A particle of mass M moves along a horizontal x axis from x = 0 to x = L . The coefficient of kinetic friction varies as a function of x as μ k ( x ) = μ 0 − αx , where μ 0 , α are constants of

Question

A particle of mass M moves along a horizontal x axis from x = 0 to x = L . The coefficient of kinetic friction varies as a function of x as μ k ​ ( x ) = μ 0 ​ − αx , 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 gL , where g is the acceleration due to gravity. The value of n is :NEET 2026 (Re-NEET)

TestHub · Accepted Answer

Given the coefficient of kinetic friction $μ_{k} (x) = μ_{0} - αx$

At $x = L$ , $μ_{k} (L) = 0 \Rightarrow μ_{0} - αL = 0 \Rightarrow α = \frac{μ _{0}}{L}$

The frictional force acting on the particle is $f_{k} = μ_{k} (x) M g = (μ_{0} - \frac{μ _{0}}{L} x) M g$

The magnitude of work done by the frictional force is given by:

$∣ W ∣ = \int_{0}^{L} f_{k} d x$

$∣ W ∣ = \int_{0}^{L} (μ_{0} - \frac{μ _{0}}{L} x) M g d x$

$∣ W ∣ = M g [μ_{0} x - \frac{μ _{0} x ^{2}}{2 L}]_{0}^{L}$

$∣ W ∣ = M g (μ_{0} L - \frac{μ _{0} L ^{2}}{2 L}) = \frac{1}{2} μ_{0} M gL$

Comparing this with $n μ_{0} M gL$ , we get $n = \frac{1}{2}$

Answer: $\frac{1}{2}$

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