Question

A marble 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 ? Please explain brefiely.

Answer 1

Mass of marble = 2 kg

Weight of Marble = 2 × 9.8 = 19.6 N

Normal force = 19.6 N

Friction = μN

Initial velocity (u) =  6 m/s

Final velocity (v) = 0 m/s

Time taken = 10 s

Let the acceleration be a.

$v = u + at$

$a = \frac{6-0}{10}$

a = 0.6 m/s^2

Also, a = μg

0.6 = μ × 9.8

μ  = 0.061

Hence, the coefficient of friction = 0.061

Answer 2

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given :

Weight of Marble = mg = 2 × 9.8 = 19.6 N

Normal force = 19.6 N

Friction = μN

initial velocity ,u =  6 m/s

Final velocity ,v = 0 m/s

Time taken = 10 s

Let the acceleration be a.

v = u + at

a = 6-10a.

[ acceleration is negative since it's deceleration]

a = 0.6 m/s^2

Also, a = μg

0.6 = μ × 9.8

μ  = 0.061

Hence, the coefficient of friction = 0.061

I hope it helps you

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