# Momentum-Work Question

1. Jan 30, 2010

### Zula110100100

I am confused about the KE = 1/2mv2

I was able to get v2=2as from it, by removing mass from each side you get as = 1/2v2 then rearrange, but I was trying to relate momentum and work in my head, and since p = mv than KE = 1/2 p * v

but starting with Work = F * s

and since s = vt

Work = F*vt

then solve for time t = W/Fv

and p = mv
v = at
so p = mat

so p = Ft(assuming no original momentum)

substitute for t and

p = FW/Fv

p = W/v

W = pv

W = mv2

So what am I doing wrong? becuase mv2 does not equals 1/2mv2

I was thinking maybe because I am not using t0 and v0, but it doesn't seem like thats it, it seems the only way is with the 2as but it should tie back in somewhere....

and even working backwards from KE = 1/2mv2
Fvt = 1/2pv
devide by V and multiply by 2
and p = 2 F*t
p = 2 m*a*t
p = 2 mv

2. Jan 30, 2010

### Staff: Mentor

Since the speed is not constant, you need to use the average velocity when calculating distance. In the KE formula, the v is the final speed.

3. Jan 30, 2010

### Zula110100100

Oh, okay, so it kinda has to do with not using v0[\SUB]...and the like

4. Jan 30, 2010

### Staff: Mentor

Right. For the simple case of a constant force applied to an object initially at rest (v0 = 0), we'll have vave = v/2. So:
W = F*s = F*v/2*t = (F*t)*v/2 = (mv)*v/2 = 1/2mv2. As expected.