What is the Total Impulse Applied to the Box?

[yesiammanu]

Homework Statement

Horizontanl force F is giving by F(x) = (2.00x + 1.00x^2) Newtons, acting on a 12.0kg box which is initially at rest and on a frictionless horizontal surface. Force is applied from x=0 to x=2.50 meters. What's the total impulse the force applies to the box?

Homework Equations

v^2 = sqr(2ax)
Impulse = mv1 - mv0
F = ma

The Attempt at a Solution

I already got the work, which was 11.5 J

2.00(2.50) + 1.00(2.50)^2 N = 11.25 N
Now I must find velocity
F = ma
a = f/m = 11.25 N/ 12kg = .9375 m/s^2

v = sqr(2 * .9375 m/s^2 * 2.50m) = 2.16 m/s
12kg * 2.16 m/s = 25.9 momentum

Since it started at rest, the momentum should be 25.9 . However the answer is 16.6 N * s which I can't understand how to get

Another attempt using joules would be , since Joules is kg * m^2/s^2

11.5 J/ 2.16 m per s = 5.32 momentum which also isn't right. Any help would be appreciated

 

[BruceW]

v = sqr(2 * .9375 m/s^2 * 2.50m) = 2.16 m/s
12kg * 2.16 m/s = 25.9 momentum

This equation is not right because it is not constant acceleration. (it will only work when the acceleration is constant).

 

[BruceW]

Your value for the work is correct. You can use this to find the answer. What is the equation for the kinetic energy?