How Does a Mass and Spring System Behave on an Inclined Plane?

AI Thread Summary
The discussion centers on a mass-spring system on an inclined plane, focusing on how far a mass moves up the slope when released from a compressed spring. It involves applying the conservation of energy theorem, incorporating spring potential energy, gravitational potential energy, and kinetic energy. The first scenario examines the distance traveled by the mass when not attached to the spring, while the second scenario considers the mass attached to the spring. The third scenario introduces kinetic friction, requiring adjustments to the energy equations to account for energy loss. Overall, the key challenge is correctly applying energy conservation principles to solve for distances and coefficients of friction.
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Homework Statement


A spring (80 N/m) has an equilibrium length of 1.00 m. The spring is compressed to a length of 0.50 m and a mass of 2.2 kg is placed at its free end on a frictionless slope which makes an angle of 41 degrees with respect to the horizontal. The spring is then released. (Intro 1 figure)


If the mass is not attached to the spring, how far up the slope will the mass move before coming to rest?

If the mass is attached to the spring, how far up the slope will the mass move before coming to rest?

Now the incline has a coefficient of kinetic friction uk. If the block, attached to the spring, is observed to stop just as it reaches the spring's equilibrium position, what is the coefficient of friction uk?

1/2mv^2+mgh=1/2mv^2+mgh
1/2kx^2

This problem has been giving me a ton of trouble.

I am sure that it involves the conservation of energy theorm, ad the spring constant.

I just need help with he correct equations to be set in the right direction.
 
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besenji said:

Homework Statement


A spring (80 N/m) has an equilibrium length of 1.00 m. The spring is compressed to a length of 0.50 m and a mass of 2.2 kg is placed at its free end on a frictionless slope which makes an angle of 41 degrees with respect to the horizontal. The spring is then released. (Intro 1 figure)


If the mass is not attached to the spring, how far up the slope will the mass move before coming to rest?

If the mass is attached to the spring, how far up the slope will the mass move before coming to rest?

Now the incline has a coefficient of kinetic friction uk. If the block, attached to the spring, is observed to stop just as it reaches the spring's equilibrium position, what is the coefficient of friction uk?

1/2mv^2+mgh=1/2mv^2+mgh

This energy equation (for conservation of energy) does not have the spring potential energy in it. It's just an extra term for each side:

1/2mv^2+mgh + 1/2 k x^2=1/2mv^2+mgh + 1/2 k x^2

where the terms on the left are for the initial point, and the terms on the right are for the final point. When you apply this equation to the first question, what does this equation become? What do you get as the answer?


(For the third question, energy is not conserved. How will the equation change for that?)
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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