Worked example 1

A 2.0 kg ball is dropped from a height of 5.0 m. (a) Calculate its gravitational potential energy at the start. (b) Assuming no air resistance, calculate its speed just before hitting the ground. (c) If, due to air resistance, its actual speed just before hitting the ground is 8.5 m/s, calculate the work done against air resistance. (Take $g = 9.81 \text{ m/s}^2$ ).

Show solution outline

(a) Initial GPE: $GPE = mgh = 2.0 \text{ kg} \times 9.81 \text{ m/s}^2 \times 5.0 \text{ m} = 98.1 \text{ J}$.
(b) By conservation of energy (no air resistance): Initial GPE = Final KE. $98.1 \text{ J} = \frac{1}{2}mv^2 = \frac{1}{2}(2.0 \text{ kg})v^2$ $98.1 = v^2$ $v = \sqrt{98.1} \approx 9.90 \text{ m/s}$ .
(c) With air resistance: Initial GPE = Final KE + Work done against air resistance. Initial GPE = $98.1 \text{ J}$ . Actual Final KE = $\frac{1}{2}mv_{actual}^2 = \frac{1}{2}(2.0 \text{ kg})(8.5 \text{ m/s})^2 = 1.0 \times 72.25 = 72.25 \text{ J}$ . Work done against air resistance = Initial GPE - Actual Final KE Work done against air resistance = $98.1 \text{ J} - 72.25 \text{ J} = 25.85 \text{ J}$ .

Worked example 2

A car of mass 1200 kg travels at a constant speed of 18 m/s up a road inclined at 6.0° to the horizontal. The car's engine works at a constant rate of 45 kW. (a) Calculate the work done against the gravitational force in 10 seconds. (b) Calculate the total work done by the engine in 10 seconds. (c) Determine the magnitude of the total resistive force acting on the car. (Take $g = 9.81 \text{ m/s}^2$ ).

Show solution outline

(a) First, find the vertical height gained in 10 seconds. Distance along the slope, $d = v \times t = 18 \text{ m/s} \times 10 \text{ s} = 180 \text{ m}$ . Vertical height gained, $h = d \sin(\theta) = 180 \text{ m} \times \sin(6.0°) = 18.81 \text{ m}$ . Work done against gravity (Gain in GPE) = $mgh = 1200 \text{ kg} \times 9.81 \text{ m/s}^2 \times 18.81 \text{ m} = 221449 \text{ J} \approx 2.2 \times 10^5 \text{ J}$.
(b) The engine's power is the rate at which it does work. Total work done by engine, $W_{engine} = P \times t = 45000 \text{ W} \times 10 \text{ s} = 450000 \text{ J} = 4.5 \times 10^5 \text{ J}$ .
(c) By the principle of energy conservation, the work done by the engine is converted into GPE and work done against resistive forces (since KE is constant). $W_{engine} = \Delta GPE + W_{resistive}$ $450000 \text{ J} = 221449 \text{ J} + W_{resistive}$ Work done against resistive force, $W_{resistive} = 450000 - 221449 = 228551 \text{ J}$ . Since $W_{resistive} = F_{resistive} \times d$ , we can find the force: $F_{resistive} = \frac{W_{resistive}}{d} = \frac{228551 \text{ J}}{180 \text{ m}} = 1269.7 \text{ N} \approx 1300 \text{ N}$ (to 2 s.f.).

Alternatively, using forces: Driving force from engine, $F_{driving} = \frac{P}{v} = \frac{45000 \text{ W}}{18 \text{ m/s}} = 2500 \text{ N}$ . At constant velocity, driving force equals opposing forces. $F_{driving} = F_{resistive} + mg\sin(\theta)$ $2500 \text{ N} = F_{resistive} + (1200 \text{ kg} \times 9.81 \text{ m/s}^2 \times \sin(6.0°))$ $2500 \text{ N} = F_{resistive} + 1230 \text{ N}$ $F_{resistive} = 2500 - 1230 = 1270 \text{ N} \approx 1300 \text{ N}$ (to 2 s.f.).

Energy conservation — practice questions

Worked example 1

Worked example 2