Why Does My Method for Calculating Spring Compression Yield a Different Result?

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SUMMARY

The discussion focuses on the discrepancy between two methods for calculating the maximum compression of a spring in a momentum problem involving a railway engine and carriage. The correct solution utilizes kinetic energy principles, yielding a maximum compression of 0.0847 m based on a spring stiffness of 320 kN/m. The user's method, which incorrectly applies average force and uniform acceleration concepts, results in a compression of 0.059 m. The key takeaway is the importance of using the correct formulas for energy conservation in spring mechanics.

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ojsimon
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Hi,

I have just tried a momentum question involving springs. I got the wrong answer and the method used in the solution is different, but i still do not see what i did wrong.

The question is: a railway engine 6.2*10^4 and a carriage 4*10^4 have coupled. They have hit a buffer spring (to stop them) from an initial speed of 0.15m/s. Assuming the buffer behaves like a spring of stiffness 320kN/m calculate the maximum compressiion of the spring.

The solution given is:

Kinetic energy of train :

0.5 × 10.2 × 104 × 0.152 = 1150 J

0.5* F e
= 0.5× (k e) × e = 0.5 × k e^2
0.5 × 320 × 103 e2 = 1150 1
gives compression e = 8.47 × 10^-2 m

My Solution
F=Change in MV/Change in time F= Ke

so:

MV/T = Ke


s= ((u+v)/2)t Therefore : t=(40/3)e

so Mv= 40/3(Ke^2)

And i calculate e=0.059 m

But this is different to the solution given, why does my method not work?

Thanks
 
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ojsimon said:
My Solution
F=Change in MV/Change in time F= Ke

so:

MV/T = Ke
Δ(MV)/Δt would give you the average force, not the maximum force.


s= ((u+v)/2)t Therefore : t=(40/3)e
The acceleration is not uniform, thus that kinematic formula will not apply.
 
Ok thank you very much for you assistance.
 

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