Verification needed - Rollar Coaster

  • Thread starter Thread starter rkslperez04
  • Start date Start date
AI Thread Summary
The discussion revolves around solving a physics problem related to a roller coaster's speed after climbing a hill. The user correctly applies the conservation of energy principle, leading to the calculation of the final speed after considering the height of the second hill as 15 meters. Feedback suggests that while the calculations are mostly accurate, the final speed should be expressed with more precision, ideally around 10.29 m/s. Clarification about the height of the hill was necessary for a complete understanding of the problem. Overall, the user demonstrates a solid grasp of the concepts but seeks confirmation and minor adjustments for accuracy.
rkslperez04
Messages
31
Reaction score
0
Bare with me guys as I enlist your help through this worksheet. I really want to get a good grade on this upcoming test and this worksheet is our study guide.

I think I have the hang off it but I want to make sure I am grasping the questions here.

A rollar coaster is traveling on a horizontal track at 20 m/s. What is its speed after climbing the second hill.

I understand this to be an Ei=Ef problem.

Here is my work:

1/2MVi^2 = 1/2MVf^2 + GH ( all the mass variables cancel out )
1/2(20)^2 = 1/2(Vf)^2 + (9.8)(15)
200 = 1/2(Vf)^2 + 147
53 = 1/2(Vf)^2
53*2 = Vf^2
106 = Vf^2
Sq rt 106 is 10 m/s = Vf



am I close? Please let me know.. I want the right answer but I also want to comperhend what's going with these. Just wanted you to know I am not looking for someone to do my work.. just nudge me along ;)

(got one for you too after this, also... a coeffencient of friction.. that one I will need someone to walk me through)
 
Physics news on Phys.org
Please provide the complete question when seeking help. Without having any info about "the second hill," we can't do much. One can only assume that it's peak is at a height of 15m above the horizontal track from your work. If that is so, your work is right. You may want to be a little more accurate with the final answer, though (to one or two decimal places).
 
LOL.. I thought the same thing about this question.. as with most..LOL

That was the entire question. You can see my fret over this problem..

With the decimals places I would say 10.0 instead of 10 ??

Looks good though??
 
rkslperez04 said:
LOL.. I thought the same thing about this question.. as with most..LOL

That was the entire question. You can see my fret over this problem..
So where did you get that 15 from?

With the decimals places I would say 10.0 instead of 10 ??

Looks good though??
I actually meant something like 10.2 or 10.29, which is a better approximation to sqrt(106) than 10 is.
 
ohh.. I see.. the second hill is 15m high.. i apolize for that.

Does my work look right?
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top