Potential Energy

[killa0587]

1. (a) Calculate the change in potential energy of 1.8 kg of water as it passes over Niagara Falls (a vertical descent of 50 m).
______ J

(b) At what rate is gravitational potential energy lost by the water of the Niagara River? The rate of flow is 5.5 106 kg/s.
______GW

(c) If 27% of this energy can be converted into electrical energy, how many households would the electricity supply? (An average household uses an average electrical power of about 1 kW.)
______Households



2. Relevant equations

on a i tried m*g*h but got the incorrect answer


3. The attempt at a solution


a. 882

[Snazzy]

Is this a web assignment? Have you tried a negative value?

[killa0587]

yeah it is. I thought about the negative but only have one attempt left and want to be really sure about it

[Snazzy]

Think about it. At the very top you have a lot of potential energy, and at the very bottom you have no potential energy if your zero point is the ground. The change in something is always the final minus the initial. Knowing this, what should your change in potential energy be then?

[killa0587]

the -882J

[killa0587]

and that was it...Thanks!

any insight on the others maybe just to get me started?

[Snazzy]

What is the rate of energy, or work? It is power. What is the work done by the fall? Don't plug any numbers in yet, though.

[killa0587]

well work is force x displacement correct

[killa0587]

so the displacement would be the length of the fall

[Snazzy]

What's the force then?

[killa0587]

1.8 x 9.8=17.64


whats GW though?
gigga watts

[Snazzy]

No numbers yet, mate. I just want you to know what the force was. But good, you know that F = mg

Now then, you also noted that W = mgd.

With just variables, what does P (power) equal to?

[killa0587]

power = work / time or
p= mgd / t

[Snazzy]

Alright, so you've got P = \frac{mgd}{t}

But they give you \frac{m}{t} in the question, so just solve away.

[killa0587]

Sweet! So I got 2.7 GW

[killa0587]

And on the last part I would take the 2.7GW and convert to kW then multiply by .27

[killa0587]

Thanks So much I got all 3 of them correct.