Work and Potential Energy Question

[Arbitrary]

Homework Statement

A 2-kg particle is acted upon by a force given by F = -100x, where F is in Newtons, while being influenced by a constant nonconservative force of 50 N. The object is released from rest at x = 5.0 m. Estimate how far the particles moves before it comes to a stop.

Homework Equations

W = $$\Delta$$K + $$\Delta$$U + Q
dW = F dx

The Attempt at a Solution

Since velocity = 0 when the object stops, I did:
W = 50*x
$$\int$$ -100x dx = 50*x

The boundaries for integration are from 5 to x. Which leads to:

-50x^2 + 50(5)^2 = 50*x

When I solve for x, though, the answer isn't right, so obviously I did something wrong. Any hinters on what? I have a feeling it's because they're asking for how far the particle moved and not just the x location, but how would I go about finding that?

 

[Dick]

F=-100x looks like the force from a harmonic oscillator. This gives it an initial potential energy that you can calculate. I'm guessing the nonconservative force you are talking about is like friction which always acts opposite to the direction of motion. If it travels a distance d, then the energy it loses to the nonconservative force is 50N*d, right? So if you equate PE to 50N*d, you'll get something. Why don't you expect it to be exact? That's a question for you to answer.

I think you have a feeling why your attempt is wrong. The x on the left side of you equation isn't a distance, it's a coordinate (and it's also a dummy variable in an integration!), the one on the right side is. Notice the question asks for an 'estimate', not an exact solution.

 

[m2003]

Your approach is on the right track, but there are a couple of errors in your calculations. First, the formula for work is not W = 50*x, but rather W = F*d, where d is the distance over which the force is applied. In this case, the force is applied over a distance of x, so the work done by the nonconservative force is W = 50*x.

Next, to calculate the work done by the conservative force, you need to use the formula dW = F dx, but you also need to take into account the fact that the force is negative. So the work done by the conservative force is actually -100x dx.

Now, to find the distance the particle moves before coming to a stop, we can set the total work done equal to the change in potential energy, since the change in kinetic energy is zero (since the particle starts and ends at rest). So we have:

W = \DeltaU
50*x + (-100x dx) = \DeltaU

To solve for x, we need to integrate both sides with respect to x. The integral of 50*x is 25x^2, and the integral of -100x is -50x^2/2, so we have:

25x^2 - 50x^2/2 = \DeltaU

Simplifying, we get:

-25x^2/2 = \DeltaU

Now, we know that the particle starts at x = 5 and ends at x = 0 (since it comes to a stop), so \DeltaU = U(0) - U(5). Plugging this into our equation, we get:

-25x^2/2 = U(0) - U(5)

To find U(0) and U(5), we can use the formula for potential energy:

U(x) = -\int F(x) dx

Plugging in the given force, F = -100x, we get:

U(x) = 50x^2 + C

To find the constant C, we can use the fact that U(5) = 0, since the particle starts at rest at x = 5. So we have:

0 = 50(5)^2 + C
C = -1250

Now we can plug this back into our equation for U(x), and we get:

-25x^