Question

A block of mass 2 kg starts moving when the angle of inclination of the inclined plane is 60o. if the coefficient of kinetic friction is 0.6, the frictional force is

Answer 1

Contact force is the resultant of Frictional force and normal reaction force. The frictional force can be calculated as = 0.5 x mgcos theta. = 0.5 x 2 x 10 x root 3 / 2normal reaction force is Mgcos theta. = 2 x 10 x root 3 / 2  They are at 90 degrees to each other therefore find out their resultant and that shall be the resultant contact force.

Answer 2

Answer:

The frictional force is 6 Newton.

Step-by-step explanation:

Given : A block of mass 2 kg starts moving when the angle of inclination of the inclined plane is 60°. If the coefficient of kinetic friction is 0.6

To find : The frictional force ?

Solution :

The formula of frictional force is $F=\mu_k\times N$

Where,

$N = m\times g\times \cos\theta$

$\mu_k = \text{coefficient of kinetic friction}$

Now, we have given that

$g = 10 m/s^2\\M = 2 \text{ kg}\\\mu_k= 0.6\\\theta = 60^\circ$

Substitute the values in the formula,

$F=\mu_k\times N$

$F=\mu_k\times m\times g\times \cos\theta$

$F=0.6\times 2\times 10\times \cos(60^\circ)$

$F=0.6\times 2\times 10\times \frac{1}{2}$

$F=0.6\times 10\times$

$F=6 N$

The frictional force is 6 Newton.