Recent content by zainbo

Thanks everyone, I have it figured out now.

Sorry about my earlier post, I had the equation written wrong on my formula sheet. But okay so now I have -22.4J for work. Ff = μ(m)(g) Ff = μ(0.200)(9.80) Is that on the right track for C? I am confused to what μ is

Update: Now I got 1/2(0.200)(10)^2 - 1/2(0.200)(18)^2 = Ek -22.4 is what I get is that the amount of WORK?

I'm aware I have to use ΔEk formula which is 1/2 mv^2 But which velocity do I use to solve for ΔEk.

ΔEk = W ΔEk = 1/2 (m)(v)^2 1/2(0.200)( ? ) For V should I use CHANGE of velocity so 10-18 or should I use one of the velocities. I think I have it figured out, just that one part is confusing me

W = F (d) I tried solving for F first by using f(t) = m (v) F = m(v)/t F = 0.200(-8)/2.50 F = -0.64 Then I tried to solve for D by multiplying final velocity and time which gave me 25m. So W = (-0.64)(25) = -16J But not to sure if that is correct. For c) I know...

Homework Statement A hockey puck of mass 0.200kg is sliding along a smooth, flat section of ice at 18.0m/s when it encounters a patch of snow. After 2.50s of sliding through the snow it returns to smooth ice, continuing at speed of 10.0m/s. a) What is the change in momentum of the puck...