Calculating Work Done by a Variable Force

[Sarah Kenney]

Homework Statement

The force exerted on a certain object varies with the object's position according to the function Fx(x)=ax^2+bx^3 where a = 3 N/m^2 and b = -0.50 N/m^3 .

What is the work done on the object by this force as the object moves from x=−0.40 m to x = 2.0 m?

Homework Equations

W=F*x

The Attempt at a Solution

Ok, I know that the answer is 6.1 J, I just don't know how they arrived at that conclusion. My attempt:
Fx(x)=(3)(-0.4)^2+(-0.5)(-0.4)^3 =0.512
Fx(x)=(3)((2)^2+(-0.5)(2)^3=8

Then I subtracted those two numbers to get 8-0.512=7.488 J
That's kind of close, but not quite. What am I doing wrong?

 

[Doc Al]

Ok, I know that the answer is 6.1 J, I just don't know how they arrived at that conclusion. My attempt:
Fx(x)=(3)(-0.4)^2+(-0.5)(-0.4)^3 =0.512
Fx(x)=(3)((2)^2+(-0.5)(2)^3=8

What you've done is calculate the value of the force at the two endpoints. Fx(x) is the value of the force as a function of x.

What's the definition of work? Hint: Since the force varies with position, you'll need a bit of calculus to find the work done.

 

[Sarah Kenney]

It makes sense now. Thank you so much!