Laws Of Motion Question 393
Question: A block of mass 10 kg is kept on the fixed incline having coefficient of friction 0.8. Find the force of friction exerted on the block by inclined:
Options:
A) 100 N
B) 64 N
C) 80 N
D) 60 N
Show Answer
Answer:
Correct Answer: D
Solution:
Friction required $ =mg\sin 37{}^\circ $
$ =10\times 10\times \frac{3}{5}=60N $
Maximum friction provided $ =\mu mg\cos 37{}^\circ $
$ =0.8\times 10\times 10\times \frac{4}{5}=64N $
So friction force will be 60 N.
BETA
