0 J internal increase = 9.0 J. The total has not decreased. Expanding the system to include the surroundings explains the destination of energy that has left the body's mechanical stores.

State the conditions before equating two stores. Starting from rest, negligible air resistance and an unchanged set of other stores justify Ep lost = Ek gained here. If a condition changes, change the account.

Optional check A 0.50 kg body is released from rest 1.8 m above a chosen zero-height level. Use g = 10 N/kg. By the time it reaches that level, 2.0 J has increased internal stores. No other energy changes occur. What is its final kinetic energy?
A 0.50 kg body is released from rest 1.8 m above a chosen zero-height level. Use g = 10 N/kg. By the time it reaches that level, 2.0 J has increased internal stores. No other energy changes occur. What is its final kinetic energy?