Question

An ideal massless spring S can be compressed by 1 m by a force of 100 N in equilibrium. The same spring is placed at the bottom of a frictionless inclined plane at 30∘ to the horizontal. A 10 kg block M is released from rest at the top of the incline and is brought to rest momentarily after compressing the spring by 2 m. If g=10 m/s2, what is the speed of mass just before it touches the spring?

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Solution

The correct option is **A** √20 m/s

F=kx

∴k=Fx=1001 N/m=100 N/m

Now from energy conservation, between natural length of spring and its maximum compression state.

12mv2+mgh=12kx2max

where h is the height of the block when it just touches the spring from the point where the maximum compression occurs.

∴v=√kx2maxm−2gh

=√(100)(2)210−(2)(10)(22)

=√20 m/s

F=kx

∴k=Fx=1001 N/m=100 N/m

Now from energy conservation, between natural length of spring and its maximum compression state.

12mv2+mgh=12kx2max

where h is the height of the block when it just touches the spring from the point where the maximum compression occurs.

∴v=√kx2maxm−2gh

=√(100)(2)210−(2)(10)(22)

=√20 m/s

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