SATHEE: Laws Of Motion Question 393

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 $= m g \sin 37^{\circ}$

$= 10 \times 10 \times \frac{3}{5} = 60 N$

Maximum friction provided $= μ m g \cos 37^{\circ}$

$= 0.8 \times 10 \times 10 \times \frac{4}{5} = 64 N$

So friction force will be 60 N.

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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:

Answer:

Solution:

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