Question

A block A is placed over block B having mass m and 2 m respectively. Block B is resting on a frictionless surface and there is friction between block A and B. The system of blocks is pushed towards a spring with a velocity $v_0$ such that A doesn't slip on B by the time the system comes to momentary rest. The correct statement is:

• A. Work done by friction on A is zero $-\frac{1}{2}mv_0^2$
• B. Work done by friction on B is $-\frac{1}{2}mv_0^2$
• C. Work done by spring on B is $-\frac{3}{2}mv_0^2$
• D. None of these

Answer 1

Answer: (C)

Work done by spring on B is $-\frac{3}{2}mv_0^2$

Answer 2

We start by noting that when block A “does not slip” on block B the friction between them is static. In problems with pure rolling or non‐slipping constraints the static friction force acts only to enforce the constraint and, importantly, does no work (its point of application does not “slide” relative to the body). Thus the work done by friction on block A (and likewise on block B) is zero.

The entire initial kinetic energy of the two–block system

$$K_i=\frac{1}{2}mv_0^2+\frac{1}{2}(2m)v_0^2=\frac{3}{2}mv_0^2,$$

is eventually stored as spring potential energy when the system comes momentarily to rest – the spring (an external agent on block B) does work

$$W_{\text{spring on B}}=-\frac{3}{2}mv_0^2.$$

Thus among the given alternatives only statement 3 is correct.