A Level Physics 9702: Energy Conservation for a Cyclist Coasting Downhill

A cyclist and her bicycle, of combined mass $78\text{ kg}$ , start from rest at the top of a straight downhill slope and free-wheel (do not pedal) to the bottom. The vertical height of the slope is $22\text{ m}$ , and the distance travelled along the slope is $140\text{ m}$ . A constant resistive force (from air resistance and rolling friction) of average magnitude $45\text{ N}$ acts on the cyclist throughout the descent.

Take $g = 9.81\text{ m s}^{-2}$ .

(a) Calculate the loss in gravitational potential energy of the cyclist and bicycle during the descent. [2]

(b) Calculate the work done against the resistive force during the descent. [2]

(c) Use the principle of conservation of energy to calculate the kinetic energy of the cyclist and bicycle at the bottom of the slope. [1]

(d) Hence calculate the speed of the cyclist and bicycle at the bottom of the slope. [2]

Part (a): Loss in gravitational potential energy

The loss in gravitational PE depends only on the vertical height dropped, $\Delta h = 22\text{ m}$ , not on the distance travelled along the slope: $\Delta E_p = mg\Delta h = 78\times9.81\times22$

Working this in two steps: $78\times9.81=765.18$ , then $765.18\times22=16833.96\text{ J}$ .

Check by recomputing the second step differently: $765.18\times22 = 765.18\times20+765.18\times2 = 15303.6+1530.36=16833.96\text{ J}$ . Both methods agree.

So the loss in gravitational PE is $\boxed{16834}\text{ J} \approx 1.68\times10^{4}\text{ J}$ (3 s.f.).

Part (b): Work done against the resistive force

The resistive force acts over the actual path length travelled, which is the $140\text{ m}$ along the slope (not the vertical height): $W_{\text{resistance}} = Fd = 45\times140 = 6300\text{ J}$

Check: $45\times140 = 45\times100+45\times40=4500+1800=6300\text{ J}$ . Confirmed.

Part (c): Kinetic energy at the bottom

By the principle of conservation of energy, the loss in gravitational PE is shared between kinetic energy gained and work done against the resistive force: $\Delta E_p = E_k + W_{\text{resistance}}$

Rearranging for the kinetic energy at the bottom: $E_k = \Delta E_p - W_{\text{resistance}} = 16833.96 - 6300 = 10533.96\text{ J}$

So $E_k \approx \boxed{10534}\text{ J} \approx 1.05\times10^{4}\text{ J}$ (3 s.f.).

Part (d): Speed at the bottom of the slope

Using $E_k = \tfrac12mv^2$ , rearranged for $v$ : $v = \sqrt{\frac{2E_k}{m}} = \sqrt{\frac{2\times10533.96}{78}}$

Working the numerator first: $2\times10533.96=21067.92$ . Dividing by the mass: $21067.92/78=270.10$ (2 d.p.).

$v = \sqrt{270.10} \approx 16.4\text{ m s}^{-1}$

Check: $16.4^2 = 268.96$ and $16.44^2=270.27$ , both close to $270.10$ , confirming $v\approx16.4\text{ m s}^{-1}$ (3 s.f.).

Final answers

(a) Loss in gravitational PE $= \boxed{16834}\text{ J} \approx 1.68\times10^{4}\text{ J}$
(b) Work done against resistance $= \boxed{6300}\text{ J}$
(c) Kinetic energy at the bottom $\approx \boxed{10534}\text{ J} \approx 1.05\times10^{4}\text{ J}$
(d) Speed at the bottom $\approx \boxed{16.4}\text{ m s}^{-1}$