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Question

Two bodies m1 and m2 are kept on a table with coefficient of friction 'μ' and are joined by a spring initially, the spring is in its relaxed state. Find out the minimum force F which will make the other block m2move. (K is the spring constant).



A

g ()

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B
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C
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D
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Solution

The correct option is C

For minimum force 

m1 will firse accelearate then decelerate and reach a distance x0 from intial point.

Kx on m1 will keep increasing with distance hence net force will be positive firse , the 0,

then negative.

This x0 will be such that spring force on m2 i.e ., Kx0μm2g   ...(1)

So m2 will just start its motion.

 Applying work energy theorem on m1 initial and final velocity of m1 = 0

Δ K.E = 0

All forces acting on m1 while it moves a distance x0 is spring force, friction, gravity, normal, force F min

WgWN = 0 (As θ 90)

WspWfrWf = Δ K.E = 0

Δ Uspμm1x0 = 0 (Wsp = ΔUsp)

- 12x02 - μm1gx0Fminx0 = 0

Fmin = μm1g + 12x0

From (i) x0 = μ m2 gk

Fmin = μm1g + μ m2 gk


Physics

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