Question

two blocks in contact with each other are being pushed by an external agent as shown. There is no friction anywhere and the blocks have an instantaneous velocity V° in the position find the instantaneous power supplied by M1 to M2​

Answer 1

The instantaneous power supplied by M1 to M2 is M2*F*v/(M1+M2)

• As they are in contact so they will move with same acceleration, i.e. a=F/(M1+M2)
• Power is defined as rate of work done (ΔW/Δt) it can also be written in terms of force and velocity as Fv
• So the force net force on the block M2 is given as M2*F/(M1+M2)
• As force in the direction of acceleration can only be provided by the component of the normal force from M1

Therefore, the instantaneous Power supplied by M1 to M2 is M2*F*v/(M1+M2)

Note :- The normal force applied to M2 by M1 = M2*F/((M1+M2)*sinθ)

Answer 2

Hence the normal force applied to M2 by M1 = M2 x F/((M1+M2)*sinθ)

Explanation:

The instantaneous power supplied by M1 to M2 = M2 x F x v/ (M1 + M2)

We know that acceleration a = F / m

As they are in contact so they will move with same acceleration, i.e.

a = F / (M1 + M2)

Power is defined as rate of doing work ΔW/Δt.

It can also be written in terms of force and velocity as F.V

So the net force on the block M2 is given as M2 x F / (M1 + M2)

As force in the direction of acceleration can only be provided by the component of the normal force from M1

Therefore, the instantaneous Power supplied by M1 to M2 = M2 x F x v/(M1 + M2)

Hence the normal force applied to M2 by M1 = M2 x F/[(M1 + M2) x sinθ]