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) 0.02 (B) 0.03 (C) 0.06 (D) 0.04

Tardigrade Admin · Accepted Answer

Retardation (u/t)=(6/10)=0.6 m/sec2 Frictional force = Î¼ mg = ma ∴ μ=(a/g)=(0.6/10)=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

$0.02$

$0.03$

$0.06$

$0.04$

Retardation $\frac{u}{t} = \frac{6}{10} = 0.6 m / se c^{2}$
Frictional force $= μ m g = ma$
$∴ μ = \frac{a}{g} = \frac{0.6}{10} = 0.06.$

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

0.02

0.03

0.06

0.04