A 4000 kg freight car rolls along rails with negligible friction. The car is bro

Question

A 4000 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 the figure below. Both springs are described by Hooke's law with k1 = 1600 N/m and k2 = 3600 N/m. After the first spring compresses a distance of 20.0 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 60.0 cm after first contacting the two-spring system. Find the car's initial speed.

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=60cm=0.6m, x2=60.0 cm – 20.0 cm = 0.4m;
thus E=E; or;
0.5m*v^2 = 0.5*k1*x1^2 +0.5*k2*x2^2, hence
v= ((k1*x1^2 +k2*x2^2)/m) =
= ((1600*0.6^2 +3600*0.4^2)/6000) =0.438 m/s;

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