An applied force varies with position

[SonRuy]

An applied force varies with position according to F = k1 xn− k2, where n = 3,
k1 = 2.4 N/m3, and k2 = 56 N.
How much work is done by this force on an object that moves from xi = 5.13 m to xf = 28.2 m?
Answer in units of kJ

i keep using the integration equation: w = ∫f(x) dx
where k1x^4/4 - 56x is the new equation which i plug 28.2m and 5.13m and subtract
i get 377736.527
but its wrong, help please?

 

[Nugatory]

I haven't checked your arithmetic, but you're attacking the problem in the right way. What's the correct answer that you aren't getting?

 

[ehild]

Answer in units of kJ

i keep using the integration equation: w = ∫f(x) dx
where k1x^4/4 - 56x is the new equation which i plug 28.2m and 5.13m and subtract
i get 377736.527
but its wrong, help please?

What is the unit of your result? You need to give the result in units of kJ.

ehild

 

[Basic_Physics]

About 1515 kJ?

 

[haruspex]

About 1515 kJ?

No, I also get 377736; but that's J, so the answer should be 377.7kJ.

 

[ehild]

No, I also get 377736; but that's J, so the answer should be 377.7kJ.

That is the correct result. Maybe it should be rounded to three digits: 378 kJ.

ehild

 

[Basic_Physics]

Ok. Left the /4 out.