Question

A ball rolled on ice with a velocity of 14ms-1 comes to rest after travelling 40m. find the ciefficient of friction. Given g=9.8ms-2​

Answer 1

Explanation:

u = 14 m/s

v = 0 m/s

S = 40 m

Using 2nd Equation of Motion,

${v}^{2} = {u}^{2} + 2as$

$a = \frac{ {v}^{2} - {u}^{2} }{2s}$

$= \frac{ - 196}{80} = -2.45 \: m {s}^{ - 2}$ ( - sign means that it is retardation )

Acceleration of the ball = friction coefficient × Acceleration due to gravity

2.45 = friction coefficient × 9.8

friction coefficient = 0.25

I hope it helps you. If you have any doubts, then don't hesitate to ask.