Force Equations and Final Speed

[odie5533]

Homework Statement

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.

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.

 

[Doc Al]

V_{f}^2 = V_{i}^2 + 2*a*x

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

Hint: How much work is done on the rock?

 

[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?

 

[Doc Al]

Looks good to me!

 

[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.