A 7,000 -kg freight car rolls along rails with negligible friction. The car is b

Question

A 7,000-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 k1 = 1,700 N/m and k2 = 3,500 N/m. After the first spring compresses a distance of x1 = 31.2 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 x2 = 48.5 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 = 48.5 m so it does work of

W1 = ( 1/2 ) k1 x12 = 0.5(1 700 N/m) (0.485 m)2 = 200 J

The second spring is compressed only after the freight car moves a distance of 0.312 m, so it is only compressed a distance of x2 = 0.173 m so it does work of

W2 = ( 1/2 ) k2 x22 = 0.5 (3 500 N/m) (0.173 m)2 = 52.3J

So both springs have done a total amount of work of

WTot = W1 + W2 = 200 J + 52.3 J = 252.3 J

[ There is an alternate way to do this. We could figure the work done during the first 0.312m with k = 1700 N/m and then find the work done during the last 0.173 m from the "area" under the graph. The total amount of work should be the same, either way ].

Actually, this total work is negative that amount, WTot = - 252.3 J, since the force and displacement are in opposite direction. This has changed the freight car's Kinetic Energy from its initial value of

KEi= ( 1/2 ) m vi2

to its final value of zero. That is

KEi= ( 1/2 ) m vi2 = ( 1/2 ) (7 000 kg) vi2 = 252.3 J = Wsprings

vi2 = 0.0720 m2 / s2

vi = 0.26m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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