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 = 1500 N/m and k2 = 3400N/m. After the first spring compresses a distance of 25.0 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 45.0 cm after first contacting the two-spring system. Find the car's initial speed.

Explanation / Answer

The first spring is compressed the full distance of x1 = 0.45 m so it does work of

W1 = ( 1/2 ) k1 x12 = 0.5 (1500 N/m) (0.45 m)2 = 151.875 J

The second spring is compressed only after the freight car moves a distance of 0.25 m, so it is only compressed a distance of x2 = 0.20 m so it does work of

W2 = ( 1/2 ) k2 x22 = 0.5 (3 400 N/m) (0.2 m)2 = 68 J

So both springs have done a total amount of work of

WTot = W1 + W2 = 151.875 J + 68 J = 219.875 J

From KEi= ( 1/2 ) m vi2

to its final value of zero. That is

KEi= ( 1/2 ) m vi2 = ( 1/2 ) (4 000 kg) vi2 = 219.875 J = Wsprings

vi2 = 0.1099375 m2 / s2

vi = 0.332 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.