A 7.40-kg block is placed on top of a 12.5-kg block that rests on a frictionless table. The coefficient of static friction between the two blocks is 0.610. What is the maximum horizontal force that can be applied before the 7.40-kg block begins to slip relative to the 12.5-kg block, if the force is applied to (a) the more massive block and(b) the less massive block?

Question

A 7.40-kg block is placed on top of a 12.5-kg block that rests on a frictionless table. The coefficient of static friction between the two blocks is 0.610. What is the maximum horizontal force that can be applied before the 7.40-kg block begins to slip relative to the 12.5-kg block, if the force is applied to (a) the more massive block and(b) the less massive block?

bobpursley · Answer

force friction between two blocks: 7.40*9.8*.610=44.3N maximum acceleration: 44.3=7.40*a or a= 5.99m/s^2 a) force=ma force=(7.4+12.5)5.99 b) force= 7.4*5.99