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 with k1- 1700 N/m and k2-3500 N/m. After the first spring compresses a distance of 15.0 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 50.0 cm after first contacting the two-spring system. Find the car's initial speed. 900 m ter thefrst soincting the twersping k-k2C6.xp Total force xx2 Distance (cm)

Explanation / Answer

Consider intial velocity of the Freight's car is V m/s

Initial energy of the 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 15.0 cm.
So, the total energy stored by the two spring system -

E = (1/2)*1700*0.15^2 + (1/2)*1700*(0.50 - 0.15)^2 +  (1/2)*3500*(0.50 - 0.15)^2

= 19.13 + 104.13 + 214.38 = 337.64 J

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

2500V² = 337.64

=> V = sqrt[337.64 / 2500] = 0.3675 m/s = 36.75 cm/s

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