## [Physics] Work

Started by Aqua, December 01, 2009, 07:17:08 pm

#### Aqua

##### December 01, 2009, 07:17:08 pmLast Edit: December 01, 2009, 07:18:14 pm by Aqua
"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...

Explain this problem please #### winkio

##### December 01, 2009, 08:00:45 pm #1 Last Edit: December 01, 2009, 08:02:13 pm by winkio
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:

Spoiler: Show

By the work-kinetic energy theorem
dW = MEf - MEi
thus
dW = Kf + Uf - (Ki + Ui)
Since we can assume that the object is at rest both before and after lifting,
Ki = Kf = 0
so
dW = Uf - Ui

Since U(s) is defined as
U(s) = mgh,
dW = mg(hf - hi)

hf - hi is how far it was lifted, so isolating that yields
hf - hi = dW/mg

Now substituting in,
hf - hi = (2.0 J) / ((180 g) * (9.8 m/s^2))
hf - hi = 1.074 m

(Remember to convert 180 g into kg) #### Aqua

##### December 08, 2009, 08:32:40 pm #2
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... .___.
Help please #### fugibo

##### December 08, 2009, 09:19:38 pm #3
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.

Yes, I'm sure D:

#### fugibo

##### December 08, 2009, 09:51:00 pm #5
I'll give you a hint: Use Hooke's Law + Law of Conservation of Energy

BTW, I'm an idiot.

#### winkio

##### December 08, 2009, 10:09:31 pm #6
Spoiler: Show
So, I'm going to make ASCII diagrams to explain this:

this is the starting diagram:
l
l
l                  _____
l         k       l          l
l-/\/\/\/\/\/\/l   m    l
L________l_____l______________
|
0

We have a spring k and a mass m in equilibrium.

now, we compress the spring by 0.100 m:
l
l
l             _____
l         k  l         l
l-/\/\/\/\/l   m    l
L______l___ _l______________
|      |
-0.100   0

The mass is still at rest, but the spring now has potential energy
U = (1/2) k x^2

now, we release the system:
l
l
l                  _____
l         k       l          l
l-/\/\/\/\/\/\/l   m    l--->v
L________l_____l______________
|
0

The spring extends until it reaches equilibrium, transferring all of it's potential energy into kinetic energy for the mass.

After this point, the spring and the mass separate, and friction starts to play a role.

Since W = integral (F ds) and F is constant, we can say that W = Fs.

F friction = F Normal * u = mgu, thus
W = mgus

Now since W = dE and dE = -U (this line is just saying that change in work is change in energy, and the energy goes from U to 0)
W = -U.

substituting in:
mgus = (-1/2)kx^2

isolating u:
u = (kx^2)/(2mgs)

then we plug in numbers and constants:
k = 105 N/m
x = -0.100 m
m = 2.00 kg
s = 0.250 m
g = 9.80 m/s

you should get u = 0.107.  It should be positive, because the only negative term is squared. #### Aqua

##### December 10, 2009, 08:04:25 pm #7
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
Um... how do I go about this? >.<

#### winkio

##### December 10, 2009, 08:38:47 pm #8
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

##### December 10, 2009, 08:55:18 pm #9
I thought of doing it like that, but I wasn't sure.

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<3