Q. 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 $\mu_k(x)=\mu_0-\alpha x$, where μ₀, α are constants of appropriate dimensions, so that $\mu_k(L)=0$. The total work done by the frictional force during the motion is $n\mu_0MgL$, where g is the acceleration due to gravity. The value of n is:
A) 1/2
B) 3
C) 1
D) 1/3
Explanation
[Solution from source] $\mu_k(L)=0$ gives $\alpha=\mu_0/L$. Work $=-\int_0^L(\mu_0-\alpha x)Mg\,dx=-Mg(\mu_0L-\alpha L^2/2)=-\mu_0MgL/2$, so the magnitude is $\frac12\mu_0MgL$. Answer (1).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.