Energy Questions? Solve Problems Now

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Homework Help Overview

The discussion revolves around energy-related problems in physics, specifically focusing on work, power, and the conservation of energy. The original poster presents two distinct scenarios involving a roller coaster and a snowmobile pulling a skier up a slope.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to calculate work and power for a roller coaster scenario and seeks help with potential energy and kinetic energy conversions. They express uncertainty about applying conservation of energy in the second scenario involving a snowmobile and skier.
  • Some participants provide calculations and suggest methods for determining energy changes, while others question the original poster's approach and understanding of the problems.
  • There are inquiries about how to properly apply the conservation of energy principle and resolve forces in the context of the second question.

Discussion Status

Participants are actively engaging with the problems, with some providing calculations and others seeking clarification on specific steps. There is a mix of responses, with some confirming the correctness of earlier calculations while others express confusion and request further assistance. No explicit consensus has been reached on the best approach to the second question.

Contextual Notes

Participants note the urgency of the original poster's request for help, indicating a time constraint for completing the homework. There is also mention of the need to consider friction in the second problem, as well as the importance of using the conservation of energy as initially suggested.

yosup231
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Energy questions??

ok i have two questions which i am completely stumped on and i need some help...

At riverside park in Agawam, Massachusetts, a ride called the cyclone is a giant roller coaster that ascends a 34.1 m hill and then drops 21.9 m before ascending the next hill. THe train of cars has a mass of 4727 kg. a)how much work is required to get an empty train of cars from the ground to the top of the hill. b)what power must be generated to bring the train to the top of the hill in 30 s. c)how much potential energy is converted into kinetic energy from the top of the hill to the bottom of the 21.9 m drop?

a) (4727 kg)(10m/s^2)(34.1m)=1611907 N.m (ithink that's right)
b) 1611907N.m/30s=53730.23 watts (i also think this is right)
c) Help please??



other question

Using her snowmobile, Midge pulls a 60.0 kh skier up a ski slope inclined at an angle of 12.0 degrees to the horizontal. The snowmobile exerts a force of 200.0 N parallel to the hill. If the coeficient of friction between the skis and hte snow is 0.120, how fast is the skier moving afster he has been pulled for 100.0 m starting from rest? (ignore the effects of the static friction that must be overcome to initially start the skier in motion.) Use the law of conservation of energy.

i know the formula for conservation of energy is KEo+PEo=KEf+PEf and i think for this problem i would have to find the net work by using the equation of KEf-KEo= net work... i also know the KE=1/2mv^2 and PE=mgh my problem is how do i use this information to solve this problem?
 
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Hi yosup
could u tell me how do i post my query?
 
Well you basically know the potential energy at the top and the potential energy at the bottom. This change in potential energy has been converted into kinetic energy.
 
okay so i guess i would do this:

(4727 kg)(10m/s^2)(21.9 m)=1035213J
1611907J-1035213J= 576694J (is that right)

can anyone help with the second question
 
im kind of new at this... why does it say my post has been moved??

CAN SOMEONE PLEASE HELP ME OUT WITH THIS QUESTION?!


EDIT: n/m i see i should have posted this in the homework question section
 
Last edited:
come on please I am sure someone here knows how to do these problems :confused:
 
C'mon its been two days; isn't there someone who could help me with these questions? Your help would be greatly appreciated.
 
Question b is correct, assuming question a is correct. Question c is also appears correct. It would be a simple question if done using Newtonian laws of motion; however, I'mnot sure how to proceed using conservation of energy.
 
ok... but how about the other question that's the one that I am having a lot of trouble with
 
  • #10
Hootenanny said:
however, I'mnot sure how to proceed using conservation of energy.
I'm talking about the second question. Perhaps someone else could give a hint?
 
  • #11
yosup231 said:
C'mon its been two days; isn't there someone who could help me with these questions? Your help would be greatly appreciated.
penguino explained how to do c) six minutes after you posted the question!

I'm talking about the second question. Perhaps someone else could give a hint?
What second question? You posted three questions. You gave answers to the first two and were told that they were correct. (You are taking g= 10 m/s2? 9.8 or even 9.81 is more accurate.)
 
  • #12
other question

Using her snowmobile, Midge pulls a 60.0 kh skier up a ski slope inclined at an angle of 12.0 degrees to the horizontal. The snowmobile exerts a force of 200.0 N parallel to the hill. If the coeficient of friction between the skis and hte snow is 0.120, how fast is the skier moving afster he has been pulled for 100.0 m starting from rest? (ignore the effects of the static friction that must be overcome to initially start the skier in motion.) Use the law of conservation of energy.

i know the formula for conservation of energy is KEo+PEo=KEf+PEf and i think for this problem i would have to find the net work by using the equation of KEf-KEo= net work... i also know the KE=1/2mv^2 and PE=mgh my problem is how do i use this information to solve this problem?


this question is kind of urgent... could someone please help!?
 
  • #13
using Newtonian motion equations; its a simple matter of resolving forces and using F=ma, v=u+at, v^2 = u^2 + 2as, s = ut + 1/2 at^2 etc.
 
  • #14
well i did:

60kg x sin 12= 12.47kg
200N x .12 x cos 12=23.47N
200N-23.47N= 176.53N
176.53N/12.47kg= 14.52 m/s^2
Vf^2=0+2(14.2m/s^2)(100m)
Vf=53.3 m/s

Now I know this is wrong could you please explain to me where i went wrong
 
  • #15
Please this is urgent someone answer
 
  • #16
yosup231 said:
well i did:
60kg x sin 12= 12.47kg
200N x .12 x cos 12=23.47N
200N-23.47N= 176.53N
176.53N/12.47kg= 14.52 m/s^2
Vf^2=0+2(14.2m/s^2)(100m)
Vf=53.3 m/s
Now I know this is wrong could you please explain to me where i went wrong

AFAIK, what you are doing with mass in the first operation is meaningless in this context. Also, you aren't finding [what I'm guessing to be] the force of friction properly in the second line. The question asked you to use the conservation of energy in the first place!

The snowmobile exerts 200N of force parallel to the slope for 100m, thus [itex]W_s = F_{||}d = (200)(100) = 200 000 J[/itex]

They give you the coefficient of friction as 0.120 and the mass (60 kg) so you can find the force of friction parallel to the displacement with some trig, and therefore [itex]W_f[/tex] should be straightforward. Friction does negative work ([itex]\cos 180 = -1[/itex]) so subtract it from [itex]W_s[/itex] to get net work.<br /> <br /> Remember [itex]\Sigma W = \Delta KE[/itex][/itex]
 
Last edited:
  • #17
c'mon i have 60 more minutes to get this done i think i figured out some stuff but I am not sure WILL SOMEONE PLEASE HELP

i did

60kg x 10m/s^2 x .12 x cos 12= 70.4
200N-70.4N=129.6 N
129.6N x 100m=12960 J

then I am not sure what else to do PLEASE PLEASE HELP ME
 
  • #18
yosup231 said:
c'mon i have 60 more minutes to get this done i think i figured out some stuff but I am not sure WILL SOMEONE PLEASE HELP
i did
60kg x 10m/s^2 x .12 x cos 12= 70.4
200N-70.4N=129.6 N
129.6N x 100m=12960 J
then I am not sure what else to do PLEASE PLEASE HELP ME
(i am somewhat rushing and assuming your answer above is correct...)

Here you found the total work done.
Now someone told you that the total work is equal to the change in kinetic energy. So if:

[tex]12960J = \Delta K =.5mv_f^2 - 0[/tex]

(0 is the initial kinetic energy.)You can solve for vf can't you?
 

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