SATHEE: Laws Of Motion Question 398

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) $1 m / s^{2}$

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

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

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

Show Answer

Answer:

Correct Answer: D

Solution:

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

Weight = 600 N

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

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

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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:

Answer:

Solution:

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