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

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) 0.01 (B) 0.02 (C) 0.03 (D) 0.06

Tardigrade Admin · Accepted Answer

v=u-a t 0=u-μ g t μ=(u/g t)=(6/10 × 10)=0.06

Q. A marble block of mass $2 k g$ lying on ice when given a velocity of $6 m / s$ is stopped by friction in $10 s$ . Then the coefficient of friction is

0.01

0.02

0.03

0.06

$v = u - a t$
$0 = u - μg t$
$μ = \frac{u}{g t} = \frac{6}{10 \times 10} = 0.06$

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