A block A of mass m kg lies on block B of mass m kg. B in turn lies on smooth horizontal plane. The coefficient of friction between A and B is µ. Both the blocks are initially at rest. A horizontal force F is applied to lower block B at t = 0 such that there is relative motion between A and B. In the duration from t = 0 second till the lower block B undergoes a displacement of magnitude L, match the statements in column-I with results in column-II .
Column-I | Column-II |
(a) Work done by friction force on block A is | (p) positive |
(b) Work done by friction force on block B is | (q) negative |
(c) Work done by friction on block A plus work done by friction on block B is | (r) less than µmgL in magnitude |
(d) Work done by force F on block B is | (s) equal to µmgL in magnitude |
Text Solution
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Step 1: Understand the forces involved. Block A moves on block B due to the applied force F. The frictional force acts between the two blocks and will dictate the motion of block A.
Step 2: Work done by the friction force on block A: The friction force acts in the direction of the displacement L of block B but opposite to the motion of block A relative to block B. Thus, this work done is negative.
Step 3: Work done by friction force on block B: Similarly, the friction force acts opposite to the displacement of B, so this work is also negative.
Step 4: The total work done by both friction forces will also be negative because they are both opposing motion.
Step 5: Work done by force F is positively contributing to the displacement, and as it is moving B, it is equal to the work done against the total friction. Therefore, the work done by F on block B is µmgL in magnitude because it is the maximum possible work used to overcome friction without sliding.
Step 6: Finally, we match the statements: (a) matches with (q), (b) matches with (p), (c) matches with (r), and (d) matches with (s).
Therefore, the correct answer combination is D.
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