JEE MAIN - Physics (2003 - No. 59)
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
$$0.02$$
$$0.03$$
$$0.04$$
$$0.06$$
คำอธิบาย
The retarding force is created by the frictional force.
Given, $$u = 6m/s,\,v = 0,\,t = 10s,$$
Retardation$$(a) = - {f \over m} = {{ - umg} \over m} = - \mu g = - 10\mu $$
$$v = u + at$$
$$0 = 6 - 10\mu \times 10$$
$$\therefore$$ $$\mu = 0.06$$
Given, $$u = 6m/s,\,v = 0,\,t = 10s,$$
Retardation$$(a) = - {f \over m} = {{ - umg} \over m} = - \mu g = - 10\mu $$
$$v = u + at$$
$$0 = 6 - 10\mu \times 10$$
$$\therefore$$ $$\mu = 0.06$$