Simple problem in calculating kinetic energy

[Laurlaur790]

Homework Statement

If you push a crate horizontally with a force of 100 N across a 10 meter factory floor, and the friction between the crate the the floor is a steady 70 N, how much kinetic energy is gained by the crate?

Homework Equations

KE=1/2mv²
Work=change in KE
Work=force*distance traveled

The Attempt at a Solution

Im not sure how to get the net force. If all you have to do is add the friciton force and force from push than the solution should be:
+100-(-70)=170 N
170 N(10 meters)=1700 J

is this correct?

 

[tiny-tim]

Welcome to PF!

Hi Laurlaur790! Welcome to PF!

If you push a crate horizontally with a force of 100 N across a 10 meter factory floor, and the friction between the crate the the floor is a steady 70 N, how much kinetic energy is gained by the crate?
…
If all you have to do is add the friciton force and force from push than the solution should be:
+100-(-70)=170 N
170 N(10 meters)=1700 J

is this correct?

Nooo … what effect do you think friction has on KE?

 

[Laurlaur790]

I think that the more friction there is, the less kinetic energy there is.

 

[tiny-tim]

I think that the more friction there is, the less kinetic energy there is.

That's right!

So what effect does that have on the numbers?