# Work Energy Pulley

1. Nov 13, 2011

### nathan8

1. The problem statement, all variables and given/known data

20 Kg box is raised 0.75m by a constant 25N force applied at the end of the cord if the pulleys are massless, determine work done on the crate. What is the power delivered at the end of the cord at this point?

Identical pulley system
http://www.oldschool.com.sg/modpub/20204469354a2bdb19c0115 [Broken]

2. Relevant equations

W=Fd I found that W=20(9.81)*0.75 = 147.15 J

P=Fv

3. The attempt at a solution

I am having trouble with the power portion, I am not sure how to find the correct velocity to apply also confused as to which force the 25N or the 20(9.81)

Thanks

Last edited by a moderator: May 5, 2017
2. Nov 14, 2011

### JHamm

Power is work divided by time, for the time consider that you can find acceleration from $F = ma$ and since you know the total displacement the time can be derived from a kinematics equation.

3. Nov 15, 2011

### nathan8

Would Power = F*v work here?

Then just find the velocity with 2Va = -Vb

And use the 25N force times this velocity?

Am I on the right track with the work done on the crate, being the change in potential energy?