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 and have spring constants with k1 = 1500 N/m and k2 = 3700 N/m. After the first spring compresses a distance of 28.2 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 53.5 cm after first contacting the two-spring system. Find the car's initial speed.

Explanation / Answer

As we can see the Graph is of Net force by spring system vs, distance .

And Area under the Fvs x curve gives total work done.

So, Total Work done = Total area from x =0 to x = x2

x1 = 0.282 m and x2 = 0.535 m

k1x1 = 1500 x 0.282 = 423 N

k1x2 + k2(x2-x1) = 1500 x 0.535 + 3700 (0.535 - 0.282) = 1738.6 N

Work done = Area = (423 x 0.282 /2 ) + ((0.535 - 0.282) x 423) + ((0.535 - 0.282) x (1738.6 - 423)/2)

          = 333.09 J

This work is done opposite to trucks motion to stop it

WOrk done = change in K.E.

- 339.09 = 0 - 4000v^2 /2

v = 0.412 m/s

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