# Kinetic Friction and box

1. Oct 7, 2008

### scurry18

1. The problem statement, all variables and given/known data
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.

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

3. 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?

2. Oct 7, 2008

### Mattowander

What does the work-energy theorem state?

3. Oct 7, 2008

### 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

4. Oct 7, 2008

### Mattowander

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

5. Oct 7, 2008

### Husker70

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

6. Oct 7, 2008

### Mattowander

That looks right to me.

7. Oct 7, 2008

### 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