Question

A piece of stone came to rest after covering a distance
of 20 m over an icy surface. If the initial velocity of the
stone was lm. s-1, what was the coefficient of friction
between the stone and ice?
(0.0026]​

Answer 1

Answer:

here u go...

Explanation:

s=20 m

u=1 m/s

v= 0 m/s

from equations of motion,

v^2-u^2= 2*a*s

-1= 2 * a * 20

a=-1/40

a=-0.025

(co-efficient of friction)*g=-a

(co-efficient of friction)=0.025/10

(co-efficient of friction)=0.0025