Question

A block of mass 10 kg is kept on a fixed rough(u=0.8) inclined plane of angle of inclination 30 degree the frictional force actong on the block is??

Answer 1

Answer:

The frictional force acting on the block is 49 N.

Explanation:

Given that,

Mass = 10 kg

incline angle = 30°

We need to calculate the friction force

A block is kept on a fixed rough inclined plane that means the block is not motion at that instant so we can not use the formula of kinetic friction and we can not use the formula for static friction because we don't know whether the friction is limiting or not.

So, we use formula of the friction

Using formula of force

$F=mg\sin\theta$

Where, m = mass

g = acceleration due to gravity

Put the value into the formula

$F=10\times9.8\times\sin30^{\circ}$

$F=49\ N$

Hence, The frictional force acting on the block is 49 N.

Answer 2

Answer:

Answer is 50 N

Explanation:

Check the pic attached for the explianation

.....Hope it helps :-)