Laws Of Motion Question 123
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 (g = 10 m/s2)[Pb. PMT 2002]
Options:
A) 1 m/s2
B) 2.5 m/s2
C) 10 m/s2
D) 5 m/s2
Show Answer
Answer:
Correct Answer: D
Solution:
Net downward acceleration $ =\frac{\text{Weight-Friction force}}{m} $
$ =\frac{(mg-\mu \ R)}{m} $
$ =\frac{60\times 10-0.5\times 600}{60} $
$ =\frac{300}{60}=5\ m/s^{2} $
BETA
