• Support PF! Buy your school textbooks, materials and every day products Here!

Work, Elastic and Kinetic Energy

1. Homework Statement

The only force acting on a 2.4 kg body as it moves along the positive x axis has an x component Fx = - 6x N, where x is in meters. The velocity of the body at x = 3.0 m is 8.0 m/s.
- What is the velocity of the body at x = 4.0 m?

2. Homework Equations
W = [tex]\Delta[/tex]0.5*m*v^2
that is for kinetic

W = [tex]\Delta[/tex]0.5kx^2

3. The Attempt at a Solution

I set the two equations equal to each other since no outside force is acting:
[tex]\Delta[/tex]0.5*m*v^2 = -[tex]\Delta[/tex]0.5kx^2 . . .
so . . . .
.5(2.4)(v^2) - .5(2.4)(8^2) = - .5(6)(3^2) - .5(6)(4^2)

is that right for the spring constant or is k = 6*x . . . .?
 

Answers and Replies

alphysicist
Homework Helper
2,238
1
Hi KMjuniormint5,

I don't see a spring in this problem so you don't need the spring expression. You have the formula that shows the effect of the work on the kinetic energy. Now what is the formula for the work done by a force?
 
w = F*d where d is the distance but it is only in the x direction. . let me try that
 
so (-6)(4^2)-(-6)(3^2) = .5(2.4)(v^2)-.5(2.4)(8^2) and solve for v?
 
alphysicist
Homework Helper
2,238
1
w = F*d where d is the distance but it is only in the x direction. . let me try that
That would apply for a constant force where the force and motion are in the same direction.

However, this force varies with x, so you need to use the integral form.
 
take the intergal of F*d? . . .
 
alphysicist
Homework Helper
2,238
1
For a particle moving in the x direction, the work done is:

[tex]
W \equiv \int\limits_{x_i}^{x_f} F_x \ dx
[/tex]
 
so the force is going to .5*(-6)(x)^2 with xi = 3 and xf = 4
 
alphysicist
Homework Helper
2,238
1
so the force is going to .5*(-6)(x)^2 with xi = 3 and xf = 4
(You said force, but I'm sure you meant work.) Yes, that will be the work done from xi to xf once you evaluate it with the limits. Once you have that you can relate it to the change in kinetic energy.

On looking back at this thread, I have to apologize. That force is the force of a spring, of course, so what we are doing will lead back to something like what you were doing in your first thread.

The work for a spring is:

[tex]
W = -\frac{1}{2} k x_f^2 + \frac{1}{2}k x_i^2
[/tex]

which is the kind of idea you had for the work done originally. You did have a sign error in that equation, but if you use the above you should get the correct sign.
 
ok ya. . .i was just confused about the spring part because they said to use the spring equation. . .all i had wrong was the sign. . . thanks a lot!
 

Related Threads for: Work, Elastic and Kinetic Energy

Replies
4
Views
4K
Replies
6
Views
6K
Replies
3
Views
2K
Replies
22
Views
2K
Replies
5
Views
3K
Replies
5
Views
682
Replies
6
Views
3K
  • Last Post
Replies
6
Views
6K
Top