A bicyclist comes to a skidding stop in 10 m. During this process, the force on the bicycle due to the road is

NCERT Exemplar · Q4

Q. A bicyclist comes to a skidding stop in 10 m. During this process, the force on the bicycle due to the road is 200N and is directly opposed to the motion. The work done by the cycle on the road is

(a) + 2000J

(b)

-

200J

(d)

-

20,000J

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Concept understanding — Work Energy Theorem

The Work-Energy Theorem: From Intuition to Precision

Imagine pushing a heavy box across the floor. The harder you push and the farther it slides, the faster it moves when you let go. That connection — the push (force) over a distance (displacement) changing the box's speed — is exactly what the Work-Energy Theorem captures.

The Intuition First

Think of work as the "currency" that buys motion. When you do work on an object, you transfer energy to it. That energy shows up as kinetic energy — the energy of motion. The more work you do, the more the object's kinetic energy changes.

If you push a stationary ball, it starts moving. If you push a moving ball in the same direction, it speeds up. If you push against its motion, it slows down. In every case, the work done equals the change in the ball's kinetic energy.

Note

Work is done by a force on an object. The object's kinetic energy changes by exactly that amount (assuming no other forces do work).

The Precise Statement

W_{\text{net}} = \Delta K = K_f - K_i

Where:

$W_{\text{net}}$ is the net work done on the object (the total work from all forces combined)
$K_f$ is the final kinetic energy
$K_i$ is the initial kinetic energy

And kinetic energy is defined as:

K = \frac{1}{2}mv^2

So the theorem can also be written as:

W_{\text{net}} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2

Why "Net" Work Matters

This is the most common point of confusion. The theorem uses net work — the work done by the net force (the vector sum of all forces). If you push a box and friction opposes it, the net work is the work you do minus the work friction does. Only that net amount changes the kinetic energy.

Watch out

If you push a box at constant speed, your work is positive, but friction does equal negative work. The net work is zero, so kinetic energy doesn't change — the box keeps moving at the same speed. Your work didn't "disappear"; it was dissipated as heat by friction.

A Simple Derivation (for constant force)

Consider a constant net force $F_{\text{net}}$ acting on an object of mass $m$ over a displacement $s$ . From Newton's second law:

F_{\text{net}} = ma

From kinematics (constant acceleration):

v_f^2 = v_i^2 + 2as

Multiply both sides by $\frac{1}{2}m$ :

\frac{1}{2}mv_f^2 = \frac{1}{2}mv_i^2 + mas

But $mas = F_{\text{net}} s = W_{\text{net}}$ , so:

\frac{1}{2}mv_f^2 = \frac{1}{2}mv_i^2 + W_{\text{net}}

Rearranging:

W_{\text{net}} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = \Delta K

Tip

The theorem holds even for variable forces and curved paths — the derivation uses calculus then, but the result is the same.

What It Tells You (and What It Doesn't) …

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The work done by a force requires the point of application of that force to undergo a displacement. Although the bicycle exerts a force on the road, the road itself does not move, so the work done by the cycle on the road is zero. The final answer is (C).

To understand this problem, we need to clearly distinguish between the work done on an object and the work done by an object, and correctly identify the displacement associated with the force in question.

Work done by a constant force $F$ is defined as $W = Fd\cos\theta$ , where $d$ is the displacement of the point of application of the force, and $\theta$ is the angle between the force vector and the displacement vector. A crucial aspect here is that the displacement $d$ must be that of the object on which the force is acting.

Let's break down the problem:

Identify the forces and their actions:

The problem states that the force on the bicycle due to the road is $200 \text{ N}$ and is directly opposed to the motion. This is the frictional force exerted by the road on the bicycle. Let's call this $F_{\text{road on cycle}}$ .
- Magnitude: $F_{\text{road on cycle}} = 200 \text{ N}$ .
- Direction: Opposes the bicycle's motion.
- Displacement of the bicycle: $d_{\text{cycle}} = 10 \text{ m}$ .
Calculate the work done by the road on the bicycle:

This is not what the question asks for, but it helps to understand the interaction.

The force $F_{\text{road on cycle}}$ acts on the bicycle, and the bicycle undergoes a displacement of $10 \text{ m}$ . Since the force opposes the motion, the angle between the force and the displacement is $180^\circ$ .

W_{\text{road on cycle}} = F_{\text{road on cycle}} \cdot d_{\text{cycle}} \cdot \cos(180^\circ)

W_{\text{road on cycle}} = (200 \text{ N}) \cdot (10 \text{ m}) \cdot (-1)

W_{\text{road on cycle}} = -2000 \text{ J}

This negative work indicates that the road removes kinetic energy from the bicycle, causing it to stop.

3. Apply Newton's Third Law to find the force by the cycle on the road:

Newton's Third Law states that for every action, there is an equal and opposite reaction. If the road exerts a force $F_{\text{road on cycle}}$ on the bicycle, then the bicycle exerts an equal and opposite force $F_{\text{cycle on road}}$ on the road.

* Magnitude: $F_{\text{cycle on road}} = 200 \text{ N}$ .

* Direction: This force acts on the road and is in the direction of the bicycle's motion (or rather, opposite to the direction of $F_{\text{road on cycle}}$ ).

Determine the displacement of the road: The question asks for the work done by the cycle on the road. For work to be done on the road, the force $F_{\text{cycle on road}}$ must act on the road, and the point of application of this force on the road must undergo a displacement. …

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Q.A bicyclist comes to a skidding stop in 10 m. During this process, the force on the bicycle due to the road is 200N and is directly opposed to the motion. The work done by the cycle on the road is

Concept understanding — Work Energy Theorem

The Work-Energy Theorem: From Intuition to Precision

The Intuition First

The Precise Statement

Why "Net" Work Matters

A Simple Derivation (for constant force)

What It Tells You (and What It Doesn't) …

Concept: Work Requires Displacement of the Point of Application

Q.A bicyclist comes to a skidding stop in 10 m. During this process, the force on the bicycle due to the road is 200N and is directly opposed to the motion. The work done by the cycle on the road is

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