Recent content by amelia2222

I think I do, that's the equation that the net work equals the change in kinetic energy correct? If that applies in this case, does that mean that I'd use this equation for this problem: W=KEf-KEi Would the equation be set up like this: 4500J=2(1/2(100)v^2)-1/2((100)v^2) Thanks for the guidance!

Thanks so much! I edited the section with my work in it, and I think it fits the guidelines now? I'm not quite sure

Here's my list of variables and things to account for: m=100kg Wnc=5000J Wfriction=-500J -Kinetic energy will be doubled (though I don't know how that plays into it exactly) -I don't think there's any PE because it's on level ground My idea of what the equation might be: Wnc +1/2mv^2initial =...