Work, Elastic and Kinetic Energy

  • #1
KMjuniormint5
67
0

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?

Homework Equations


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

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

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

  • #2
alphysicist
Homework Helper
2,238
2
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?
 
  • #3
KMjuniormint5
67
0
w = F*d where d is the distance but it is only in the x direction. . let me try that
 
  • #4
KMjuniormint5
67
0
so (-6)(4^2)-(-6)(3^2) = .5(2.4)(v^2)-.5(2.4)(8^2) and solve for v?
 
  • #5
alphysicist
Homework Helper
2,238
2
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.
 
  • #6
KMjuniormint5
67
0
take the intergal of F*d? . . .
 
  • #7
alphysicist
Homework Helper
2,238
2
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]
 
  • #8
KMjuniormint5
67
0
so the force is going to .5*(-6)(x)^2 with xi = 3 and xf = 4
 
  • #9
alphysicist
Homework Helper
2,238
2
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.
 
  • #10
KMjuniormint5
67
0
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!
 

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