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Ilia_Sergeevich [38]
3 years ago
8

HELP PLSSS

Physics
1 answer:
Andrews [41]3 years ago
8 0

Given :

A 13.3 kg box sliding across the ground  decelerates at 2.42 m/s².

To Find :

The coefficient of kinetic friction.

Solution :

Frictional force applied to the box is :

f = ma    ....1)

Also, force of friction is given by :

f = \mu mg  ....2)

Equating equation 1) and 2), we get :

\mu mg = ma\\\\\mu = \dfrac{a}{g}\\\\\mu = \dfrac{2.42}{9.8}\\\\\mu = 0.247

Therefore, the coefficient of kinetic friction is 0.247 .

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