1. Feb 23, 2008

jj8890

1. The problem statement, all variables and given/known data
An applied force varies with position according to F = k_1x^n − k_2 , where n = 3,
k_1 = 5.6 N/m, and k_2 = 68 N. How much work is done by this force on an object that moves from x_i = 4.56 m to x_f = 7.88 m? Answer in units of kJ.

2. Relevant equations

Integral from x_i to x_f of the function

3. The attempt at a solution
I attempted to take the integral from 4.56 to 7.99 and got 4310.17 but this is not right. I then thought that I had not converted to kJ so i divided by 1000 and got 4.31017 but that is also wrong. I would appreciate any help.

2. Feb 23, 2008

blochwave

I'm whipping this up on window's calculator, but it doesn't look like you got the right result from integrating

Show your work for the actual integration and we'll make sure you did it right

3. Feb 23, 2008

jj8890

Okay...

The antiderivative of 5.6x^3-68 would be 1.4x^4-68x which is then:

=[(1.4(7.99)^4)-(68*7.99)]-[(1.4(4.56)^4)-(68*4.56)]
=5020.973-295.243
=4725.73

Hmm....got a different answer this time...must have done something wrong the first..does this look right now?

4. Feb 23, 2008

jj8890

Then to make it into kJ I would just divide by 1000 right?

=4.72573 kJ

5. Feb 24, 2008

jj8890

Can I get a confirmation that my work looks correct?

6. Feb 24, 2008

blochwave

Yah, as long as you punched in numbers right

7. Feb 24, 2008

jj8890

i apologize I made a mistake in the question and put 7.99. It was actually 7.88 like stated at first. I re-integrated and got 4.56692 kJ and this was correct. Thanks for your help.