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Question

Two blocks of masses m1 and m2 inter connected by a spring of stiffness k are placed on a horizontal surface. If a constant horizontal force F acts on the block m1 it slides through a distance x whereas m2 remains stationary. If the coefficient of friction between all contacting surfaces is μ, find the speed of the block m1 as the function x.

A
2x(Fm1μg)km1x2
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B
2(Fm1μg)km1x2
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C
2x(Fm12μg)km1x2
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D
2x(Fm1μg)k2m1x2
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Solution

The correct option is A 2x(Fm1μg)km1x2

Since the block m2 does not move force acting on m2 do not perform work.
As blocks do not move in vertical direction, hence work done by normal reaction and weight will be zero.
Then, due to the forces Fsp, F and fk, work is done on the block.

Let v= speed of m1

Applying WE theorem,
Wsp+WN+Wgr+Wfk+Wext=ΔK
(12kx2)+(0)+(0)μNx+Fx=m1v22
where N=m1g

v=2x(Fm1μg)km1x2

Hence option A is the correct answer

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