Laws Of Motion Question 407
Question: A 40 kg slab rests on a frictionless floor as shown in the figure. A 10 kg block rests on the top of the slab. The static coefficient of friction between the block and slab is 0.60 while the kinetic friction is 0.40. The 10 kg block is acted upon by a horizontal force 100 N. If $ g=9.8m/s^{2}, $ the resulting acceleration of the slab will be
Options:
A) $ 1m/s^{2} $
B) $ 1.5m/s^{2} $
C) $ 2m/s^{2} $
D) $ 6m/s^{2} $
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Answer:
Correct Answer: A
Solution:
Limiting friction between block and slab
$ =m _{S}m _{A}g=0.6\times 10\times 1=60N $
But applied force on block A is 100 N. So the block will slip over the slab.
Now, kinetic friction works between block and slab, $ F _{k}=m _{k}m _{A}g=0.4\times 10\times 10=40N $
This kinetic friction helps to move the slab.
$ \therefore $ Acceleration of slab $ =\frac{40}{m{{} _{B}}}=\frac{40}{40}=1m/s^{2} $
