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A 40kg slab rests on a frictionless floor. A 10kg block rests on top of the slab. The coefficient of kinetic friction between the block and slab is 0.4. A horizontal force of 100N applied on 10kg block . Find the resulting acceleration of the slab.


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Solution

Step 1: Given Data

Mass of the slab ms=40kg

Mass of the block mb=10kg

The coefficient of kinetic friction between the block and slab μ=0.4

The horizontal force of F=100N on -x direction.

Step 2: Calculate the Common acceleration

The total mass of the system,

m=ms+mb

m=40+10=50kg

Since acceleration is action on -x direction, the pseudo force will be acting on +x direction with a force of 50a.

50a=100

a=2m/s2

Step 3: Check acceleration of slab and block

Let us consider both the blocks move with the same acceleration.

Pseudo force on the slab fs=msa=40×2=80N

Pseudo force on the block fb=mba=10×2=20N

Again, the normal force N=100N

Therefore, limiting force, fL=μN=0.4×100=40N

Since fs>fL, this is not the case here.

Step 4: Calculate resultant acceleration of slab

Therefore, the block and the slab do not move with the same acceleration.

If as is the acceleration of the slab, then

40as=40

as=1m/s2

Hence, the resulting acceleration of the slab is 1m/s2.


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