A cyclist comes to rest on the application of brakes in 10 metre. During the process the road opposes the cyclist by a force of 200 N. How much work is done by the road on the cyclist?

Question

Accepted Answer

The correct option is (3):

To determine the work done by the road on the cyclist, we need to understand the definition of work in physics. Work done by a constant force is given by the formula $W = Fd \cos 	heta$, where $F$ is the magnitude of the force, $d$ is the magnitude of the displacement, and $	heta$ is the angle between the force vector and the displacement vector. In this problem, the cyclist comes to rest after applying brakes, covering a distance of $10$ meters. The road opposes the cyclist with a force of $200$ N. This "opposing" nature of the force is crucial. It means the force exerted by the road (friction) is acting in the direction opposite to the cyclist's displacement.

Given:  * Force ($F$) = $200$ N
  * Displacement ($d$) = $10$ mSince the road opposes the cyclist, the force applied by the road is in the opposite direction to the displacement. Therefore, the angle ($	heta$) between the force and displacement is $180^\circ$.

Now, we can calculate the work done:
$W = Fd \cos 	heta$
$W = (200 	ext{ N}) 	imes (10 	ext{ m}) 	imes \cos(180^\circ)$
Since $\cos(180^\circ) = -1$,
$W = 2000 	ext{ J} 	imes (-1)$
$W = -2000 	ext{ J}$

The negative sign indicates that the work done by the road on the cyclist is negative, meaning the force acts to slow down or stop the motion. This aligns with the concept of dissipative forces like friction, which remove kinetic energy from a system.

The incorrect option is (1):
$2000$ J would be the work done if the force and displacement were in the same direction, meaning the road was helping the cyclist, which is contrary to the problem statement.

The incorrect option is (2):
$4000$ J is an incorrect value, possibly arising from a calculation error or misunderstanding of the values provided.

The incorrect option is (4):
$-4000$ J is an incorrect value; while it has the correct sign (negative work), the magnitude of the work done is incorrect.