$A$ 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?

(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 $W = F \times 0 = 0 \; J$.