Recent content by Baran

Thank you for your help, as well as all the other people who helped me solve this question. I really do appreciate all the help that I've received!

I think it is Ek=1/2(m)(v)^2 so would I sub in the values like this? 3.1x10^6 J=1/2(750kg)(v)^2 then rearranged I think it would be v=2Ek/m (the right hand side should have a square root, but I don't know how to make that symbol)Also sorry about the late reply, I was busy with some other IRL...

I'm not sure how to

The total distance, so W=402mx7700N =3.1x10^6 J

Would the total work be +7 700N?

I guess that would be the formulas, W=F(Total Distance) Ek=1/2(m)(v)^2 Wnet=[(m)(vf)^2]/2 - [(m)(vi)^2]/2 Em intial=Em final Eg=mghThat's everything that I could think of.

I don't know what to do.

Thanks for all your help!

I think I got it: a= vf-vi/t a= 0 m/s- 45 m/s [N] / 0.10 s a= -450 m/s^2 [N] a= 450m/s (this is the paintball) Fnet= ma Fnet= (0.15 kg) x(450 m/s^2 ) Fnet= 68 N (this is the paintball gun) Fnet= ma a= Fnet/m a= 68 N / 5.5 kg a= 12 m/s^2 can you check if I did this right please...

I'm confused, in order to find the net force you would need acceleration, wouldn't you?

I've been stuck on this for sometime now, and am unsure how to approach this question.