IB Physics SL A.3.5 Work-energy change

Practise using the work–energy principle to connect force, displacement and change in kinetic energy in multi-step motion.

Exam points

use the area under a force–distance graph as work done, then relate that area to the car's kinetic energy change
apply net work over successive stages, including friction after a push, to find a stopping distance or final speed
compare initial KE with work done by a resisting force along a path, then decide whether motion can continue

[Maximum number: 4]

This question is about the motion of a bicycle.

A cyclist is moving up a slope that is at an angle of $19^{\circ}$ to the horizontal. The mass of the cyclist and the bicycle is 85 kg .

Figure for Question A.3.5—Work-energy change question 1 — IB Physics SL

At the bottom of the slope the cyclist has a speed of $5.5 \mathrm{~m} \mathrm{~s}^{-1}$ . The cyclist stops pedalling and applies the brakes which provide an additional decelerating force of 250 N . Determine the distance taken for the cyclist to stop. Assume air resistance is negligible and that there are no other frictional forces.

total decelerating force $=271+250(=521 \mathrm{~N})$ ;
acceleration $=(-) \frac{521}{85}\left(=-6.13 \mathrm{~ms}^{-2}\right)$ ;
$s=\frac{v^{2}-u^{2}}{2 a} ;$ $2.47(\mathrm{~m}) ;\} \begin{aligned} & \text { (signs must be consistent for this mark, ie: if acceleration assumed } \\ & \text { positive, look for negative distance) }\end{aligned}$

Allow use of g=10. Answers are $527 \mathrm{~N}, 6.2 \mathrm{~m} \mathrm{~s}^{-2}, 2.44 \mathrm{~m}$ .
or
total decelerating force $=271+250(=521 \mathrm{~N})$ ;
initial kinetic energy $=\frac{1}{2} m v^{2}=1290 \mathrm{~J}$ ;
distance $=\frac{\text { energy lost }}{\text { force }}=\frac{1290}{521}$ ;
2.47 (m);

All question bank results loaded

A.3.5—Work-energy change question 1