# Force Equations and Final Speed

1. Oct 18, 2007

### odie5533

1. The problem statement, all variables and given/known data
A 7.0-kg rock is subject to a variable force given by the equation
$$F(x) = 6.0N - (2.0N/m)x + (6.0N/m^2)x^2$$
If the rock initially is at rest at the origin, find its speed when it has moved 9.0m.

3. The attempt at a solution
$$\sum F_{x} = 6 - 2x + 6x^2$$
$$ma=6 - 2x + 6x^2$$
$$a = \frac{6}{7} - \frac{2}{7}x + \frac{6}{7}x^2$$
$$V_{f}^2 = V_{i}^2 + 2*a*x$$
$$V_{f}^2 = 0 + 2(\frac{6}{7} - \frac{2}{7}x + \frac{6}{7}x^2)x$$
$$V_{f}^2 = 2(\frac{6}{7} - \frac{2}{7}(9) + \frac{6}{7}(9)^2)(9)$$
$$V_{f} = 1219m/s$$

That seems waaay too fast, and I think I should have used an integral/derivative somewhere since the force changes as x changes. I'm rather lost on this one.

2. Oct 18, 2007

### Staff: Mentor

This kinematic equation assumes constant acceleration--not a good assumption for this problem.

Hint: How much work is done on the rock?

3. Oct 18, 2007

### odie5533

I think I have it now.
$$W = \int^{9}_{0} 6 - 2x + 6x^2 dx = 1431J$$
$$W = \frac{1}{2}mv_{f}^2 - \frac{1}{2}mv_{0}^2$$
$$1431 = \frac{1}{2}7v_{f}^2$$
$$v_{f} = 20.22m/s$$

Any improvement?

4. Oct 18, 2007

### Staff: Mentor

Looks good to me!

5. Oct 18, 2007

### odie5533

Thanks for the help. I've never done a problem like that, all the ones I've completed so far have used the kinematics equations. I guess I'll have to learn how to apply the work equations better.