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 $
BETA
