Laws Of Motion Question 398

Question: A fireman of mass 60 kg slides down a pole. He is pressing the pole with a force of 600 N. The coefficient of friction between the hands and the pole is 0.5. With what acceleration will the fireman slide down?

Options:

A) $ 1m/s^{2} $

B) $ 2.5m/s^{2} $

C) $ 10m/s^{2} $

D) $ 5m/s^{2} $

Show Answer

Answer:

Correct Answer: D

Solution:

Friction =$ \mu R=0.5\times 600=300N $

Weight = 600 N

$ ma=W-F\Rightarrow a=\frac{W-F}{m}=\frac{600-300}{60} $

$ a=5m/s^{2} $



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