First slide

Third law

Question

A boy is hanging from the free end of a massless rope of length 40 m. The mass of the boy is 20 kg and the maximum tension that the rope can bear is 300 N. Then starting from rest, the minimum time in which the boy can climb the other end of the rope is $(g = 10 m / s^{2})$

Moderate

A

2 s
B

4 s
C

6 s
D

8 s

Solution

$m a x i m u m t e n s i o n T = m (g + a) 300 = 20 (10 + a) a = 5 m / s^{2} t i m e t a k e n t = \sqrt{\frac{2 S}{a}} = \sqrt{\frac{2 (40)}{5}} = 4 s$

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