A 6800 kg freight car rolls along rails with negligible friction. The car is bro

Question

A 6800 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 42.0 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)(6800)V² = 3400V².......................................1

The two springs act in parallel after the first one is compressed 30.0 cm.
Spring #1: P.E. = 1/2kx² = 0.5*1600*x^2

=800*x^2
Spring #2: P.E. = 1/2k² = 0.5*3400*x^2

=1700*x^2

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

Spring # 1 absorbs P.E. = 800(0.42)^2 = 141.12 J

Spring # 2 absorbs (parallel w Spring # 1) = 1700(0.42 - 0.30)² = 24.48 J

Total Energy absorbed by Springs (#1 & # 2) = 141.12+24.48 = 165.6 J

From 1
Car's initial K.E. = 3400V²
3400V² = 165.6
V = 0.220 m/s

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