Question

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?

Answer 1

Answer:

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:

$\sum F = ma$

where

m is the mass

a is the acceleration

In this problem, we have:

- The net force can be written as

$\sum F = F - F_f$

where

F = 40 N is the horizontal forward push

$F_f$ is the force of friction

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

$a=0$

By combining all the equations together, we find:

$F-F_f=0\\F_f = F = 40 N$

So, the force of friction is 40 N.