Question

A force of 98N is just able to move a block of mass 20 kg on a rough horizontal surface. Calculate the coefficient of friction and the angle of friction . ( g = 9.8 m/sec^2)

Answer 1

U=0.5
Angle=cos^-1 (2÷(5)^.5)
Hope so

Answer 2

Answer:The coefficient of friction is $\mu = \dfrac{1}{2} = 0.5$ and the angle of friction is $\theta = 26.57^{0}$

Explanation:

Given that,

Force F = 98 N

Mass m = 20 kg

We know that,

The frictional force is the product of the coefficient of friction, mass and acceleration due to gravity.

The coefficient of friction is

$F = \mu mg$

$98 = \mu \times20\times 9.8$

$\mu = \dfrac{1}{2} = 0.5$

The angle of friction is

$\theta = tan^{-1}\mu$

$\theta=tan^{-1}\dfrac{1}{2}$

$\theta = 26.57^{0}$

Hence, The coefficient of friction is $\mu = \dfrac{1}{2} = 0.5$ and the angle of friction is $\theta = 26.57^{0}$