# Mechanics problem.

**1. The problem statement, all variables and given/known data**

Find the minimum Force required to move M2.

M1=3kg.M2=5kg,[itex]\mu[/itex]1=0.4and[itex]\mu[/itex]2=0.6.

**2. Relevant equations**

Net f =Ma

**3. The attempt at a solution**

For a displacement ‘x’ force applied on the block M1 by the spring=kx.

From free body diagrams we can get

F=[itex]\mu[/itex]1*M1*g+kx.

When spring applies kx on block 1 it will apply kx to block 2 but in opposite direction.The block 2 will move when kx=[itex]\mu[/itex]2*m2*g.Combining Two equations we get f= 41.16.

It is not right,somebody told me that I could not do this question in this way because force exerted by the spring is not constant.But maximum force that spring can exert on block is [itex]\mu[/itex]2*m2*g and if F exceeds than that then block 2 should start moving.This thing does not have anything to do with the nature of force.

http://ift.tt/UePciU

## Leave a comment