Q: A 40 kg slab rests on a frictionless floor.A 10 kg block rests on the top of the slab,the acc.of slab is?

Question

The 10 kg block is acted upon by a horizontal force of 100 N. Coefficent of static friction=0.60, coefficient of kinetic fricton =0.40, g=9.8 m/s2.

Answer 1

the force that the block can apply to the slab is limited by friction
m x g x coef friction. get the values for static and kinetic.
now consider the two masses as one on the frictionless floor. the mass is 50kg and the applied force is 10N. from that you get the theoretical accel if the two blocks stay connected. now consider the force necessary to accelerate the slab at that rate, m x a = F. is the force needed less than the force that can be applied by static friction which you looked at in the first case?
If yes, then the a of the slab is the same as the two blocks together.
If no, then the accel of the slab is force applied by the block by k friction divided by it's mass.

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.

Answer 3

normal reaction on the slab = 40 x 9.8 N = 392N
frictional force between the slab and the block
= coefficient of static friction x normal reaction = 0.40 x 392 =156.8 N
since this force is greater than 100 N, the applied force, the block and the slab will move together

common acceleration = applied force/total mass
= 100 N/ (10kg+40kg)
= 2ms^-2

Answer 4

There could be 2 cases:
1. both slab and block accelerate with the same acceleration
2. block accelerates faster than slab

Let's consider both cases:

1. In first case total force F=100N accelerates total mass m+M=50kg with acceleration a=F/(m+M)=2m/s^2. It means that net force on block is ma=20N and net force on slab is Ma=80N. Net force on slab is equal to static friction force between block and slab. This force can not be larger than 0.60*20*10=120N. So this case is feasible.

2. If block and slab moves relatively to each other than friction force between them is equal to 0.40*20*10=80N. That force accelerates slab with acceleration 80/40=2m/s^2. Slab can not be accelerated faster than block, so block's acceleration is higher than 2m/s^2 and net force should be larger than 20N, but net force is equal to 100-80=20N, so this case is unfeasable.

As a result: block and slab accelerate with the same acceleration 2m/s^2.