How Far Will the Ice Cube Travel Up the Slope After Being Released?

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The discussion revolves around a physics problem involving a 49 g ice cube sliding up a 31° slope after being released from a compressed spring. The user attempts to calculate the distance the ice cube travels using energy conservation principles but encounters issues with their calculations. Key points include the need to correctly account for the angle of the slope and the initial compression of the spring in the equations. There is confusion regarding whether to measure the distance from the compressed position or the relaxed position of the spring. Ultimately, the user is advised to ensure their calculations incorporate the correct parameters to resolve the problem accurately.
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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?

Homework Equations


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?
 
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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...

Physics13.jpg


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?
 
  • #10
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.
 
  • #11
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?
 
  • #12
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! :smile:

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

You need sin31º times (x + .119),

and (assuming the ice stays on the spring) you need kx2 - k(.119)2 :wink:
 
  • #13
he didnt have his C++ language right, aka he forgot to incorporate the degrees into the anweser. Thanks eveyrone
 
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