A spring is compressed by 90 cm

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A spring is compressed by 90 cm at the bottom of a frictionless hill, releasing a 2 kg block that reaches a height of 18 m with a speed of 6 m/s at the top. The energy conservation equation used is mgy = 1/2kx^2 + 1/2mv^2, but the correct interpretation is that the elastic energy equals the sum of potential and kinetic energy. The discussion also highlights a second part of the problem regarding the minimum compression needed for the block to reach the top without sliding down, which requires a different approach. Attempts to solve for the spring constant and the minimum compression faced challenges, indicating a need for clarity in energy conservation principles. The conversation emphasizes the importance of correctly applying energy equations to solve for spring constants and compression.
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Homework Statement



A spring is compressed by 90 cm at the bottom of a frictionless hill of height 18m. the spring is realeased and propels a 2 kg block over the hill. At the top of the hill, the block's speed is 6 m/s. Find the force constant (spring constant) of the spring.

Homework Equations



1/2kx^2, 1/2mv^2 and mgy

The Attempt at a Solution


mgy=1/2kx^2+1/2mv^2 I solved for K and it did no give me the right answer. what I'm I doing wrong is this the right set up?
 
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ScienceGeek24 said:

Homework Statement



A spring is compressed by 90 cm at the bottom of a frictionless hill of height 18m. the spring is realeased and propels a 2 kg block over the hill. At the top of the hill, the block's speed is 6 m/s. Find the force constant (spring constant) of the spring.

Homework Equations



1/2kx^2, 1/2mv^2 and mgy

The Attempt at a Solution


mgy=1/2kx^2+1/2mv^2 I solved for K and it did no give me the right answer. what I'm I doing wrong is this the right set up?

At the beginning, the whole energy of the system is in the elastic energy of the spring. When the block is at the top of the hill, the energy is in the potential energy of the block and the kinetic energy of the block. Thus, it's NOT the potential energy which equals the sum of the elastic energy and the kinetic energy as you have written BUT the elastic energy which equals the sum of the potential energy and the kinetic energy.
 
thanks man it worked!
 
ScienceGeek24 said:
thanks man it worked!

Glad to help.
 
But i forgot also to emphazie the second part of the question where is asking waht is the smallest compression of the spring that will allow the block to reach the op of the hill and not slide down?

I tried Fs=-kdeltax and it did not give me the right answer.

than i tried the same thing from part 1 solving for x and nope.
 
ScienceGeek24 said:
But i forgot also to emphazie the second part of the question where is asking waht is the smallest compression of the spring that will allow the block to reach the op of the hill and not slide down?

I tried Fs=-kdeltax and it did not give me the right answer.

than i tried the same thing from part 1 solving for x and nope.

The only difference to part 1 of the question is that now the speed of the block at the top of the hill is not 6 m/s but... how much?
 
om/s yess! thanks man!
 
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