How Fast Can You Raise a Bucket with 104 W Power?

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To determine the speed at which a bucket can be raised using a power output of 104 W and a mass of 6.00 kg, the relevant equations for power and work are discussed. The work done is primarily a change in potential energy, calculated as W = mgΔh. The relationship between power, force, and velocity is emphasized, suggesting that using the equation P = Force * velocity simplifies the problem. The discussion highlights the importance of understanding energy changes and the role of displacement in calculating speed. Ultimately, the participants agree that this approach is more straightforward for solving the problem.
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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?
 
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Just be careful, Wnet = deltaKE.
I will eventually learn latex.:approve:
 
Theres another equation for power: Power = Force * velocity
 
PhyzicsOfHockey said:

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?

W = \Delta K + \Delta U

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 W = mg\Delta h

where h = vertical displacement


ALso what is displacement divded by time??
 
turdferguson said:
Theres another equation for power: Power = Force * velocity

this would be the easier way :biggrin:
 
turdferguson said:
Theres another equation for power: Power = Force * velocity

yep that will do it, thanks.
 
easier yes, better for learning...:confused:
 
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