A 5250 kg freight car rolls along rails with negligible friction. The car is bro

Question

A 5250 kg freight car rolls along rails with negligible friction. The car is brought to rest by a combination of two coiled springs as illustrated in Figure P6.12. Both springs obey Hooke's law with k1 = 1600 N/m and k2 = 3400 N/m. After the first spring compresses a distance of 30.0 cm, the second spring acts with the first to increase the force as additional compression occurs as shown in the graph. The car comes to rest 45.0 cm after first contacting the two-spring system. Find the car's initial speed.



Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. m/s

Explanation / Answer

This uses conservation energy. Energy initial (Ei) = Energy final (Ef) There is only kinetic energy (K=.5mv^2) in the beginning and there is only potential energy in the end. The potential energy in the end is in the spring, and Usp=.5kx^2, where Usp=Spring Potential Energy, k=spring constant, and x=magnitude of displacement from the spring's equilibrium position. In this case, Ef=Usp1 + Usp2 Usp1=(.5)(k1)(x1^2) with k1=1600N/m and x1=45cm+30cm=75cm=.75m (Assuming the spring that is compressed the most is the spring with k1), thus Usp1=(.5)(1600N/m)(.75m^2)=450.0J Usp2=(.5)(k2)(x2^2) with k2=3400N/m and x2=45cm=.45m, thus Usp2=(.5)(3400N/m)(.45m^2)=344.25J thus from the beginning, where m=5250kg Ei=(.5)(5250kg)(v^2)=Usp1+Usp2 (2625kg)(v^2)=450.0J+344.25J=794.25J v^2=0.30257J/kg, thus v=sqrt(0.30257J/kg)=0.55006m/s Of course you had no diagram but this makes sense, assuming you understand energy conservation. In this case I had the spring with k1, be the spring it hits first. I might also have got the wrong picture or made a mistake so if you are sure you are right and something is completely illogical in my steps you might be right. Good luck.

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