
#1
Mar1709, 05:09 AM

P: 11

Hi
Question is In the assesment of a buffers performance railway wagon of mass 6 tonne is allowed to roll freely down an incline of 1 in 20 Sine for a distance of 50M into a horizontal yard. At the end of yard is brought to rest by a pair of parrallel springs. The stiffness in each spring is 30kN/m and the intial resistance is 4.5kN. Calculate comprssion of springs. I can to the final part once I know speed of inpact on springs. I cant work out speed on wagon as point of impact. The length of slope is 50m and height will be (50/20 which is 2.5m). I just need to know how to work out speed? Cheers Sheldon 



#2
Mar1709, 05:22 AM

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P: 4,301

The easiest way to solve it is using energy balance.
What is the gravitational potential energy and the kinetic energy at the top of the slope? What is the potential energy and the kinetic energy at the bottom of the slope? 



#3
Mar1709, 05:33 AM

P: 11

PE1  mgh
KE1  MV(Squared) PE2  0 KE2  Same as PE1 Correct? How does this help tho? 



#4
Mar1709, 05:36 AM

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P: 4,301

Velocity down a slope
Almost correct. The formula for KE1 and KE2 is the same, but it is not true that KE1 = KE2.
What velocity do you need to take for V in KE1 and which velocity in KE2? What can you say about the total energy PE1 + KE1 in relation to PE2 + KE2? 



#5
Mar1709, 05:57 AM

P: 11

I dont know what velocity I need to take for V in KE1 or KE2 do I?
PE1 + KE 1 = PE2 + KE2 . There was me hoping there was a simple formular to use, I spent hours looking on net, no wonder I cant find one. Doh 



#6
Mar1709, 06:09 AM

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P: 4,301

Well the question is not very clear, but usually the object is at rest at the top of the incline, so there V = 0.
The velocity at the bottom is precisely what you want to calculate. So you can call it V. You will get PE1 = KE2 Plug in what you know, and solve the equation you wrote down for V. 



#7
Mar1709, 06:12 AM

P: 11

(Square route)2gh Would give me Velocity but how do I take into account slope?




#8
Mar1709, 06:17 AM

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P: 4,301

It's in h.
The higher the slope, the larger the final velocity. How long (steep) the slope is doesn't matter, when there is no friction involved. 



#9
Mar1709, 06:21 AM

P: 11

Ah ok. Thanks
√2x9.81x2.5 =7.00357 m/s 



#10
Mar1709, 06:30 AM

P: 11

mv²/2 = 2(sx²/2)
X= √mv²/2s 6000x7²/2x30000  4.9m Thanks soooo much!!! 



#11
Mar1709, 06:59 AM

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P: 4,301

You're welcome.
Just as an aside, technically the question is talking about "rolling" down the slope, so you should be using rotational formulas (angular momentum, moment of inertia, etc.) However, as nothing is said about the shape of the wagon, for example, I suppose you'll have to assume that the wagon is sliding down instead of rolling down, which will give you the answer you calculated. 



#12
Mar1809, 03:34 AM

P: 11

Thanks. Wouldnt I need the friction to work out it angular momentem?
P=ma x f So Im hoping I have done it correctly 



#13
Apr310, 10:46 AM

P: 1





#14
Apr310, 01:44 PM

Sci Advisor
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P: 4,301

Yes, in both cases all potential gravitational energy is converted to kinetic energy.
Only when it slides down an inclined slope, the velocity will decompose into a horizontal and a vertical component. 


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