A 5.00 kg block is placed on top of a 12.0 kg block that rests on a frictionless table. The coefficient of st

Question

A 5.00 kg block is placed on top of a 12.0 kg block that rests on a frictionless table. The coefficient of static friction between the two blocks is 0.600. What is the maximum horizontal force that can be applied before the 5.00 kg block begins to slip relative to the 12.0 kg block, if the force is applied to (a) the more massive block and (b) the less massive block?

Answer 1

First, consider the force applied to the 5 kg block. Looking at FBDs of each and using F=m*a

Find an F such that the acceleration of the 5 kg block is greater than the acceleration of the 12 kg block

F-5*g*0.6=5*a1

5*g*0.6=12*a2

solve for a2 = 2.4525

plug into the first equation
F>5*2.4525+5*g*0.6
F> 41.7 N

Similarly when the force is applied to the 12 kg block, F such that the acceleration of the 12 kg block is greater than the acceleration of the 5 kg block
F-5*g*0.6=12*a2
5*g*0.6=5*a1
a1= 9.81*0.6
F>12*9.81+5*9.81*0.6

F> 100

j

Answer 2

Apply Newton's law : f = ma, where f=force in newtons, m=mass in kg, a=acceleration in m/s2. Since the ground is frictionless, then, a= f/m = 100/(40+10) = 2 m/s2 So, the answer has nothing to do with the friction between the slab and block, both will act as one mass of 50 kg. Regardless of that friction, the answer always will be the same.