  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 (  )  B   C   D   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 ., Kx0 = μm2g   ...(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 Wg = WN = 0 (As θ =  90∘) Wsp + Wfr + Wf = Δ K.E = 0 Δ Usp - μm1g x0 = 0 (Wsp = ΔUsp) - 12 K x02 - μm1gx0 + Fminx0 = 0 Fmin = μm1g + 12x0 From (i) x0 = μ m2 gk Fmin = μm1g + μ m2 gk Physics

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