Kinetic Friction and box

  • Thread starter scurry18
  • Start date
  • #1
12
1

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?
 

Answers and Replies

  • #2
159
0
What does the work-energy theorem state?
 
  • #3
90
0
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
159
0
That's correct. So you have the distance over which the force acts. What is the net force in this situation?
 
  • #5
90
0
It is the Work force of 650J - the frictional force of 588J to get 62J?
Kevin
 
  • #6
159
0
That looks right to me.
 
  • #7
90
0
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
 

Related Threads on Kinetic Friction and box

  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
4
Views
6K
  • Last Post
Replies
1
Views
2K
Replies
6
Views
2K
Replies
3
Views
17K
Replies
2
Views
4K
  • Last Post
Replies
1
Views
3K
Replies
1
Views
4K
Top