Physics Q1 – NEET-UG 2026

Home Physics Work, Energy & PowerQuestion #661

A particle of mass

M

moves along a horizontal

x

axis from

x = 0

x = L

. The coefficient of kinetic friction varies as a function of

x

μ_{k} (x) = μ_{0} - α x

, where

μ_{0}

α

are constants of appropriate dimensions, so that

μ_{k} (L) = 0

. The total work done by the frictional force during the motion is

n μ_{0} M g L

, where

g

is the acceleration due to gravity. The value of

n

is :

Clue 1 of 3: Work Done by a Variable Force

When a force is not constant, the work done by it is calculated by integrating the force over the displacement.

🥷

Ninja Strategy•Average Friction Coefficient Estimation

Since the friction coefficient decreases linearly from $μ_{0}$ to $0$ , the average effective coefficient is $μ_{0} / 2$ . This suggests the work done will be proportional to $\frac{1}{2} μ_{0} M g L$ , making $n = 1 / 2$ the most plausible answer.

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● Video Walkthrough

Step 1: Determine the Frictional Force as a Function of Position✦ Active

The coefficient of kinetic friction is given by $μ_{k} (x) = μ_{0} - α x$ . For a particle moving on a horizontal surface, the normal force is $N = M g$ . Therefore, the kinetic frictional force is $f_{k} = μ_{k} (x) N = (μ_{0} - α x) M g$ . The work done by friction is negative as it opposes motion.

We are given that $μ_{k} (L) = 0$ . Using this condition:

μ_{0} - α L = 0 ⟹ α = \frac{μ_{0}}{L}

Substitute $α$ back into the expression for $f_{k}$ :

f_{k} = (μ_{0} - \frac{μ_{0}}{L} x) M g = μ_{0} M g (1 - \frac{x}{L})

💡 Teacher's Secret Hint

Remember that work done by friction is always negative relative to the direction of displacement.

Step 2: Calculate the Work Done by Frictional Force○ Expand

The work done by a variable force is given by the integral of the force over the displacement. Since friction opposes motion, the work done by friction is $W_{f} = \int_{0}^{L} - f_{k} d x$ .

W_{f} = \int_{0}^{L} - μ_{0} M g (1 - \frac{x}{L}) d x

Take the constants out of the integral:

W_{f} = - μ_{0} M g \int_{0}^{L} (1 - \frac{x}{L}) d x

Evaluate the integral:

W_{f} = - μ_{0} M g {[x - \frac{x^{2}}{2 L}]}_{0}^{L}

W_{f} = - μ_{0} M g [(L - \frac{L^{2}}{2 L}) - (0 - 0)]

W_{f} = - μ_{0} M g [L - \frac{L}{2}]

W_{f} = - μ_{0} M g (\frac{L}{2}) = - \frac{1}{2} μ_{0} M g L

💡 Teacher's Secret Hint

Pay attention to the negative sign for work done by friction, as it indicates energy dissipation.

Step 3: Determine the Value of n○ Expand

The total work done by the frictional force is given as $n μ_{0} M g L$ . Comparing this with our calculated work done:

n μ_{0} M g L = - \frac{1}{2} μ_{0} M g L

Since the options are positive, $n$ is likely referring to the magnitude factor. Therefore, we compare the magnitudes:

| n μ_{0} M g L | = | - \frac{1}{2} μ_{0} M g L |

n μ_{0} M g L = \frac{1}{2} μ_{0} M g L

Dividing both sides by $μ_{0} M g L$ (assuming it's non-zero):

n = \frac{1}{2}

💡 Teacher's Secret Hint

In physics problems, 'work done' can sometimes refer to the magnitude of work, especially when comparing with a positive constant like $n$ here.

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Physics Question 1 – NEET-UG 2026

Clue 1 of 3: Work Done by a Variable Force

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