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PolarNik [594]
3 years ago
14

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?
Chemistry
1 answer:
irga5000 [103]3 years ago
5 0

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.

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