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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