The force on a crate increases uniformly from 30.0N to 40.0N as the crate is pushed from the 0.50m mark to the 0.70m mark. What is the total amount of work done.

Question

The force on a crate increases uniformly from 30.0N to 40.0N as the crate is pushed from the 0.50m mark to the 0.70m mark.  What is the total amount of work done.

mathmate · Answer

The crate moved horizontally or vertically?

anonymous · Answer

horizontally

mathmate · Answer

The force F(x) as a function of x is
50x+5 N (from 30 at x=0.5 to 40 at x=0.7)
So work done, $$W = \int\limits_{0.5}^{0.7}F(x)dx = \int\limits_{0.5}^{0.7}(50x+5)dx =[25x^2+5x]_{0.5}^{0.7}$$

anonymous · Answer

answer?

mathmate · Answer

Are you able to evaluate the integral on the right hand side?

anonymous · Answer

Yes.  Not really supposed to be using calculus to do this problem, though.

mathmate · Answer

In that case, you just take the average force and multiply by the distance.

anonymous · Answer

Use W=Fdcos(theta)

mathmate · Answer

Cos(theta)=1, assuming force is acting in the direction of the displacement.
F varies with position uniformly (linearly), so the product
Fav.D is the same as integral of F(x)dx

anonymous · Answer

Indeed.  Got it now....thanks!!!

mathmate · Answer

Great!
You're welcome!   :)