Rotational Motion: Energy Conservation

1. Nov 20, 2012

SpringWater

1. The problem statement, all variables and given/known data
the sliding block has a mass of .850kg the counterweight has a mass of .420kg , and the pulley has a mass of .350kg with outer and inner radius .030m, .020m, the coefficient of friction is .250. the pulley turns the axle. the light cord does not stretch and does not slip on the pulley. The block has velocity of .820m/s towards the pulley when passes by photogate . use the energy method to predict the velocity it moved to a second photogate. .700m away?

2. Relevant equations

3. The attempt at a solution

I understand how to set up the problem using
ΔKE (initial)= ΔKE (final)

however i am wondering if this is possible...
Ke(m1)+KE(m2)+KE(rot)=(uk)(m1)(g)(X)

(V)^(2)*((.5)(m1)+(.5)(m2)+(.5)(I)(1/(r)^(2)))=(μk)(m1)(g)(X)

Where I=2.275E-4 & r=.02m & X=.7m By doing this i get V=1.5855 m/s

This is the correct answer! Did i get the correct answer just by coincidence or is this actually possible????

2. Nov 20, 2012

stonecoldgen

Use the fact that W=ΔEk

So you get:

h=separation between photogates
a=acceleration of the system
m=mass of the block moving through the photogate
M=mass of the other block
v=velocity of the system

mah=0.5m(v22-v12)

So the only thing you need to know in order to solve for v2 in the equation above is a, which you can get using rotational formulas and Newton's laws.