A little power thingy...

[thedarkshade]

Well this is not really a typical homework it's just something that just came to my mind!

 

Anyways! I was helping my brother to solve some problems is physics and few of the problems were related to power ([math]P=\frac{A}{t}[/math]). And there was this other formula [math]P=F\times v[/math]. I mean, I'd seen this formula for a hundred times, but never really paid attention to it. Then my brother asked me a question :"What is the power of an object whose mass is 5kg and which is falling from a distance of 10 meters?". OK, it's not any hard question and I just want to make sure I gave to him the right answer. This is what I did!

 

In [math]P=F\times v[/math] instead of F I put [math]Q=mg[/math] (as the unit is correct and it still fits the equation), and about v, I measured that from the formula for finding velocity in free fall. So I had the following equations:

 

[math]P=Q\times v[/math]

[math]Q=m\times g = 5kg \times 9.81\frac{m}{s^2}=49.05kg\frac{m}{s^2}[/math]

so I got [math]Q=49N[/math]

 

and for the velocity:

[math]v=\sqrt{2gh}= \sqrt{2\times 9.81\frac{m}{s^2} \times 10m} = \sqrt {196.2\frac{m^2}{s^2}}[/math]

so approximately [math]v=14\frac{m}{s}[/math]

 

And then we'd get:

 

[math]P=Q\times v=49N \times 14\frac{m}{s} =686\frac{Nm}{s}=686\frac{J}{s}=[/math]

so the final score would be: [math]P=0.68kW[/math]

 

Is there anything I've missed:doh:!

[swansont]

Is there anything I've missed:doh:!

 

The calculation seems fine, but the unanswered question is "what does this mean?"

 

The question is worded awkwardly, since an object doesn't "have" power. That's the power gravity is generating by acting on the object. But if you tried to do work with the object, the power would change quite rapidly.

 

P=Fv is more useful when the system is under load and moving at constant speed.

[thedarkshade]

OK, thanks swansont!

He'd found that on his problems books, but today morning he didn't have the key to it, and just now he brought the key book and it seems that is correct;)!

 

Thanks again!