How Do You Calculate the Spring Constant in a Frictionless System?

[CollegeStudent]

Homework Statement

A low friction cart has a spring plunger. The plunger and spring are compressed 6.80 cm and locked in place. The cart is launched from rest up an inclined track tipped at an angle of 13.5°. When the spring is released the cart travels 76.4 cm up the incline from its start position before rolling back down. (a) Assuming frictionless conditions, what is the spring constant of the spring if the cart has a mass of 246 g? (b) The cart rolls back down and is stopped by the spring. Will the spring compress more than, less than, or equal to the original compression of 6.80 cm. Give reasons for your answer. (c) Describe all the energy transformations that take place from start to finish.

Homework Equations

W = ΔK + ΔU_g + ΔU_s
K = F/Δx
1/2mv²
1/2kx²

The Attempt at a Solution

(a) Assuming frictionless conditions, what is the spring constant of the spring if the cart has a mass of 246 g?

So I started by using the work equation...we're assuming this system is frictionless so W = 0

there is no kinetic energy in this one so

1/2kx² = mgh

Since I'm solving for k I rearranged this like

k = (2mgh)/x²

the thing is...when I solved for h (height) I did the .764m * sin(13.5) ...but wouldn't that be what "x" is as well?

I know that when using

K = F / Δx ...this Δx would be the amount the spring is compressed here...but in that other equation I think it's the distance traveled


I'm just getting really confused here

any hints?

 

[CollegeStudent]

or actually...wouldn't "x" be the sum of the distance and the amount the spring is compressed? Because isn't "x" the distance from where the spring is relaxed?

 

[Doc Al]

or actually...wouldn't "x" be the sum of the distance and the amount the spring is compressed? Because isn't "x" the distance from where the spring is relaxed?

In this case, x is just the amount the spring was initially compressed. It's given. Not directly related to the distance traveled up the incline.

 

[CollegeStudent]

In this case, x is just the amount the spring was initially compressed. It's given. Not directly related to the distance traveled up the incline.

That's not clicking in my head...is there a reasoning behind that? Because every other time I've used that equation, the "x" was used as the distance. Is it because of the angle and the fact that there is another energy (Gravitational P.E) in the equation?

But okay so then it would be

K = 2(.178N) / .0680m = 5.25 N/m?

 

[Doc Al]

That's not clicking in my head...is there a reasoning behind that? Because every other time I've used that equation, the "x" was used as the distance. Is it because of the angle and the fact that there is another energy (Gravitational P.E) in the equation?

Realize that the cart is not attached to the spring. Once the spring uncompresses, the cart will keep on going without it.

But okay so then it would be

K = 2(.178N) / .0680m = 5.25 N/m?

Double check that calculation.

 

[CollegeStudent]

Realize that the cart is not attached to the spring. Once the spring uncompresses, the cart will keep on going without it.

Oh, that makes sense now!

Double check that calculation.

Well what I did was I actually got .178352258 for mgh...my professor taught us to write down the answers with significant figures but use the original calculation in equations ...so I did

K = 2(.178352258N) / .0680m

which gave me the 5.25N/m

 

[CollegeStudent]

Nvm sorry about that...only used the height in that mistake...so now adding in the mg portion...

K = 2(.4299716236)/.0680

K = 6.32N/m look better?

 

[CollegeStudent]

So now

(b) The cart rolls back down and is stopped by the spring. Will the spring compress more than, less than, or equal to the original compression of 6.80 cm.

While part (a) says assume frictionless conditions...the original question says "a low friction cart has a spring plunger" ...if the system is completely frictionless then the spring will compress more than it did with just the cart resting on the spring correct? Because now the fact that Kinetic energy would have played a factor in the system with some velocity...the spring would be compressed more. ?

(c) Describe all the energy transformations that take place from start to finish.
Well it goes from Spring P.E in the very beginning because all the energy is in the spring, to Kinetic Energy because the object will be moving, then it will go to Gravitational Potential Energy because the system will no longer be moving (at a point) and the cart is not connected to the spring...is this correct?

 

[Doc Al]

Nvm sorry about that...only used the height in that mistake...so now adding in the mg portion...

K = 2(.4299716236)/.0680

K = 6.32N/m look better?

Try it one more time. Go back to your original equation.

 

[CollegeStudent]

Try it one more time. Go back to your original equation.

argh x²!

K = 2(.4299716236)/(.0680²)

K = 92.99N/m

 

[Doc Al]

argh x²!

K = 2(.4299716236)/(.0680²)

K = 92.99N/m

Redo that arithmetic once more!

 

[CollegeStudent]

Redo that arithmetic once more!

Okay

K = (2mgh)/x²

mg = 2.4108
h = .764sin(13.5)

mgh = .4299716236
2mgh = .8599432472


***forgot the (*2) part***

x = .0680
x² = .004624

(2mgh) / x²

.8599432472 / .004624

185.97N/m

Sorry for those terrible mistakes!

 

[Doc Al]

OK, now you've got it.

 

[Doc Al]

So now

(b) The cart rolls back down and is stopped by the spring. Will the spring compress more than, less than, or equal to the original compression of 6.80 cm.

While part (a) says assume frictionless conditions...the original question says "a low friction cart has a spring plunger" ...if the system is completely frictionless then the spring will compress more than it did with just the cart resting on the spring correct? Because now the fact that Kinetic energy would have played a factor in the system with some velocity...the spring would be compressed more. ?

Think about it. Whatever energy the spring gives the cart is exactly the energy needed to compress the spring.

(c) Describe all the energy transformations that take place from start to finish.
Well it goes from Spring P.E in the very beginning because all the energy is in the spring, to Kinetic Energy because the object will be moving, then it will go to Gravitational Potential Energy because the system will no longer be moving (at a point) and the cart is not connected to the spring...is this correct?

Not bad. Let me tweak it a bit.

The gravitational PE increases continuously as the cart rises. The KE increases, then decreases.