# Energy Transformation

• Alexi

#### Alexi

1. A skier has an initial velocity of 12.0 m/s, after a while reaches a slope with 18° of inclination and he slides upwards 12.2m before reaching 0
¿What is the coefficient of friction? "μ"
Vi= 12.0 m/s
Vf= 0 m/s
Distance = 12.2 m

2. Ec = 1/2 m(v²)
Wt = Ec(final) - Ec (initial)

3. Wt = (F)(d) = (μ)(N) (D) = (μ)(m)(g)(D)
With this in mind:
(μ)(m)(g)(D) = Ec(final) this is zero - 1/2 m (v²)
The masses go away
(μ)(g)(D) = - 1/2 (12²)
You clear (μ)
(μ)= -72/119.68
That equals
(μ) = 0.6

This was way logical when I did it but then I realized I didn't used the degrees of inclination, and I think I have to use them like "(μmgcos18°)(D)" but I'm not that sure.

Help?
Thanks

And I'm from mexico, some terms perhaps are not to clear but I hope to have made myself clear.

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You can use the energy relationship to determine the force on the skier, but the force on the skier is composed of 2 components. The g*sinθ down the slope which yields the increase in potential energy as well as the μ*m*g*cosθ*d which represents the work taken away by friction.

At the highest point where speed is zero, it has energy, potential energy.
using trigonometry, you can find the height(you have angle and...).
than, work done =friction*distance = Uk *mg*cos18= change in total energy.

good luck,
seems like Lowly beat me XD

At the highest point where speed is zero, it has energy, potential energy.
using trigonometry, you can find the height(you have angle and...).
than, work done =friction*distance = Uk *mg*cos18= change in total energy.

good luck,
seems like Lowly beat me XD

the slope is above the ground, haha, so he goes up 12.2 before reaching the top... mmm, I guess I'm understanding.

Thanks, I have several problems but I guess with this explanation I will try to answer them again correctly this time.

yes he goes 12.2 , but it is not his height above the ground!
if you go 12.2 meters on an inclined plane with very small angle, you won't reach a very high height, but if You go on an inclined plane with big angle, You'll reach a very high point,
if X is the distance You went than:
sin 18 = h/x

If You can't manage the other problems We will be more than happy to help You!

1. A skier parts from 0 and slides down an inclined plane of 22° with a length of 75 meters.

a)If μ is 0.09 ¿What's the velocity at the end of the plane
b)If the snow is equal to the friction of the plane ¿How far will he go before stopping?

2.Ep = mgh
Ec = 1/2 m(v²)
Wt = Ec - Ep

3.
I got the height first, like 75sin22° =28.09
So, with this in mind
Wt = 1/2 m(v²) - mgh
Wt = (μ)(m)(g)(D)
eliminate masses
(μ)(g)(D) = 1/2 (v²) - gh
(μ)(g)(D) + gh = 1/2(v²)
sqrt(((μ)(D) + h)(g))2) = v
SO after all of this you get v is 20.46 m/s
Fd = -1/2m(v²)
eliminate masses
(μ)(g)(D) = 1/2 (v²)
substitute with the velocity you just got and get D that equals to 311.94m

It sounds logical BUT THEN AGAIN I completely got rid of the angle, dammit.

What You're missing is that the friction equals to
Us*g*N

while N is the normal force.
the normal force is not always equal to mg, in an inclined plane it equals to 1 of hos components.
one is responsible for accelerating downward and when is on pushing on the surface
mgsina and mg cosa respectively, try to prove that for yourself, remember mg is always acting down(straight down) and always perpendicular to the surface.
good luck!

I've understood everything now, thanks Dweirdo!

But now I've got one small question and one big one! (Hope I'm not getting on your nerves :P)

1.- An elevator(920 kg) is being hanged from a rope until it breaks from a height of 28 m upon a big spring (k= 2200000N/m) at the bottom.
a)Work done by gravity?
b)Speed just before it lands on the spring?
c)How much does it compress?
2.- Wd = Changes in energy
Ek = 1/2 mv²
Ee= 1/2kx²
Ep= mgh
3.-
-Wd= - mgh = - 252705.6 (a) (can this be done?)
252705.6 = 1/2 m v²
You take out v:
v= 23.43 m/s
And then you compare this to the energy that is transferred to the spring
-Wd= 1/2 k x² -1/2 mv²
You've got work done, the velocity, the mass so you simply substitute some numbers
x= 1.5 m

I think I'm correct on this one although I would like you guys to check it out.

