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Question

# In the arrangement shown, the block is released from rest from x=0 (when the spring was in its natural length), on a rough incline surface having μ=0.5x. The maximum elongation in the spring will be,

A
1 m
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B
2.5 m
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C
2 m
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D
1.5 m
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Solution

## The correct option is C 2 mLet the maximum elongation in the spring be xm Using Work-Energy therorem, Wnet=ΔK Wg+Wf+WN+Wspring=ΔK WN=0 Wf=∫xm0f.dx Also, f=μN=μmgcosθ (50sin37∘)xm−∫xm0(0.5x(50)(cos37∘))dx+0−12kx2m=12m(v2f−v2i) At the point of maximum elongation v=0 Hence, vf=0 and vi=0 30xm−10x2m−5x2m=0 ⇒30=10xm+5xm ⇒15xm=30 ⇒xm=2 m Hence, option (C) is the correct answer.

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