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Question

An ideal massless spring S can be compressed 1.0 m in equilibrium by a force of 100 N. This same spring is placed at the bottom of a friction less inclined plane which makes an angle θ=30o with the horizontal. A 10 kg mass m is released from the rest at the top of the inclined plane and is brought to rest momentarily after compressing the spring by 2.0 m. The distance through which the mass moves before coming to rest is (Take g=10 m/s2):

A
8 m
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B
6 m
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C
4 m
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D
5 m
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Solution

The correct option is B 4 m
Spring is compressed 1 m with the action of 100N force
from this we can calculate the spring constant k.
F=kx100=k×1k=100N/m

By applying law of conservation of energy
l = length of incline plane x = 2m compression in spring

12kx2=mglsinθ

12×100×22=10×10×lsinθ

lsinθ=2θ=300l=4m
Therefore the distance covered by mass is equal to length of incline plane
L=4m

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