$A$ cyclist comes to a skidding stop in $10 \; m$. During this process,the force on the cycle due to the road is $200 \; N$ and is directly opposed to the motion.
$(a)$ How much work does the road do on the cycle?
$(b)$ How much work does the cycle do on the road?

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(A) The work done on the cycle by the road is the work done by the stopping (frictional) force on the cycle due to the road.
$(a)$ The stopping force and the displacement make an angle of $180^{\circ} \; (\pi \; \text{rad})$ with each other. Thus,the work done by the road,$W_{r} = F d \cos \theta$
$W_{r} = 200 \times 10 \times \cos(180^{\circ})$
$W_{r} = 2000 \times (-1) = -2000 \; J$
It is this negative work that brings the cycle to a halt in accordance with the Work-Energy theorem.
$(b)$ From Newton's Third Law,an equal and opposite force acts on the road due to the cycle. Its magnitude is $200 \; N$. However,the road undergoes no displacement $(d = 0)$. Thus,the work done by the cycle on the road is $W = F \times 0 = 0 \; J$.

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