Irodov Problem 1.196

Problem Statement

A small block of mass m slides down an inclined plane of length l and angle α. The coefficient of kinetic friction is μk. Find the velocity at the bottom.

Given

m, l, α, μk. Starts from rest.

Concepts & Formulas

Work-energy: ½mv² = mgl sinα − μk·mg cosα·l.

Step-by-Step Solution

Step 1: Work by gravity: W_g = mgl sinα.
Step 2: Work by friction: W_f = −μk·mg cosα·l.
Step 3: ½mv² = mgl(sinα − μk cosα).
Step 4: v = √(2gl(sinα − μk cosα)).

Worked Calculation

v = √(2gl(sinα − μk cosα)).

Boxed Answer

v = √[2gl(sinα − μk cosα)]

Physical Interpretation

The velocity at the bottom depends on the net work — friction reduces the available energy. If μk ≥ tanα, the block doesn’t slide at all.