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ExamsJEE AdvancedPhysics

A box of mass m initially at rest on a horizontal surface is pushed by a constant horizontal force F = mg/2. The coefficient of kinetic friction varies with distance as mu = mu0 * x, where x is the distance from the starting point. Over what distance is the box pushed before it comes to rest again?

  1. 2 / mu0
  2. 1 / mu0
  3. 1 / (2*mu0)
  4. 1 / (4*mu0)

Correct answer: 1 / mu0

Solution

Since the box starts and ends at rest, the net work done on it is zero. Work by F: W_F = (mg/2)*d. Work by friction: W_f = integral from 0 to d of (mu0*x)*mg dx = mu0*mg*d²/2. Setting W_F = W_f: (mg/2)*d = mu0*mg*d²/2 => 1/2 = mu0*d/2 => d = 1/mu0.

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