How Fast Was the Freight Car Moving Before Hitting the Springs?

[klmartin]

Homework Statement

A 8000 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 = 1600 N/m and k2 = 3000 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.

Homework Equations

U=1/2kx^2
KE=1/2mv^2

The Attempt at a Solution

i tried to set U and KE equal and solve for v but its not really working out. i tried the equation 1/2(k1)(x1)^2+1/2(k2)(x2)^2=1/2mv^2. the problem is confusing because k2 and x2 are not explicitly expressed.

 

[dynamicsolo]

You aren't working out of Serway & Jewett, by any chance? (I just went through this problem with students last week... ;-) )

You have to break this problem into two parts. (You can use conservation of mechanical energy because we are ignoring friction and air resistance.)

At the beginning of part 1, the freight car just begins to make contact with the longer spring. What is the kinetic energy of the car and the potential energy in the springs at that moment? (We can skip gravitational potential energy, since the rails are horizontal, so there is no change in this PE.)

At the end of part 1/start of part 2, the car has compressed spring 1 by 15 cm. and is just beginning to contact spring 2. What are the energies at this point?

At the end of part 2, the car has been brought to rest. By what distance has each spring been compressed? What are the energies now?

How would you use the information about the amount and kind of energy at the end to tell you how fast the freight car was moving at the beginning?

 

[ogb p]

I would approach this problem by first identifying the key variables and equations that are relevant to the situation. In this case, we have the mass of the freight car (m = 8000 kg), the spring constants (k1 = 1600 N/m and k2 = 3000 N/m), and the distances that the springs compress (x1 = 15.0 cm and x2 = 50.0 cm).

Next, I would consider the forces acting on the freight car. In this case, we have the force of gravity (mg) and the force exerted by the springs (F1 and F2). Since the car is coming to rest, we can assume that the net force is zero.

Using Hooke's law (F = -kx), we can write the following equations for the forces exerted by the springs:

F1 = -k1x1
F2 = -k2(x1 + x2)

Since the net force is zero, we can set these equations equal to each other:

-k1x1 = -k2(x1 + x2)

Solving for x2, we get:

x2 = (k1/k2)x1 - x1

Plugging in the values for k1, k2, and x1, we get:

x2 = (1600/3000)(0.15 m) - (0.15 m) = 0.05 m

Now, we can use the equation for potential energy (U = 1/2kx^2) to calculate the total potential energy stored in the springs:

U = 1/2(k1x1^2 + k2(x1 + x2)^2)

Plugging in the values, we get:

U = 1/2(1600 N/m)(0.15 m)^2 + 1/2(3000 N/m)(0.05 m)^2 = 7.5 J

This potential energy must be equal to the initial kinetic energy of the freight car, which we can calculate using the equation KE = 1/2mv^2. Setting these two equations equal to each other and solving for v, we get:

v = √(2U/m) = √(2(7.5 J)/(8000 kg)) = 0.05 m/s

Therefore, the initial speed of the freight car was