CAIE A-Level Mathematics AS 4.5.3 Power and Efficiency

[Maximum number: 6]

A cyclist is travelling along a straight horizontal road. The total mass of the cyclist and her bicycle is 80 kg . There is a constant resistance force of magnitude 32 N to the cyclist's motion. At an instant when she is travelling at $7 \mathrm{~ms}^{-1}$ , her acceleration is $0.1 \mathrm{~ms}^{-2}$ .

The cyclist later descends a straight hill of length 32.2 m , inclined at an angle of $\sin ^{-1}\left(\frac{1}{20}\right)$ to the horizontal. Her power output is now 120 W , and the resistance force now has variable magnitude such that the work done against this force in descending the hill is 1128 J . The time taken to descend the hill is 4 s .

Given that the speed of the cyclist at the top of the hill is $7.5 \mathrm{~m} \mathrm{~s}^{-1}$ , find her speed at the bottom of the hill.

$120 \times 4 \quad[480]$
Work done by cyclist.

$\frac{1}{2} \times 80 v^{2}$ $\left[40 v^{2}\right]$

or $\frac{1}{2} \times 80 \times 7.5^{2}$

[2250]
For at least one KE term.

$80 g \times \frac{1}{20} \times 32.2 \quad[1288]$ $[80 \times 10 \times 1.61]$
Change in PE.

Attempt at work-energy equation
Attempt at work energy equation with five relevant terms (four relevant terms plus work done against resistance);
dimensionally
correct. Allow sign errors. Allow sin/cos mix.

$120 \times 4+80 g \times \frac{1}{20} \times 32.2-1128=\frac{1}{2} \times 80\left(v^{2}-7.5^{2}\right)$ $\left[480+1288-1128=40 v^{2}-2250\right]$
For correct equation.

Speed $=8.5[0] \mathrm{m} \mathrm{s}^{-1}$ or $\frac{17}{2}$
OE

Use of constant acceleration scores M0 and cannot score B marks if the method leading to their answer only uses constant acceleration.

CAIE A-Level Mathematics AS 4.5.3 Forces Questions

Exam points

CAIE A-Level Mathematics AS 4.5.3 Forces Questions question 1