A particle of mass M moves along a horizontal x-axis from x=0 to x=L. The coefficient of kinetic friction varies as µ_k(x)=µ₀-α x, where µ₀,α are constants of appropriate dimensions, so that µ_k(L)=0. The total work done by the frictional force during the motion is nµ₀ MgL. The value of n is:

Solution & Explanation

Governing idea: friction is a variable force here, so the work done is the integral of force over the path, not force × distance. Step 1 — Fix the constant α using the given condition µ_k(L)=0. µ_k(x)=µ₀-αx, and at x=L it is zero: µ₀-αL=0 → α=µ₀/L. Step 2 — Write the friction force at position x. The normal force on a horizontal surface is N=Mg, so f(x)=µ_k(x)·Mg=(µ₀-αx)Mg. Step 3 — Integrate from x=0 to x=L to get the work done by friction. W=∫₀ᴸ f(x)dx=Mg∫₀ᴸ(µ₀-αx)dx =Mg[µ₀x-(αx²)/2]₀ᴸ =Mg[µ₀L-(αL²)/2]. Step 4 — Substitute α=µ₀/L. W=Mg[µ₀L-(µ₀/L)·L²/2]=Mg[µ₀L-(µ₀L)/2]=(1/2)µ₀MgL. Comparing with W=n·µ₀MgL gives n=1/2, which matches option D. Trap: if you treat µ as constant at µ₀ you get n=1; the linear decrease to zero halves the work, giving n=1/2.