IB Physics SL A.3.5 Work–Energy Change Question Bank

Practise using work–energy and force–distance graphs to calculate speed, stopping distance, frictional loss and multi-step motion.

Exam points

Use work–energy and force–distance graphs to calculate speed, energy change or stopping distance.
Calculate frictional work and stopping distance for ramps, brakes and rough surfaces using an energy balance.
Analyse multi-step motion by tracking work, kinetic, gravitational potential energy and losses between stages.

[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);

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A.3.5—Work-energy change question 1