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A 40 kg slab rests on a frictionless floor. A 10 kg block rests on top of the slab. The static coefficient of friction between the block and the slab is 0.60 while the kinetic coefficient is 0.40. The 10 kg block is acted upon by a horizontal force of 100 N. If g = 9.8m/s2, then what will be the resulting acceleration of the slab?


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Solution

Step1: Given data

Given that a 40kg slab rests on a frictionless floor and a 10kg block rests on top of the slab.
Also, the static coefficient of friction between the block and the slab is μs=0.60

Step2: Formula used

F=ma[whereF=force,m=mass,a=acceleration]

Step3: Calculating the acceleration

Let us first analyze if these blocks move together.
For that, we will be finding the acceleration of the total mass, which means the massofslab+massofblock, i.e., (40+10)kg.
Now using the formula F=ma

or we can write- a=Fm

Now, given the value of F=100N in the question and the total mass is 40+10=50kg.
So, a=10050=2m/s2

Now the maximum frictional force is ffm=μs×N

Also, N=m1g so the maximum frictional force will be : fs=μs×m1g

So, putting the values of friction coefficient μs=0.60, mass of block is m1=10kg and g=9.8m/s2
we get- fs=(0.60)(10)(9.8)=58.8N

Thus, we can see frictional force is smaller than the applied force and proves that the block and the slab do not move relative to each other.
Now, for the block we have F=μm1g

Putting the values of the known terms, we get- F=0.4×10×9.8=39.2N

Now, the resulting acceleration of the slab will be, a=Fm=39.240=0.98m/s2

Hence, the resulting acceleration of the slab is 0.98m/s2.


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