A 5000-kg freight car rolls along rails with negligible friction. The car is bro

Question

A 5000-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 and have spring constants with k_1 = 1700 N/m and k_2 = 3200 N/m. After the first spring compresses, a distance of 27.0 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 50.5 cm after first contacting the two-spring system. Find the car's initial speed.

Explanation / Answer

Freight car's K.E.(just prior to Spring #1) = 1/2mV² = (0.5)(5000)V² = 2500V²

The two springs act in parallel after the first one is compressed 27cm.
Spring #1: P.E. = 1/2kx² = 850(x)²
Spring #2: P.E. = 1/2k² =1600(x)²

The freight car's K.E. is totally transferred to Spring # 1 and Spring # 2

Spring # 1 absorbs P.E. = 850(0.505)² = 216.77 J

Spring # 2 absorbs (parallel w Spring # 1) = 1600(0.505 - 0.27)² = 88.36 J

Total Energy absorbed by Springs (#1 & # 2) = 216.77 + 88.36 = 305.13J
Car's initial K.E. = 2500V²
2500V² = 305.13
V = 0.35 m/s

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