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} $



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक