A block of mass m lies on wedge of mass M. The wedge in turn lies on smooth horizontal surface. Friction is absent everywhere. The wedge block system is released from rest. All situation given in column-I are to be estimated in duration the block undergoes a vertical displacement 'h' starting from rest (assume the block to be still on the wedge). Match the statement in column-I with the results in column-II. (g is acceleration due to gravity)
Column I | Column II |
(a) Work done by normal reaction acting on the block is | (p) positive |
(b) Work done by normal reaction (exerted by block) acting on wedge is | (q) negative |
(c) The sum of work done by normal reaction on block and work done by normal reaction (exerted by block) on wedge is | (r) zero |
(d) Net work done by all forces on block is | (s) less than mgh in magnitude |
Text Solution
Verified by ExpertsA
Thus, the work done by the normal reaction on the block is zero (since work done = force x displacement x cos(theta), where theta is the angle between the force and the displacement vector). Hence, (a) matches with (p) as positive.
Step 2: The work done by the normal reaction (exerted by the block) on the wedge is also zero for similar reasons, so (b) matches with (q) as negative.
Step 3: The sum of work done by the normal reactions on both block and wedge is therefore zero, matching with (r).
Step 4: The total work done on the block is equal to the change in kinetic energy; however, it is less than mgh due to the energy being used to move the wedge. Thus, (d) matches with (s).
Therefore, the matching is
(a) -> (p)
(b) -> (q)
(c) -> (r)
(d) -> (s).
Hence, the correct matches are (a, p), (b, q), (c, r), (d, s).
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