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} $
BETA
