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).
g ( + )
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,
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