JEE Advance - Physics (2011 - Paper 2 Offline - No. 9)
A block of mass 0.18 kg is attached to a spring of force-constant 2 N/m. The coefficient of friction between the block and the floor is 0.1. Initially the block is at rest and the spring is un-stretched. An impulse is given to the block as shown in the figure. The block slides a distance of 0.06 m and comes to rest for the first time. The initial velocity of the block in m/s is V = N/10. Then N is
Explanation
Loss of kinetic energy $$ = {1 \over 2}m{V^2}$$
Work done against friction $$ = \mu mgx$$
Gain in potential energy $$ = {1 \over 2}k{x^2}$$
From work-energy principle,
$${1 \over 2}m{V^2} = \mu mgx + {1 \over 2}k{x^2}$$
$$ \Rightarrow {1 \over 2} \times 0.18 \times {V^2} = 0.1 \times 0.18 \times 10 \times 0.06 + {1 \over 2} \times 2 \times {(0.06)^2}$$
$$ \Rightarrow V = 0.4 = {4 \over {10}}$$ ms$$-$$1. Hence, N = 4.
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