A 5300 kg freight car rolls along rails with negligible friction. The car is bro

Question

A 5300 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 57.0 cm after first contacting the two-spring system. Find the car's initial speed.

    m/s

Explanation / Answer

car’s initial kinetic energy is E=0.5m*v^2;
final potential energy of spring system is E=E1+E2, where
E1=0.5*k1*x1^2 is energy of spring 1,
E2=0.5*k2*x2^2 is energy of spring 2,
x1=57cm=0.57m, x2=57.0 cm – 30.0 cm = 0.27m;
thus E=E; or;
0.5m*v^2 = 0.5*k1*x1^2 +0.5*k2*x2^2, hence
v= v((k1*x1^2 +k2*x2^2)/m) =
= v((1600*0.57^2 +3400*0.27^2)/5300) =0.38 m/s

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