What Was the Initial Speed of the Freight Car?

[Meagan]

Homework Statement

A 5000-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 = 1700 N/m and k2 = 3000 N/m. After the first spring compresses a distance of 28.5 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

Homework Equations

I do not know.

The Attempt at a Solution

Completely wrong so please help!

 

[Nathanael]

The Attempt at a Solution

Completely wrong so please help!

Show your work anyway. Did you try using conservation of energy?

 

[Meagan]

I tried using F_s=-kx, then adding all of the forces together to find the work with W_s=∑F*Δr, and then using the work-kinetic energy theorem, I was not exactly sure how to set this equation up correctly, however, to find the initial velocity.

 

[Nathanael]

I tried using F_s=-kx, then adding all of the forces together to find the work, and then using the work-kinetic energy theorem to find the initial velocity.

It's a valid approach, but the force is not constant. You need to show your work so I can see where you go wrong.
Do it in two pieces; for the first 28 cm there is a certain force acting, and for the rest of the distance there is another force acting.

 

[Meagan]

I was thinking there was technically three forces, the first spring by itself, the second spring, and then the first spring again but over a different distance. I did: F₁=-1700 * 0.285m, F₂=-3000*0.165m and then F₃=-1700*0.165m. I left out the units for k to avoid confusing that anymore than necessary.

 

[Meagan]

Using this approach I found the net work to be 567 J. I am not really sure how to transition from this information to the work-kinetic energy theorem. I tried W=1/2 (k₁+k₂)v_f²-1/2 k₁v_i² and solving for initial velocity.

 

[lightgrav]

because the Force is not constant, the Work is ½ F_max x_max = ½ k s² ... where s is the spring stretch (or compression)

 

[Nathanael]

I was thinking there was technically three forces, the first spring by itself, the second spring, and then the first spring again but over a different distance. I did: F₁=-1700 * 0.285m, F₂=-3000*0.165m and then F₃=-1700*0.165m. I left out the units for k to avoid confusing that anymore than necessary.

You have the right idea. Using 0.165*(k₁+k₂) is the same as breaking it into two pieces like you did 0.165*k₁+0.165*k₂

But 1700*0.285 and (3000+1700)*0.165 is only the force at those distances! The force varies with distance... The formula Work = Force * distance only applies to constant forces... The correct formula is work = ∫F.dx

 

[Meagan]

because the Force is not constant, the Work is ½ F_max x_max = ½ k s² ... where s is the spring stretch (or compression)

And in this situation

You have the right idea. Using 0.165*(k₁+k₂) is the same as breaking it into two pieces like you did 0.165*k₁+0.165*k₂

But 1700*0.285 and (3000+1700)*0.165 is only the force at those distances! The force varies with distance... The formula Work = Force * distance only applies to constant forces... The correct formula is work = ∫F.dx

Ok. I will try again after my next class. Thank you for your help and sorry for my initial mess up not showing my work. This was my first post lol.

 

[Meagan]

You have the right idea. Using 0.165*(k₁+k₂) is the same as breaking it into two pieces like you did 0.165*k₁+0.165*k₂

But 1700*0.285 and (3000+1700)*0.165 is only the force at those distances! The force varies with distance... The formula Work = Force * distance only applies to constant forces... The correct formula is work = ∫F.dx

I do not understand how to use that formula in this situation. I only remember how to integrate when there is a function.

 

[Nathanael]

I do not understand how to use that formula in this situation. I only remember how to integrate when there is a function.

The function is F=-k₁x for the first 28.5 cm and F=-(k₁+k₂)x for the remaining distance.

In other words, the integral (for the work) is the area under the graph in post #1 from zero to 45 cm.

 

[Meagan]

Well I got it wrong . That's ok though, I think I know what I did wrong! Thank you for all of your help!