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Question

A block of mass m is pulled by horizontal constant force F=5 μmg over a rough surface of coefficient of friction μ as shown. Initially spring was at its natural length, the position where block will finally comes to rest will be


A
μmgk
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B
4μmgk
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C
6μmgk
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D
8μmgk
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Solution

The correct option is B 4μmgk
Block will return after maximum elongation
i.e F.xmax1/2Kx2maxμmgxmax=0

xmax=2(Fμmg)k=8μmgk
So block will finally comes to rest returning i.e. v=0 & a=0.
By work energy theorem while returning
(12kx212kx2max)(F+μmg)(xmaxx)=0x=4μmgk


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