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
Text Solution
Verified by ExpertsThe correct answer is:
CHECK THE SOLUTION.
(4)
μ = 0.1
= μ mg × 0.06 + 
kx 2
× 0.18 u 2 = 0.1 × 0.18 × 10 × 0.06
0.4 = 
⇒ N = 4
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