A man, of mass m, standing at the bottom of the staircase, of height L climbs it and stands at its top. (Note:

NCERT Exemplar · Q19

m

(a) Work done by all forces on man is equal to the rise in potential energy

mgL

(b) Work done by all forces on man is zero.

mgL

(d) The reaction force from a step does not do work because the point of application of the force does not move while the force exists.

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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 man starts and ends at rest, so the net work of all forces on him is zero; gravity does $-mgL$ and the step's reaction does no work. The correct statements are (B) and (D).

Reasoning via the Work-Energy Theorem

The man is at rest at the bottom and at rest at the top, so his kinetic energy is unchanged:

W_{\text{net}} = \Delta KE = 0.

Now examine each force and each statement.

Gravity. As the man rises a height $L$ , gravity (downward) opposes the displacement:

W_{\text{gravity}} = -mgL.

Reaction from the steps. The normal reaction acts on the foot, which stays fixed on a step while it presses down (no slipping). Its point of application does not move while the force acts, so it does no work on the man. …

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Q.A man, of mass $m$ , standing at the bottom of the staircase, of height $L$ climbs it and stands at its top. (Note: more than one of the given options may be correct.)

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) …

Reasoning via the Work-Energy Theorem

Concept: Work-Energy Theorem Applied to Each Force Separately

Q.A man, of mass mmm, standing at the bottom of the staircase, of height LLL climbs it and stands at its top. (Note: more than one of the given options may be correct.)

Reasoning via the Work-Energy Theorem

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Q.A man, of mass $m$ , standing at the bottom of the staircase, of height $L$ climbs it and stands at its top. (Note: more than one of the given options may be correct.)

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