Work-energy theorem — calculate work done by friction on a sliding block

Question

A block of mass 4 kg slides on a rough horizontal surface with an initial speed of 10 m/s and comes to rest after travelling 20 m. Using the work-energy theorem, find the work done by friction and the coefficient of kinetic friction. Take $g = 10$ m/s².

(NEET 2022, similar pattern)

Solution — Step by Step

The work-energy theorem says: the net work done on an object equals the change in its kinetic energy.

W_{net} = \Delta KE = \frac{1}{2}mv^2 - \frac{1}{2}mu^2

This is always true — whether forces are constant or variable, conservative or non-conservative.

Initial KE: $\frac{1}{2} \times 4 \times 10^2 = 200$ J

Final KE: $\frac{1}{2} \times 4 \times 0^2 = 0$ J (block comes to rest)

\Delta KE = 0 - 200 = -200 \text{ J}

On a horizontal surface, the only force doing work along the direction of motion is friction (gravity and normal reaction are perpendicular to displacement, so their work is zero).

Therefore: $W_{friction} = W_{net} = \Delta KE$

\boxed{W_{friction} = -200 \text{ J}}

The negative sign confirms that friction removes energy from the block (it decelerates the block).

Friction force: $f = \mu_k N = \mu_k mg$ (on a horizontal surface, $N = mg$ )

Work done by friction: $W_{friction} = -f \times s = -\mu_k mg \times s$

-200 = -\mu_k \times 4 \times 10 \times 20

-200 = -800\mu_k

\boxed{\mu_k = 0.25}

Why This Works

The work-energy theorem is a direct consequence of Newton’s second law. When friction acts opposite to motion, it does negative work — it takes kinetic energy away from the object and converts it to heat. The block slows down precisely because its kinetic energy is being drained by friction.

This approach is powerful because we don’t need to find acceleration separately. We go directly from forces and displacement to energy — skipping the intermediate kinematics step entirely.

Alternative Method — Using kinematics + Newton’s second law

Find deceleration using $v^2 = u^2 + 2as$ :

0 = 100 + 2a(20) \Rightarrow a = -2.5 \text{ m/s}^2

Then $f = ma = 4 \times 2.5 = 10$ N, and $\mu_k = f/(mg) = 10/40 = 0.25$ .

Work by friction: $W = -f \times s = -10 \times 20 = -200$ J.

The work-energy theorem is often faster than the kinematics approach, especially when displacement is given but time is not. If a NEET question gives you mass, initial speed, and distance, go straight to work-energy — don’t waste time finding acceleration first.

Common Mistake

Students frequently forget the negative sign in the work done by friction. Friction opposes motion, so the angle between friction force and displacement is 180°. Since $\cos 180° = -1$ , the work is negative. Writing $W_{friction} = +200$ J would imply friction is adding energy to the block — which is physically absurd for kinetic friction.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using kinematics + Newton’s second law

Common Mistake

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