Finding the Change in Kinetic Energy of a Box on a Rough, Horizontal Floor

[scurry18]

Homework Statement

A 40.0kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 130 N. The coefficient of friction between box and floor is .300.
Find the change in kinetic energy of the box.

Homework Equations

(delta)K=K(final)-K(initial)
K(final)=1/2mv^2

The Attempt at a Solution

Work=650. J
(change)Internal Energy=588 J
Fk=118

How do you find the change in Kinetic energy without the velocity?

 

[Mattowander]

What does the work-energy theorem state?

 

[Husker70]

When work is done on a system and the only change in the system is in its speed,
the work done by the net force equals the change in kinetic energy of the system.

Wnet = Kf - Ki = deltaK

In this situation Ki = 0, Kf = 1/2mv^2

Right?
This is a classmate of mine and we are working on this together.
Kevin

 

[Mattowander]

That's correct. So you have the distance over which the force acts. What is the net force in this situation?

 

[Husker70]

It is the Work force of 650J - the frictional force of 588J to get 62J?
Kevin

 

[Mattowander]

That looks right to me.

 

[Husker70]

Now we will try and work on her other one with the skier that she posted and we
should be done.
Thanks for the help,
Kevin