A small block starts slipping down from a point B on an inclined plane AB, which is making an angle θ with the horizontal section BC is smooth and the remaining section CA is rough with a coefficient of friction μ. It is found that the block comes to rest as it reaches the bottom (point A) of the inclined plane. If BC = 2AC, the coefficient of friction is given by μ = ktan θ . The value of k is _________.

Solution

Physics sir · 4 steps

Starts at rest, ends at rest: gravity's work over all of BA equals friction's work over CA alone.

1. Set the lengths. Let AC = L. Then BC = 2L, so the whole slope BA = 3L. Friction acts only on the rough part CA, of length L.
2. Work-energy theorem. The block starts at rest at B and stops at A, so the total work done on it is zero. Gravity's component along the slope, mg sin θ, acts over the full 3L. Friction, μ mg cos θ, acts only over L.

   mg sin θ × 3L = μ mg cos θ × L

3. Solve for μ. Cancel mgL on both sides to get 3 sin θ = μ cos θ, so μ = 3 tan θ.
4. So. Comparing with μ = k tan θ gives k = 3.

The common slip: 2. Counting gravity's work only on the smooth part BC, and forgetting that gravity keeps pulling on CA too, gives k = 2 instead of 3.

The idea to keep. When a body starts and ends at rest, the net work is zero. Each force does work only over the stretch where it acts, but gravity acts along the whole path.

Answer from the NTA final answer key. Solution by PrepWiser.