Laws Of Motion Question 118

Question: A marble block of mass 2 kg lying on ice when given a velocity of 6 m/s is stopped by friction in 10s.Then the coefficient of friction is [AIEEE 2003]

Options:

A) 0.01

B) 0.02

C) 0.03

D) 0.06

Show Answer

Answer:

Correct Answer: D

Solution:

$ v=u-at\Rightarrow u-\mu gt=0 $

$ \therefore $ $ \mu =\frac{u}{gt}=\frac{6}{10\times 10}=0.06 $



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