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Q.

An ideal massless spring S can be compressed 1  m by a force of 100N in equilibrium .The same spring is placed at the bottom of a frictionless inclined plane at 30o to the horizontal. A 10kg block M is releases from rest at the top of the incline and is brought to rest momentarily after compressing the spring by 2m. If g = 10 ms-2, what is the speed of mass just before it touches the spring?

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a

20ms−1

b

30ms−1

c

10ms−1

d

40ms−1

answer is A.

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Detailed Solution

restoring force of a spring is F=-kx∴spring constant is k=Fx=1001=100N/maccording to principle of conservation of energy, when mass=10kg slide down inclineEinitial=EfinalKEMass+PEMass=energy stored in a spring∴12×10×v2+10102sin300=12×100×22solving we get velocity of mass is V = 20m/s
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