A 10.0 kg crate is pushed with a horizontal force of 40 N. The crate moves at a constant velocity of 3.0 m/s. What is the value of the friction force on the crate?

40 N

Explanation:

Newton's second law of motion states that the net force acting on the crate is equal to the product between its mass and its acceleration:

where

m is the mass

a is the acceleration

In this problem, we have:

- The net force can be written as

where

F = 40 N is the horizontal forward push

is the force of friction

Also, we know that the crate moves at constant velocity, so its acceleration is zero:

By combining all the equations together, we find:

So, the force of friction is 40 N.

i think   the answer is c hope i'm right