A single, non-constant force acts in the +x direction on an 3.81 kg object that is constrained to move along the x axis. As a result the object's position as a function of time is:

x(t) = A+Bt^2+Ct^4, A = 7.88m, B = 4.07m/s^2, C=0.478 m/s^4

How much work is done by this force from t = 0s to t = 2.59s?

Question

A single, non-constant force acts in the +x direction on an 3.81 kg object that is constrained to move along the x axis. As a result the object's position as a function of time is:

x(t) = A+Bt^2+Ct^4, A = 7.88m, B = 4.07m/s^2, C=0.478 m/s^4

How much work is done by this force from t = 0s to t = 2.59s?

anonymous · Answer

@Kainui

anonymous · Answer

So from my understanding we need to get the velocity and acceleration right? So then F = ma, and W = integral F(x) dx with the limits.

anonymous · Answer

I guess basically my question is the mass that trips me out lol.

kainui · Answer

The mass is just 3.81kg isn't it? I'm not sure I understand

anonymous · Answer

Oh I didn't even notice...wow.

anonymous · Answer

Well while you're here check out my integral I guess, 

$$W = \int\limits_{0}^{2.59} (2B+12Ct^2)m dt$$

kainui · Answer

How did dx become dt?

anonymous · Answer

Oh I'm missing the velocity, dx = vdt so I should edit the integral and multiply it by (2Bt+4Ct^3)

kainui · Answer

Yeah right on, that sounds good to me. =)

anonymous · Answer

Alright, thanks, good thing I made sure haha.