A 6000 kg freight car rolls along rails with negligible friction. The car is brought to rest by a...

Question:

A {eq}6000 \ kg {/eq} freight car rolls along rails with negligible friction. The car is brought to rest by a combination of two coiled springs. Both springs are described by Hooke's law with {eq}k_1 = 1500 N m^{-1} {/eq} and {eq}k_2 = 3700 N m^{-1} {/eq}. After the first spring compresses a distance of {eq}30.0 \ cm{/eq}, 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 {eq}60.0 \ cm {/eq} after first contacting the two-spring system. Find the car's initial speed.

Energy conservation

The energy of the car that is kinetic will convert or store in the spring as the potential energy of the spring due to the collision of the car with the spring. We know that from the law of energy conservation that energy can neither be created nor be destroyed it can only change its forms from one to another. Therefore, the total energy of the system would remain constant at every instant.

Given

Mass of the car (M) = 6000 kg

The velocity of the car = v m/s

Now, the initial energy of the system

{eq}E_{i} = 0.5Mv^{2} {/eq}

Now, the spring 1 is compressed by the x = 30 cm, therefore energy stored would be

{eq}E_{1} = 0.5k_{1}x^{2} \\ E_{1} = 0.5*1500*(0.3)^{2} \\ E_{1} = 67.5 \ J {/eq}

Now, both spring comes together in action and compressed by the other 30 cm, therefore

{eq}E_{2} = 0.5(k_{1}+k_{2})x^{2} \\ E_{2} = 0.5(1500+3700)*0.3^{2} \\ E_{2} = 234 \ J {/eq}

Now, applying the energy conservation

{eq}E_{i} = E_{1}+E_{2} \\ 0.5*Mv^{2} = 67.5+234 \\ 0.5*6000*v^{2} = 301.5 \\ v = 0.317 \ m/s {/eq} 