A block of mass ′m′ is pulled by horizontal constant force F=5μmg 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 B4μmgk Block will return after maximum elongation
i.e F.xmax−1/2Kx2max−μmgxmax=0
xmax=2(F−μmg)k=8μmgk
So block will finally comes to rest returning i.e. v=0 & a=0.
By work energy theorem while returning −(12kx2−12kx2max)−(F+μmg)(xmax−x)=0⇒x=4μmgk