Question 151: Laws of Motion

Question

A marble block 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

Accepted Answer

The retarding force is created by the frictional force.
Given, $$u = 6m/s,\,v = 0,\,t = 10s,$$
Retardation$$(a) =  - {f \over m} = {{ - umg} \over m} =  - \mu g =  - 10\mu $$
$$v = u + at$$
$$0 = 6 - 10\mu  	imes 10$$
$$	herefore$$ $$\mu  = 0.06$$

Answer

$$0.02$$

Answer

$$0.03$$

Answer

$$0.04$$

Answer

$$0.06$$

Laws of Motion

A marble block 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

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