Raising a Bucket out of a Well

  • #1

Homework Statement



You raise a bucket of water from the bottom of a deep well. If your power output is 104 W, and the mass of the bucket and the water in it is 6.00 kg, with what speed can you raise the bucket? Ignore the weight of the rope.

Homework Equations



P=W/t
W=P*t
W=1/2m*vf^2-1/2m*vi^2

The Attempt at a Solution



P*t=1/2m*vf^2
2*P*t/m=vf^2


I don't know what the time is can this problem be done without it?
 

Answers and Replies

  • #2
340
0
Just be careful, Wnet = [tex]delta[/tex]KE.
I will eventually learn latex.:approve:
 
  • #3
312
0
Theres another equation for power: Power = Force * velocity
 
  • #4
1,444
2

Homework Statement



You raise a bucket of water from the bottom of a deep well. If your power output is 104 W, and the mass of the bucket and the water in it is 6.00 kg, with what speed can you raise the bucket? Ignore the weight of the rope.

Homework Equations



P=W/t
W=P*t
W=1/2m*vf^2-1/2m*vi^2

The Attempt at a Solution



P*t=1/2m*vf^2
2*P*t/m=vf^2


I don't know what the time is can this problem be done without it?
[tex] W = \Delta K + \Delta U [/tex]

where K is the kinetic energy and U is the potential energy
Work is a change in the kinetic energy or a change in the potential energy.
In th problem which energy is changing?? Keep in mind that they have asked what speed at which you can pull it up... which would be constant.

So is there a change in the kinetic energy??

P = W/t
the work done is a change in the potential energy ONLY.
so [itex] W = mg\Delta h [/itex]

where h = vertical displacement


ALso what is displacement divded by time??
 
  • #7
340
0
easier yes, better for learning...:confused:
 

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