# Finding kinetic energy and initial velocity of a cart over time

• amelia2222

#### amelia2222

Homework Statement
During a period of time, a 100kg cart’s KE doubles. During that time, friction does -500J of work and the vehicle that is pulling the cart does 5000J of work on the cart. This scenario occurred on level ground. Find the final kinetic energy of the cart and the cart’s initial velocity
Relevant Equations
Worknc + KEinitial = KEfinal +Workfriction
Energy(initial) = Energy(final)
-I think there’s a equation for the initial velocity, but 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 = 2(1/2mv^2final) + Wfriction
5000+ 1/2(100)v^2 = 2(1/2(100)v^2) + 500

-From here, I'm not quite sure how to find both velocities and subsequently the final KE

Last edited:

Hello @amelia2222 , ##\qquad##!​
That's not enough to get started. It probably isn't enough for PF to allow assistance (see guidelines).
Perhaps you want to set up a list of variables involved (##t_0,\ t_1,\ v_0,\ v_1, \ ## etc.) where the 0 is for initial and the 1 for final.
And link them to the ones you have already (like ##KE_0 = {1\over 2} mv_0^2\ ## etc).

Don't forget the friction loss.

##\ ##

• berkeman
Hello @amelia2222 , ##\qquad##!​
That's not enough to get started. It probably isn't enough for PF to allow assistance (see guidelines).
Perhaps you want to set up a list of variables involved (##t_0,\ t_1,\ v_0,\ v_1, \ ## etc.) where the 0 is for initial and the 1 for final.
And link them to the ones you have already (like ##KE_0 = {1\over 2} mv_0^2\ ## etc).

Don't forget the friction loss.

##\ ##
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

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
Do you know the Work - Energy Theorem? (That's a hyphen, not a minus sign.)

I agree with @SammyS, you need to review and understand the Work - Energy theorem. In addition, you need to understand what Wnc is all about, how it differs (if it does) from Wfriction and where the 5000 J fits in all this.

Do you know the Work - Energy Theorem? (That's a hyphen, not a minus sign.)
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!

• BvU
That would be correct.