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Physics · PYQ NEET UG 2026
हिन्दीA 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 \( \mu_k(x) = \mu_0 - \alpha x \), where \( \mu_0 \), \( \alpha \) are constants of appropriate dimensions, so that \( \mu_k(L)=0 \). The total work done by the frictional force during the motion is \( n\mu_0 MgL \), where g is the acceleration due to gravity. The value of n is :
Correct answer · Option D
\( \frac{1}{2} \)
Explanation
The work done by friction is \( W = -\int_0^L \mu_k(x) Mg dx \). Using \( \mu_k(L) = 0\Rightarrow \alpha = \mu_0/L \), integrating gives \( W = -\frac{1}{2}\mu_0 MgL \). Thus, \( n = 1/2 \).
