# Ice pressed against a spring

• talaroue
In summary, a 49 g ice cube can slide without friction up and down a 31° slope. The ice cube is pressed against a spring at the bottom of the slope, compressing the spring 11.9 cm. The spring constant is 25 N/m. When the ice cube is released, it will travel up the slope before reversing direction.

## Homework Statement

A 49 g ice cube can slide without friction up and down a 31° slope. The ice cube is pressed against a spring at the bottom of the slope, compressing the spring 11.9 cm. The spring constant is 25 N/m. When the ice cube is released, what distance will it travel up the slope before reversing direction?

Pi=Pf
U=mgx, kx^2/2
K=mv^2/2

## The Attempt at a Solution

Initially U=k*x^2/2
then at the end U=mgLsin31
so then i combined the 2 equations and solve for L and got L=(k*x^2)/(2*.049*9.8*sin(31))
it is saying its wrong...what am i doing wrong?

Looks OK to me, assuming the distance is measured from the compressed position.

Yeah that is right if you are considering from compressed pos. If it is from relaxed position, you might have to subtract 0.119. Is the answer 0.597m?

yeah thas my anwser but its wrong.

I tried both .7157 and when you subtract .597 and they both are wrong.

Last edited:
bump... I still don't know what I am doing wrong

I have one try left? FML

here is a copy of my work...

Ui=(delta x^2)*k/2
Uf=mgL

solved for L= (delta x^2)*k/2mg

As I said earlier, there's nothing wrong with your work. But online systems can be flaky. Since they gave the compression in units of cm, perhaps that's the units they want.

Did you post the entire problem word for word exactly as it was given?

talaroue said:
Ui=(delta x^2)*k/2
Uf=mgL

solved for L= (delta x^2)*k/2mg
For some reason you chose to find the vertical distance instead of the distance up the slope. I'd say your first post was correct.

o ok I am almost positive here ill copy and paste it again...

A 49 g ice cube can slide without friction up and down a 31° slope. The ice cube is pressed against a spring at the bottom of the slope, compressing the spring 11.9 cm. The spring constant is 25 N/m. When the ice cube is released, what distance will it travel up the slope before reversing direction?

talaroue said:
A 49 g ice cube can slide without friction up and down a 31° slope. The ice cube is pressed against a spring at the bottom of the slope, compressing the spring 11.9 cm. The spring constant is 25 N/m. When the ice cube is released, what distance will it travel up the slope before reversing direction?

Initially U=k*x^2/2
then at the end U=mgLsin31
so then i combined the 2 equations and solve for L and got L=(k*x^2)/(2*.049*9.8*sin(31))
it is saying its wrong...what am i doing wrong?

Hi talaroue!

(try using the X2 tag just above the Reply box )

You need sin31º times (x + .119),

and (assuming the ice stays on the spring) you need kx2 - k(.119)2

he didnt have his C++ language right, aka he forgot to incorporate the degrees into the anweser. Thanks eveyrone

## 1. What happens to a spring when ice is pressed against it?

When ice is pressed against a spring, the spring will compress and become shorter in length. This is because the ice exerts a force on the spring, causing it to deform and store potential energy.

## 2. Will the spring return to its original length when the ice is removed?

Yes, the spring will return to its original length when the ice is removed. This is because the potential energy stored in the spring is released and the spring will regain its original shape.

## 3. How does the temperature of the ice affect the behavior of the spring?

The temperature of the ice does not affect the behavior of the spring significantly. The main factor that affects the behavior of the spring is the force applied by the ice, which is dependent on the weight and pressure of the ice.

## 4. Can the spring break when ice is pressed against it?

It is possible for the spring to break when ice is pressed against it, depending on the strength and durability of the spring. If the force applied by the ice is greater than the spring's breaking point, then the spring may break.

## 5. Is there a limit to how much the spring can compress when ice is pressed against it?

Yes, there is a limit to how much the spring can compress when ice is pressed against it. This limit is determined by the strength and elasticity of the spring. If the force applied by the ice exceeds this limit, the spring will either break or reach a maximum compression point and will not compress further.