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Question

A body moving along the ground with a velocity of $$14 m/s $$ comes to rest due to friction after travelling a distance of $$50 cm $$ Determine the coefficient of friction between the body and the ground $$ g = 9.8 m/s^2 $$


Solution

By applying the law of motion

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

$${\left( {14} \right)^2} =  - 2 \times 0.5 \times a$$

$$a =  - 196\;m/{s^2}$$

The coefficient of friction is given as

$$\mu  = \dfrac{{ma}}{{mg}}$$

The coefficient of friction is$$20$$

 


Physics

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