### Author Topic: [Physics] Work  (Read 6200 times)

#### Aqua

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##### [Physics] Work
« on: December 02, 2009, 02:17:08 AM »
"If 2.0 J of work is done in raising a 180 g apple, how far is it lifted?"

How does mass relate to work...? O.o
That's all the given info...

« Last Edit: December 02, 2009, 02:18:14 AM by Aqua »

#### winkio

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##### Re: [Physics] Work
« Reply #1 on: December 02, 2009, 03:00:45 AM »
Work is defined as the integral of force with respect to distance.  (Integral[F(s) ds])
Since gravity is a conservative force, we know that the work is independent of the path taken.

Variables:
s - position
F - force
K - kinetic energy
U - potential energy
ME - mechanical energy

There is a slight ambiguity in this problem, as I cannot tell if it is asking for the net work done on the apple, or the work done by the supplying force of whatever is lifting it.  The question only gives enough information to work it out as net work done on the apple, so:

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« Last Edit: December 02, 2009, 03:02:13 AM by winkio »

#### Aqua

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##### Re: [Physics] Work
« Reply #2 on: December 09, 2009, 03:32:40 AM »
A light horizontal spring has a spring constant of 105 N/m. A 2.00 kg block is pressed against one end of the spring, compressing the spring 0.100 m.
After the block is released, the block moves 0.250 m to the right before coming to a rest.

What is the coefficient of kinetic friction between the horizontal surface and the block?

The answer is 0.107 (back of the book... XD), but I don't know HOW to get it... and that's what's important... .___.

#### fugibo

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##### Re: [Physics] Work
« Reply #3 on: December 09, 2009, 04:19:38 AM »
Are you sure the figures are 0.1m and 0.25m? I could be wrong, but I don't think the block would cause the spring to expand.

#### Aqua

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##### Re: [Physics] Work
« Reply #4 on: December 09, 2009, 04:40:04 AM »
Yes, I'm sure D:

#### fugibo

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##### Re: [Physics] Work
« Reply #5 on: December 09, 2009, 04:51:00 AM »
I'll give you a hint: Use Hooke's Law + Law of Conservation of Energy

BTW, I'm an idiot.

#### winkio

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##### Re: [Physics] Work
« Reply #6 on: December 09, 2009, 05:09:31 AM »
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#### Aqua

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##### Re: [Physics] Work
« Reply #7 on: December 11, 2009, 03:04:25 AM »
Water flows over a section of Niagra Falls at the rate of 1.2 x 10 ^ 6 kg/s and falls 50.0 m.  How much power is generated by the falling water?

I don't think we're supposed to use Calculus for this one... XD

#### winkio

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##### Re: [Physics] Work
« Reply #8 on: December 11, 2009, 03:38:47 AM »
Simplified explanation:

Equation for power:
P = W/t
Work-Energy theorem:
W = dE
Definition of total mechanical energy:
dE = (K + U)i - (K + U)f
Water starts and ends at same speed:
Ki = Kf = A
Potential energy before and after:
Ui = mgh
Uf = mg(h-50)
Substituting into the equation:
dE = mgh + A - mg(h-50) - A = 50mg
Substituting back into the power equation
P = 50mg / t

Since we don't have t or m, the give us the ratio of m/t
m/t = 1.2 x 10 ^ 6 kg / s

P = 50g * (m/t)

P = 50 (9.8) (1.2 x 10 ^ 6) = 588 MW

#### Aqua

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##### Re: [Physics] Work
« Reply #9 on: December 11, 2009, 03:55:18 AM »
I thought of doing it like that, but I wasn't sure.

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