And HOPEFULLY the last one:

1.- An object (6 kg) is pushed upwards a slope (37°) 8 meters by a force of 75 N. If Vi= 2.2 m/s and the force of friction is 25 N:
A) Kinetic energy of the object
B)Work done by the force of 75 N
c) Work done by the force of 25 N
D)Work done by gravity
E)Work done by N
F)Final Kinetic Energy.
2.- Wd = FD
Ek = 1/2 mv²
3.-
Ek = 1/2 (6) (2.2²) =14.52 (a)
Wd= FD
Wd = 75sin37(8)= 361.08 (b)
-Wd = 25sin37(8)= -120.36 (c)
-Wd = 9.81sin37(8)= -47.23 (d)
-Wd = mgcos37(8) = -283.38 (e)
You simply substract the negatives and all I've got was
Wd= -89.89
So
-89.89 = 1/2 m v² - 14.52 (the one from the a)
75.37 = Final kinetic energy (f)

Something must be wrong there, I doubt it would end up negative.

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I've understood everything now, thanks Dweirdo!

But now I've got one small question and one big one! (Hope I'm not getting on your nerves :P)

1.- An elevator(920 kg) is being hanged from a rope until it breaks from a height of 28 m upon a big spring (k= 2200000N/m) at the bottom.
a)Work done by gravity?
b)Speed just before it lands on the spring?
c)How much does it compress?
2.- Wd = Changes in energy
Ek = 1/2 mv²
Ee= 1/2kx²
Ep= mgh
3.-
-Wd= - mgh = - 252705.6 (a) (can this be done?)
252705.6 = 1/2 m v²
You take out v:
v= 23.43 m/s
And then you compare this to the energy that is transferred to the spring
-Wd= 1/2 k x² -1/2 mv²
You've got work done, the velocity, the mass so you simply substitute some numbers
x= 1.5 m

I think I'm correct on this one although I would like you guys to check it out.

And HOPEFULLY the last one:

1.- An object (6 kg) is pushed upwards a slope (37°) 8 meters by a force of 75 N. If Vi= 2.2 m/s and the force of friction is 25 N:
A) Kinetic energy of the object
B)Work done by the force of 75 N
c) Work done by the force of 25 N
D)Work done by gravity
E)Work done by N
F)Final Kinetic Energy.
2.- Wd = FD
Ek = 1/2 mv²
3.-
Ek = 1/2 (6) (2.2²) =14.52 (a)
Wd= FD
Wd = 75sin37(8)= 361.08 (b)
-Wd = 25sin37(8)= -120.36 (c)
-Wd = 9.81sin37(8)= -47.23 (d)
-Wd = mgcos37(8) = -283.38 (e)
You simply substract the negatives and all I've got was
Wd= -89.89
So
-89.89 = 1/2 m v² - 14.52 (the one from the a)
75.37 = Final kinetic energy (f)

Something must be wrong there, I doubt it would end up negative.
"(Hope I'm not getting on your nerves :P)"
You're not, I'm happy to help, cause I'm studying as well for the second stage olympics in my country, so this helps me as well.
"-Wd= - mgh = - 252705.6 (a) (can this be done?)"
be careful with the signs, I'm not really sure It's negative, depends where is your zero point.
But let's say The zero point is where the spring , than work done by gravity equals to the change in kinetic energy=positive=mg(h).
than mgh=mv^2/2
V=sqrt(2gh) You have the speed hitting the spring,You've done that. so this is cool.
now,here i think something is messed up,
we have a spring , and mass hitting the spring with certain velocity, I'm attaching a picture which may help u solve it, but I'm not really sure myself , Do You have final answers?

anyway some questions to get You through the attached file;

What Kind of energy does it have in position A?
and after compressing X what kind of energy it has in position B?
HINT:
ITS NOT ONLY KINETIC ENERGY IN POINT A!
That's how I'd solve it...
About the second question, Let's first finish this one.
good luck!

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No, the teacher didn't give me the answers, and I don't know how to take in consideration the weight since I'm already using the energy of the spring minus the speed, I thought that gravity was done for? I think:
in A it has potential energy mgX (X is the how much the spring will be compressed)
and kinetic energy.
in B it has zero kinetic energy zero gravitational potential energy and k*X^2 /2 elastic potential energy.
by conservation of energy ...
try to solve it.
good luck
P,S
There are other ways to solve this.

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