A particle of mass M moves from x = 0 to x = L on a horizontal surface. Its kinetic-friction coefficient varies as μₖ(x) = μ₀ - αx, and μₖ(L) = 0. The magnitude of total work done by friction is nμ₀MgL. Find n. [The source omits 'magnitude', creating a sign ambiguity.]
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Question

A particle of mass M moves from x = 0 to x = L on a horizontal surface. Its kinetic-friction coefficient varies as μₖ(x) = μ₀ - αx, and μₖ(L) = 0. The magnitude of total work done by friction is nμ₀MgL. Find n. [The source omits 'magnitude', creating a sign ambiguity.]

Options

G1 Physics · Accepted Answer

Step 1: μₖ(L) = μ₀ − αL = 0, so α = μ₀/L. Step 2: Friction opposes motion: Fₓ = −μₖ(x)Mg = −μ₀Mg(1 − x/L). Step 3: Integrate from 0 to L: W_f = ∫₀ᴸ Fₓ dx = −μ₀Mg [x − x²/(2L)]₀ᴸ. Step 4: W_f = −μ₀MgL/2. Its magnitude is μ₀MgL/2. Final answer: n = 1/2 (Option D) when work magnitude is intended. Signed work would give n = −1/2, absent from the options.

A particle of mass M moves from x = 0 to x = L on a horizontal surface. Its kinetic-friction coefficient varies as μₖ(x) = μ₀ − αx, and μₖ(L) = 0. The magnitude of total work done by friction is nμ₀MgL. Find n. [The source omits 'magnitude', creating a sign ambiguity.]