Question

a box having 50 kg mass is placed on a rough horizontal surface having mu=0.6. What is the value of force of friction between box and the surface.

Answer 1

Hey there!

$\mu \: = 0.6$
Here, The box is not said to be in motion , So it's resting on a horizontal surface.

The forces acting on it are normal force and frictional force.

We know that,
Coefficient of static friction ( $\mu$ ) = $= \frac{ \: frictional \: force \: }{normal \: force}$

We know that,
Normal force
= mg
= 50 Kg * 9.8m/s²
= 490 N

Now,
Let the force of friction be f.

$\mu \: = \frac{f}{490} \\ \\0.6 \times 490 = f \\ \\ f = 294 N$

Therefore, The force of friction acting on the body is 294 N

Answer 2

Answer:

this will help you.....