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 opposite to the motion. How much work does the road do on the cycle?

Question

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 opposite to the motion. How much work does the road do on the cycle? (A) 2000 J (B) -2000 J (C) -1000 J (D) 0

Tardigrade Admin · Accepted Answer

Displacement travelled by cyclist to reach skidding stop,

s = 10 m

Force on the cyclist due to road,

F = 200 N

Since, force is applied directly opposite to motion (or displacement),
hence

θ = 18 0^{\circ}

∴

Work done by the road on the cycle

W = F s cos θ = 200 \times 10 \times cos 18 0^{\circ}

= - 2000 J

Q. A cyclist comes to a skidding stop in 10m. During this process, the force on the cycle due to the road is 200N and is directly opposite to the motion. How much work does the road do on the cycle?

2000 J

-2000 J

-1000 J

0

Displacement travelled by cyclist to reach skidding stop, s=10m Force on the cyclist due to road, F=200N Since, force is applied directly opposite to motion (or displacement), hence θ=180∘ ∴ Work done by the road on the cycle W=Fscosθ=200×10×cos180∘ =−2000J

Q. 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 opposite to the motion. How much work does the road do on the cycle?