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Question

A block of mass m slides from rest at a height H on a frictionless inclined plane as shown in the figure. It travels a distance d across a rough horizontal surface with coefficient of kinetic friction μ, and compresses a spring of spring constant k by a distance x before coming to rest momentarily. Then the spring extends and the block travels back attaining a final height of h. Then:
631328_99c82878bb2a459eaf804534aa4a6a2a.png

A
h=H2μ(d+x)
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B
h=H+2μ(dx)
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C
h=H2μd+kx2/mg
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D
h=H2μ(d+x)+kx2/2mg
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Solution

The correct option is D h=H2μ(d+x)
Let us split the motion of the body into 3 parts. Let the mass of the body be m

Part 1: Climbing down the inclined plane

External Non conservative forces = 0
KEi=0;PEi=mgH
KEf=K;PEf=0
By conservation of Mechanical energy, we have K=mgH

Part 2: Forward and Reverse motion on Rough Surface

External Non conservative force = Friction = μmg
Distance travelled = 2×(d+x) (forward and backward)
Work done = FΔx=2μmg(d+x)

KEi=mgH;PEi=0
KEf=K;PEf=0

By work energy theorem, we have
KmgH=2μmg(d+x)K=mgH2μmg(d+x)

Part 3: Climbing up the inclined plane

External Non conservative forces = 0
KEi=mgH2μmg(d+x);PEi=0
KEf=0;PEf=mgh
By conservation of Mechanical energy, we have mgh=mgH2μmg(d+x)

h=H2μ(d+x)

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