Work Question - 4.31 kJ Answer

[jj8890]

[SOLVED] Work question please help

Homework Statement

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.

Homework Equations

Integral from x_i to x_f of the function

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.

 

[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

 

[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?

 

[jj8890]

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

=4.72573 kJ

 

[jj8890]

Can I get a confirmation that my work looks correct?

 

[blochwave]

Yah, as long as you punched in numbers right

 

[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.