Question

A marble block of mass 2 kg lying on ice when given a velocity of 6 m/s is stopped by friction in 10 s. then the coefficient of friction is

• A. 0.02
• B. 0.03
• C. 0.06
• D. 0.01

Answer 1

Answer: (C)

0.06

Using equation of motion,

$v=u+at$
$0=6+a\left(10\right)$

$⟹a=-0.6m/s^2$

This retardation is caused by a force of friction=$m\left|a\right|$

$=1.2N$

$⟹\mu N=1.2$

$⟹\mu mg=1.2$
$⟹\mu =\frac{1.2}{2\times 10}$$=0.06$