Relationship Between Work Done and Energy Transfered (A level Physics)

In science, doing work is means transferring energy from one object to another. Work and energy are interrelated in the sense that work is a measurement of the amount of energy transfered from one object to another or from one form to another. For example, when lifting a rock from the ground, the following energy conversions occur:

• chemical potential energy in your muscles is converted to kinetic energy of the moving hand carrying the rock.
• as the rock is moving upwards, you are doing work against gravity in the sense that you are the kinetic energy of the rock into gravitational potential energy.
• since the rock moved a distance, work has been done and the work done is equal to the kinetic energy converted to gravitational potential energy.

Energy is the ability to do work.

Or we can say:

Energy is that which is transferred when a force moves through a distance.

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No movement no work done

The everyday sense of doing work is slightly misleading when we look at it from a scientific point of view. For example, if you support a heavy load above your head without moving it for some time, your muscles will get tired and you will feel like you are doing work. If there is no movement, you are not doing any work on the load, because you are not transferring any energy to the load.

Your muscles get tired and ache because they are constantly providing an equal and opposite reaction to the load and this uses energy though none of the energy is being transferred to the load.

Here are some examples to show you the scientific concept of work:

Doing workNot doing work
When pushing a car to start it moving the force you apply transfers energy to the car. If the car moves, its kinetic energy increases.When pushing a car but it does not move, no energy is transferred, because your force does not move it. The car’s kinetic energy does not change and so no work is done.
When lifting a load you are doing work as the load moves upwards because the gravitational potential energy of the load increases.When holding the load above your head you are not doing work on the load because the force you apply is not moving it. Therefore the gravitational potential energy of the load is not changing.

Calculating work done

The work done by a force is defined as the product of the force and the distance moved in the direction of the force.

Mathematically, W = F × s

whereby W is the work done in joules(J), F is the force in newtons(N) and s is the distance moved in the direction of the force in metres(m).

For example, in the diagram above, a person pushes a box a distance of 5 m by applying a force of 200 N.

Work done = F × s = 200 N × 5 m = 1 000 J.

The units of work done and energy transferred

As we mentioned earlier, doing work is a way of transferring energy. The SI unit of both energy and work the is the joule (J). Calculating work done using W = F × s, shows us the amount of energy transferred.

Work done = energy transferred

From the equation W = F × s we can dimension the unit of work done in terms of the unit of force (the newton) and the unit of distance (the metre) as follows:

• W = F × s
• this means, 1 joule = 1 newton × 1 metre
• and therefore 1J = 1Nm

That gives us the definition of the joule.

The joule is defined as the amount of work done when a force of 1 newton moves a distance of 1 metre in the direction of the force.

Gravitational potential energy

If you lift a load, you do work against gravity since you are providing an upward force to overcome the downward force of gravity on the object (weight of the object). The force you apply moves the object upwards, so the force is doing work, which means energy is being transferred from you to the load. The chemical potential energy you lose is gained by the object in form of gravitational potential energy Ep.

How to calculate gravitational potential energy

To calculate the gain in the gravitational potential energy, Ep of an object when it is lifted to a height, h, we use the fact that gravitational potential energy gain is equal to work done.

We the formula Work done = Force × distance

However, we know that:

• work done in lifting a load = gravitational potential energy gained by the load
• force needed to lift the load = weight of the load = mg, whereby m is the mass of the load and g is the acceleration due to gravity
• distance = height reached by the load

Therefore, Ep = weight × change in height

Ep = (mg)×h

Ep = mgh

Kinetic energy

One of the effects of a force is that it can make an object accelerate. When an object accelerates, work is done by the force and energy is transferred to the object in form of kinetic energy, Ek. The faster an object is moves, the greater its
kinetic energy.

How to calculate kinetic energy

To derive the formula for calculating the kinetic energy of an object, first we imagine an object being accelerated from rest (u = 0) to a velocity v by a force F for a distance s.

• Since u = 0, we can write the equation of motion v2 = u2+2as as v2 = 2as
• Then we multiply both sides by $latex \frac{1}{2}m$ to get:

$latex \displaystyle \frac{1}{2}mv^2 = mas$

Using F = ma, we get:

$latex \displaystyle \frac{1}{2}mv^2 = Fs$

Since work done = Fs

$latex \displaystyle work\ done = \frac{1}{2}mv^2$

Work done in accelerating the car = kinetic energy gain by the car.

