A cyclist comes to a skidding stop at $50 m$ . During this process, the force on the cycle due to the road is $1000 N$ and is directed opposite to the motion. How much work is done by the road on the cycle in this process?

10 kJ

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50 kJ

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- 50 kJ

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- 10 kJ

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Solution

The correct option is C $- 50 kJ$
Work done by the road on the cycle in this process,

$W = F d = - 1000 \times 50 = - 5 \times 10^{4} J = - 50 kJ$

Hence, option $(C)$ is the correct answer.

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Similar questions

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 oppose 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?

Q. A bicyclist comes to a skidding stop in

10 m

. During this process, the force on the bicycle due to the road is

200 N

and is directly opposed to the motion. The work done by the cycle on the road is:

Q. When a cyclist moves on the road, work done by the road on the cycle is negative.

Q. A cyclist moving on a level road at

4

m/s stops pedalling and lets the wheel come to rest. The retardation of the cycle has two components: a constant

0.08 {m/s}^{2}

due to friction in the working parts and a resistance of

0.02 v^{2}

/m where

v

is speed in meters per second. The distance traveled by cycle before it comes to rest is: (Consider

ln 5 = 1.61

Q. Solve the following problems.
(a) A force of 1000 N was applied to stop a car that was moving with a constant velocity. The car stopped after moving through 10 m. How much is the work done?
(b) A cart with mass 20 kg went 50 m in a straight line on a plain and smooth road when a force of 2 N was applied to it. How much work was done by the force?

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A cyclist comes to a skidding stop at 50 m. During this process, the force on the cycle due to the road is 1000 N and is directed opposite to the motion. How much work is done by the road on the cycle in this process?

The correct option is C −50 kJWork done by the road on the cycle in this process, W=Fd=−1000×50=−5×104 J=−50 kJ Hence, option (C) is the correct answer.

A cyclist comes to a skidding stop at $50 m$ . During this process, the force on the cycle due to the road is $1000 N$ and is directed opposite to the motion. How much work is done by the road on the cycle in this process?

The correct option is C $- 50 kJ$
Work done by the road on the cycle in this process,

$W = F d = - 1000 \times 50 = - 5 \times 10^{4} J = - 50 kJ$

Hence, option $(C)$ is the correct answer.