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

• CollegeStudent
In summary, the conversation discusses a low friction cart with a compressed spring being launched up an inclined track. The spring constant is found assuming frictionless conditions, and the energy transformations from start to finish are described.
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?

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?

CollegeStudent said:
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.

Doc Al said:
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?

CollegeStudent said:
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.

Doc Al said:
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!

Doc Al said:
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

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?

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?

CollegeStudent said:
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.

Doc Al said:
Try it one more time. Go back to your original equation.

argh x²!

K = 2(.4299716236)/(.0680²)

K = 92.99N/m

CollegeStudent said:
argh x²!

K = 2(.4299716236)/(.0680²)

K = 92.99N/m
Redo that arithmetic once more!

Doc Al said:
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!

OK, now you've got it.

CollegeStudent said:
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.

## What is work?

Work is a measure of energy transfer that occurs when a force is applied to an object and the object moves in the direction of the force. It is calculated by multiplying the magnitude of the force by the distance the object moves.

## What is spring energy?

Spring energy, also known as elastic potential energy, is the energy stored in a stretched or compressed spring. It is due to the spring's ability to return to its original shape after being deformed.

## How is work related to spring energy?

When a force is applied to a spring and the spring is compressed or stretched, work is done on the spring and it stores potential energy in the form of spring energy. As the spring returns to its original shape, that potential energy is converted back into kinetic energy, which can then be used to do work.

## What is Hooke's Law?

Hooke's Law is a physical law that states that the force needed to stretch or compress a spring is directly proportional to the distance the spring is stretched or compressed. This relationship is expressed mathematically as F = -kx, where F is the force, k is the spring constant, and x is the distance the spring is stretched or compressed.

## How is spring energy calculated?

Spring energy is calculated using the equation E = 1/2kx^2, where E is the spring energy, k is the spring constant, and x is the distance the spring is stretched or compressed. This equation is derived from Hooke's Law and represents the amount of work done on the spring to stretch or compress it.

• Introductory Physics Homework Help
Replies
17
Views
394
• Introductory Physics Homework Help
Replies
3
Views
406
• Introductory Physics Homework Help
Replies
5
Views
906
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
20
Views
2K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
5
Views
9K
• Introductory Physics Homework Help
Replies
6
Views
2K
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
6
Views
380