Work (VCE SSCE Physics): Revision Notes

Worked Example 1: Pushing a Box with Friction

A man pushes a box of mass 46 kg horizontally along the ground with a force of 500 N. The box encounters a frictional force of 50 N. Calculate the work done to push the box 32 m.

Part a: Work done by the pushing force

The pushing force acts in the direction of motion, so $\theta = 0°$:

$W_{\text{push}} = F_{\text{push}} s \cos\theta$

$W_{\text{push}} = (500)(32) \cos 0°$

$W_{\text{push}} = 16\,000 \text{ J}$

This is positive work - energy is being transferred to the box.

Part b: Work done by the friction force

The friction force acts opposite to the direction of motion, so $\theta = 180°$:

$W_{\text{friction}} = F_{\text{friction}} s \cos\theta$

$W_{\text{friction}} = (50)(32) \cos 180°$

$W_{\text{friction}} = -1600 \text{ J}$

The negative sign indicates that friction removes energy from the box.

Part c: Total energy given to the box

$W_{\text{total}} = W_{\text{push}} + W_{\text{friction}}$

$W_{\text{total}} = 16\,000 + (-1600)$

$W_{\text{total}} = 14\,400 \text{ J}$

The net work represents the actual energy gained by the box.

Worked Example 2: Towing at an Angle

A cable-towed kneeboarder is being pulled with a force of 250 N. The towing cable makes an angle of 22° to the horizontal. If the kneeboarder travels 163 m in a straight line, calculate the work done by the cable.

Solution:

The force acts at 22° to the direction of travel:

$W = Fs \cos\theta$

$W = (250)(163) \cos 22°$

$W = 3.78 \times 10^4 \text{ J}$

Note that the work done is less than it would be if the cable were horizontal (which would give $250 \times 163 = 40\,750$ J) because only the horizontal component of the force contributes to the work.

Worked Example 3: Car Being Pushed with Changing Force

A broken-down car is pushed with a force of 0.95 kN in the direction of displacement for 20 m. The person pushing then tires, and their force reduces at a constant rate to zero over the next 10 m.

Solution:

First, convert kN to N: $0.95 \text{ kN} = 0.95 \times 10^3 \text{ N} = 950 \text{ N}$

The graph consists of two shapes:

• A rectangle from 0 to 20 m
• A triangle from 20 to 30 m

Calculate the area of each:

Rectangle area: $\text{Area}_1 = (0.95 \times 10^3)(20) = 19\,000 \text{ J}$

Triangle area: $\text{Area}_2 = \frac{1}{2}(0.95 \times 10^3)(10) = 4750 \text{ J}$

Total work: $W_{\text{total}} = 19\,000 + 4750 = 2.38 \times 10^4 \text{ J}$

This shows how the area method handles changing forces elegantly - each geometric shape corresponds to a different phase of the push.

Worked Example 4: Pulling a Toy Cart

A child pulls a toy cart with a string. The tension in the string is 150 N and the string forms an angle of 30° to the horizontal. Calculate the work done if the child pulls the toy for 122 m.

Step 1: Resolve the force into components

Break the force into horizontal and vertical components using trigonometry:

• Horizontal component: $F_{\text{horizontal}} = 150 \cos 30°$
• Vertical component: $F_{\text{vertical}} = 150 \sin 30°$

Step 2: Identify the component needed

Since the cart moves horizontally, we need the horizontal component.

Step 3: Calculate the work

$W = Fs \cos\theta$

$W = (150)(122) \cos 30°$

$W = 1.58 \times 10^4 \text{ J}$

Notice that this gives the same result as if we had calculated the horizontal component first ($150 \cos 30° = 129.9$ N) and then multiplied by displacement: $129.9 \times 122 = 15\,848$ J.