$latex \displaystyle E_k = \frac{1}{2}mv^2$

Questions and Answers on Work and Energy

Question 1 – Explaining whether work is done or not

In each of the following examples, explain whether or not work is done by the mentioned forces

1. When you pull a heavy sack along rough ground.
2. When the force of gravity pulls you downwards as you fall off a wall.
3. The tension in a string pulls on a stone when you whirl it around in a circle at a steady speed.
4. The reaction force of the bedroom floor stops you from falling into the room below.
5. An apple falling towards the ground.
6. A car decelerating when the brakes are applied.

Solution 1

1. When you pull a heavy sack along rough ground – work is done because kinetic energy of the sack increases.
2. When the force of gravity pulls you downwards as you fall off a wall – work is done because as you fall your speed increases which means kinetic energy increases.
3. The tension in a string pulls on a stone when you whirl it around in a circle at a steady speed – no work is done because the the kinetic energy of the stone is not changing.
4. The reaction force of the bedroom floor stops you from falling into the room below – no work is done because the reaction force of the floor keeps you stationary which means no kinetic energy or potential energy is gained or lost.
5. An apple falling towards the ground – work is done because the potential energy will be converted to kinetic energy.

Question 2 – Calculating speed and kinetic energy

A proton (mass m = 1.67 × 10-27kg) is being accelerated along a straight line at 3.6 × 1015m/s2 in a machine. If the proton has an initial speed of 2.4 × 107 m/s and it travels 3.5 cm, what then is (a) its speed and (b) the increase in its kinetic energy?

Solution 2(a)

• m = 1.67 × 10-27 kg
• a = 3.6 × 1015 m/s2
• u = 2.7 × 107 m/s
• s = 3.5 cm = 0.035 m

v = ?

To find its final velocity we use the equation of motion, v2 = u2 + 2as

v2 = (2.7 × 107)2 + 2(3.6 × 1015)(0.035)

v2 = 9.81 × 104

v = 3.13 × 107 m/s

Solution 2(b)

$latex \displaystyle Change\ in\ kinetic\ energy\ = \frac{1}{2}m(\Delta v)^2$

$latex \displaystyle \Delta E_K = \frac{1}{2}m(v – u)^2$

$latex \displaystyle \Delta E_K = \frac{1}{2}\times 1.67 \times 10^{-27}(3.13 \times 10^{7} – 2.7 \times 10^{7})^2$

$latex \displaystyle \Delta E_K = 1.54 \times 10^{-14}J$

Question 3 – Calculating Work done

A man of mass 70 kg climbs stairs of vertical height 2.5 m. Calculate the work done against the force of gravity, if g = 9.81 m/s2).

Solution 3

• m = 70 kg
• h = 2.5 m
• g = 9.81 m/s2

Work done against gravity = gravitation potential energy gained

W = mgh

W = 70 × 9.81 × 2.5

W = 1 716.75 J

Question 4

A rocket has a combined mass of 2.9 × 105 kg. If it reaches a speed of 11.2 km/s, how much kinetic energy would it then have?

Solution 4

• m = 2.9 × 105 kg
• v = 11.2 km/s = 11 200 m/s

$latex \displaystyle E_K = \frac{1}{2}mv^2$

$latex \displaystyle E_K = \frac{1}{2}\times 2.9 \times 10^5 \times 11200^2$

$latex \displaystyle E_K = 1.82 \times 10^{13} J$

Question 5 – Calculating Work done and Energy Transfered

A stone of mass 1 kg falls from the top of a 250 m high cliff. Take g to be 9.81 m/s2 and air resistance to be negligible.

1. Calculate how much work is done by the force of gravity in pulling the stone to the foot of the cliff.
2. At what speed does the stone hit the bottom of the cliff?

Solution 5(a)

• m = 1 kg
• g = 9.81 m/s2
• h = 250 m

Work done by gravity = mgh

W = 1(9.81)(250)

W = 2 452,5 J

Solution 5(b)

As the stone falls, the work done by gravity is converted to kinetic energy of the stone. Therefore at the foot of the cliff the stone has 2 452,5 J.

$latex \displaystyle E_K = \frac{1}{2}mv^2$

$latex \displaystyle 2452.5 = \frac{1}{2}(1)v^2$

$latex \displaystyle 4905 = v^2$

$latex v = 70\ m/s$

Sydney Chako

Mathematics, Chemistry and Physics teacher at Sytech Learning Academy. From Junior Secondary School to Tertiary Level Engineering Mathematics and Engineering Science.