Laws Of Motion Question 57
Question: A 40 kg slab rests on a frictionless floor . 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 [NCERT 1982]
Options:
A) $ 0.98m/s^{2} $
B)$ 1.47m/s^{2} $
C) $ 1.52m/s^{2} $
D) $ 6.1m/s^{2} $
Show Answer
Answer:
Correct Answer: A
Solution:
Limiting friction between block and slab$ ={{\mu } _{s}}m _{A}g $
$ =0.6\times 10\times 9.8=58.8N $
But applied force on block A is 100 N. So the block will slip over a slab.
Now kinetic friction works between block and slab $ F _{k}={{\mu } _{k}}m _{A}g $
$ =0.4\times 10\times 9.8=39.2\ N $
This kinetic friction helps to move the slab $ \therefore $ Acceleration of slab$ =\frac{39.2}{m _{B}}=\frac{39.2}{40}=0.98\ m/s^{2} $
