Work, Energy, and Power (AQA A-Level Further Maths): Revision Notes

Worked Example: Variable Force Integration

A force varies according to $F = (10 - 2x)$ N, where $x$ is in metres. Find the work done as the object moves from $x = 0$ to $x = 4$ m.

Solution:

$W = \int_0^4 (10 - 2x) \, dx$

$W = \left[10x - x^2\right]_0^4$

$W = (10 \times 4 - 4^2) - (0)$

$W = 40 - 16 = 24 \text{ J}$

The work done is 24 joules.

Worked Example: Force at an Angle

A force of 10 N acts at 60° to the horizontal and moves an object 4 m horizontally.

Solution:

$W = F \cos \theta \times x = 10 \times \cos 60° \times 4$

$W = 10 \times 0.5 \times 4 = 20 \text{ J}$

The work done is 20 joules.

Worked Example: Average Power

A person pulls a crate with a rope parallel to a horizontal floor with constant tension of 15 N. The crate travels 4 m in 1.6 s.

Solution:

Work done = $15 \times 4 = 60$ J

Average power = $\frac{\text{work done}}{\text{time taken}} = \frac{60}{1.6} = 37.5$ W

The average power is 37.5 watts.

Worked Example: Instantaneous Power

The same crate passes point B with speed 3 m s$^{-1}$. The power at this instant is:

Solution:

$P = F \times v = 15 \times 3 = 45 \text{ W}$

The instantaneous power is 45 watts.

Worked Example 1: Object Falling with Air Resistance

A 4 kg object falls 12 m from rest under gravity. Take $g = 9.81$ m s$^{-2}$.

Part (a): Find the final speed if air resistance is negligible.

Solution:

Energy equation: GPE lost = KE gained

$mgh = \frac{1}{2}mv^2$

$4 \times 9.81 \times 12 = \frac{1}{2} \times 4 \times v^2$

$470.88 = 2v^2$

$v^2 = 235.44$

$v = \sqrt{235.44} = 15.3 \text{ m s}^{-1}$

The final speed is 15.3 m s$^{-1}$.

Part (b): If air resistance is modelled by a constant force of 5 N, find the final speed.

Solution:

Energy equation: GPE lost = KE gained + Work done against air resistance

$4 \times 9.81 \times 12 = \frac{1}{2} \times 4 \times v^2 + 5 \times 12$

$470.88 = 2v^2 + 60$

$2v^2 = 410.88$

$v^2 = 205.44$

$v = \sqrt{205.44} = 14.3 \text{ m s}^{-1}$

The final speed is 14.3 m s$^{-1}$.

Note: A limitation of this model is that air resistance actually increases with speed, not remains constant. A more realistic model would have air resistance dependent on velocity.

Worked Example 2: Car Climbing a Slope

A car of mass 800 kg starts from rest at the foot of a slope. A constant tractive force $T$ N acts on it. The car climbs the slope, travelling 500 m in 50 s to reach a speed of 20 m s$^{-1}$ at a vertical height of 4 m.

Assume negligible energy is lost to friction or air resistance.

Part (a): Find the gain in GPE and KE.

Solution:

Gain in GPE: $\text{GPE} = mgh = 800 \times 10 \times 4 = 32\,000 \text{ J}$

Gain in KE: $\text{KE} = \frac{1}{2}mv^2 = \frac{1}{2} \times 800 \times 20^2 = 160\,000 \text{ J}$

Total energy gain = 32,000 J (GPE) + 160,000 J (KE)

Part (b): Find the tractive force.

Solution:

Work done by engine = Gain in GPE + Gain in KE

$T \times x = 32\,000 + 160\,000$

$T \times 500 = 192\,000$

$T = \frac{192\,000}{500} = 384 \text{ N}$

The tractive force is 384 N.

Part (c): Find the average power of the engine.

Solution:

$\text{Average power} = \frac{\text{work done}}{\text{time taken}} = \frac{192\,000}{50} = 3840 \text{ W}$

The average power is 3.84 kW.

Part (d): Find the instantaneous power at the end of the journey.

Solution:

$P = T \times v = 384 \times 20 = 7680 \text{ W}$

The instantaneous power is 7.68 kW.

Worked Example 3: Metal Ingot with Friction

A metal ingot of weight 80 N is pulled at constant speed 0.2 m s$^{-1}$ for 3 m across a rough horizontal floor. There is a constant frictional force of 30 N.

Part (a): Find the work done by each force.

Solution:

Since there is no vertical motion, no vertical acceleration. Therefore, vertical forces do no work.

Resolving vertically: Reaction $R =$ Weight $= 80$ N

Since horizontal motion is at constant speed, there is no horizontal acceleration. Therefore, resultant horizontal force is zero.

Resolving horizontally: Tension $T =$ Friction $F = 30$ N

The horizontal distance moved = 3 m

• Work done by tension: $T \times x = 30 \times 3 = 90$ J
• Work done against friction: $F \times x = 30 \times 3 = 90$ J
• Work done by weight and reaction: 0 J (perpendicular to motion)

Part (b): Find the power exerted by the tension.

Solution:

$P = T \times v = 30 \times 0.2 = 6 \text{ W}$

The power generated by the tension is 6 watts.

Worked Example 4: Car on Slope with Resistance

A car of mass 1 tonne (1000 kg) travels from rest up a slope at 30° to the horizontal with constant acceleration against a constant resistance of 400 N. At the top of the slope, it has speed 10 m s$^{-1}$ and the engine is working at 58 kW. Take $g = 9.8$ m s$^{-2}$.

Find the length of the slope.

Solution (Method 1: Using work and energy):

At top of slope: Power = $T \times v$

$58\,000 = T \times 10$

$T = 5800 \text{ N}$

Let the length of the slope be $x$ metres.

Using right-angled triangle: vertical height gained = $x \sin 30° = 0.5x$ m

KE gained: $\frac{1}{2}mv^2 = \frac{1}{2} \times 1000 \times 10^2 = 50\,000$ J

Work done by engine: $T \times x = 5800x$ J

Work done against resistance: $R \times x = 400x$ J

Energy equation:

Work done by engine = KE gained + GPE gained + Work done against resistance

$5800x = 50\,000 + 1000 \times 9.8 \times 0.5x + 400x$

$5800x = 50\,000 + 4900x + 400x$

$5800x = 50\,000 + 5300x$

$500x = 50\,000$

$x = 100 \text{ m}$

Solution (Method 2: Using Newton's second law):

Component of weight down slope: $mg \sin 30° = 1000 \times 9.8 \times 0.5 = 4900$ N

Equation of motion parallel to slope:

$T - R - mg \sin 30° = ma$

$5800 - 400 - 4900 = 1000a$

$500 = 1000a$

$a = 0.5 \text{ m s}^{-2}$

Using $v^2 = u^2 + 2as$ with $u = 0$, $v = 10$:

$10^2 = 0 + 2 \times 0.5 \times x$

$100 = x$

$x = 100 \text{ m}$

The length of the slope is 100 metres.

Note: Both methods give the same answer, demonstrating the consistency of physics principles.