PDA

View Full Version : 2 boxes with energy attempt


Johnny_99
Jan4-07, 02:09 PM
1. The problem statement, all variables and given/known data
Two boxes are connected over a pulley and hel AT REST. Box A has a mass of 15kg and box B has a mass of 12kg. If the bottom of box A is originally 85cm above the floor, with what speed will it contact the floor when the system is released? Use conservation of energy and assume that friction is negligible. (Hint: Think abou the total energy of the system before and after the block A is released)


2. Relevant equations
ET= mgh + 1/2mv squared
Ek=1/2mv squared
Ep=mgh


3. The attempt at a solution
Before released: Eta= (15)(9.8)(.85)
Eta= 124.95J
After released: Etb= mgh
Etb= 12(9.8)(.85)

up to this point, the test says I have it write (as seen from the teachers marking)

then this is where i somehow went wrong: 124.95-99.96= Change in Energy
*at this point, the marker said, "B HAS Ek TOO!"

then i did: 24.99= Change in Energy
24.99= 1/2mv squared
24.99= 1/2(15)v squared
3.332m/s=v
3.3m/s=v

This is unfortunately not the correct answer. If anyone can assist me in where i went wrong that would be great!

Dorothy Weglend
Jan4-07, 02:51 PM
I think block B is still moving when A hits the floor. That's probably what your teacher was talking about (there's no floor to stop B).

So

U_A = K_A+K_B+U_B

Dorothy

Johnny_99
Jan4-07, 02:54 PM
what does U stand for Dorothy?

Doc Al
Jan4-07, 03:02 PM
Not only is B moving, but what is the relationship between the speeds of A and B? (They are connected by a rope!)

A useful way to view conservation of energy is in terms of changes:
\Delta{E} = \Delta{KE} + \Delta{PE} = 0

Remember: You want the change in energy of the system--both boxes.

(Dorothy is using U to stand for potential energy.)