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a 40-kg slab rests on a frictionless floor. a 10-kg block rests on top of the slab. the coefficient of static friction between the block and the slab is 0.60 whereas the kinetic coefficient is 0.40. the 10 kg block is pushed by a horizontal force with a magnitude of 100 n. what are the resulting acceleration magnitudes of



Answer :

qwarrow

The resulting acceleration of the slab is 0.98m/s2

Let's first examine whether these blocks move in unison.

We shall do that by determining the whole mass's acceleration.

(40+10) kg first.

Using formula F=ma or a= F/m

= 100N/50kg

The acceleration is 2m/s 2

and the maximum frictional force =μ s ×N

=μs ×m1g

=(0.60)(10)(9.8)=58.8N

Thus, we see frictional force is smaller than the applied force and proves that the block and the slab do not move relative to each other.

For the block we have F=μmg

=0.4×10×9.8=39.2N

The resulting acceleration of the slab, a=F/m=39.2/40=0.98m/s2

Thus, the resulting acceleration of the slab is 0.98m/s2

Learn more about acceleration, here:

brainly.com/question/3046924

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