What is the Force Constant k of a Spring?

[ImmaGrump]

Homework Statement

A 4.50 ✕ 10^5 kg subway train is brought to a stop from a speed of 0.500 m/s in 0.900 m by a large spring bumper at the end of its track. What is the force constant k of the spring?

M- 4.50E5 kg
IV- .5m/s
FV- 0m/s

Homework Equations

k=mg/x
F=-kx
PE = (.5)kx^2

The Attempt at a Solution

k=(450000kg)(9.81m/s^2)/ .9 = 4900000, which isn't the right answer. Can someone help me out?

 

[Princu]

Have you read anything about the concept of potential energy of a spring?This question is directly based on that..
The kinetic energy of the train is being converted in the potential energy of spring?

 

[ImmaGrump]

Well I know PE = (.5)kx^2, but I am afraid I don't know how to plug in the numbers.

 

[Hollumber]

What's the intial kinetic energy (KE) of the train?
KE will = 0 when v = 0 (why is this true?)
Due to conservation of potential energy, initial(KE+PE) = final(KE+ E)

 

[Nickie]

The force constant, k, of a spring is a measure of its stiffness or how much force is required to stretch or compress the spring by a certain distance. It is typically measured in units of Newtons per meter (N/m).

To solve this problem, we can use the formula F=-kx, where F is the force exerted by the spring, k is the force constant, and x is the distance the spring is stretched or compressed. In this scenario, the force exerted by the spring is equal to the force required to bring the train to a stop, which is the product of its mass and acceleration. So we can write:

F = ma = (4.50E5 kg)(0.5 m/s^2) = 225000 N

Substituting this into the formula F=-kx, we get:

225000 N = -k(0.900 m)

Solving for k, we get:

k = 225000 N / 0.900 m = 250000 N/m

So the force constant of the spring is 250000 N/m.