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.



NCERT Chapter Video Solution

Dual Pane