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Question

# A block of mass 5 kg is released from rest when compression in spring is 2 m. Block is not attached with the spring of spring constant k, natural length of the spring is 4 m. The maximum height of the block from the ground is: (g=10 m/s2)

A
5.5 m
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B
4.5 m
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C
6 m
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D
7.5 m
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Solution

## The correct option is A 5.5 mGiven: m=5 kg; x=2 m; k=300 m/s Since, the surface of wedge is smooth, so we can apply conservation of mechanical energy between initial and final points. (PE)i+(KE)i=(PE)f+(KE)f Taking ground as a reference, [12kx2+mg(2sin30∘)+0]i=[0+mg(6sin30∘)+12mv2]f [12300(2)2+50(2sin30∘)]i=[50(6sin30∘)+125v2]f ∴v2=200 .......(1) Now, block will follow the projectile path. The maximum height of a projectile from the ground surface is given by H=3+hm=3+v2sin230∘2g From equation (1) equation v2=200 H=3+2.5=5.5 m So, the maximum height of the block from the ground is 5.5 m. Therefore, option (A) is correct.

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