Final speed and compression in a kinetic-potential energy problem?

AI Thread Summary
The discussion focuses on solving a kinetic-potential energy problem involving a package sliding down an incline and compressing a spring. The participants initially struggle with calculating the speed of the package just before it reaches the spring, with one user incorrectly arriving at 8.50 m/s. They also discuss the maximum compression of the spring, with attempts yielding incorrect results due to miscalculations involving friction. Ultimately, the user corrects their errors and successfully finds the right answers, emphasizing the importance of accurately calculating work done against friction. The conversation highlights the application of energy methods in solving physics problems.
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A 2.00-kg package is released on a 53.1 ^\circ incline, 4.00 m from a long spring with force constant 120 N/m that is attached at the bottom of the incline. The coefficients of friction between the package and the incline are\mu_{s} \;=\; 0.40 and \mu_{k} \;=\; 0.20. The mass of the spring is negligible.

1. What is the speed of the package just before it reaches the spring?

2. What is the maximum compression of the spring?

3. The package rebounds back up the incline. How close does it get to its initial position?
 

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I forgot that I violated the sticky rule. Sorry!

I started off by breaking the force of weight into x and y components. I used the equation K_1+U_1+W_f=K_2+U_2. But, U_1=K_1=0. I'm looking for the final velocity, which can be found in the final kinetic energy.

My answer was 8.50 m / s, but that was incorrect.

For the second problem, I used K\Delta X=f_k+mgsin(\Theta). Solving for \Delta X, I got .151 m, which is also incorrect.

I don't know how to do part C.
 
erik-the-red said:
I started off by breaking the force of weight into x and y components. I used the equation K_1+U_1+W_f=K_2+U_2. But, U_1=K_1=0. I'm looking for the final velocity, which can be found in the final kinetic energy.

My answer was 8.50 m / s, but that was incorrect.
The idea of using energy methods is fine. Show exactly what you did.

For the second problem, I used K\Delta X=f_k+mgsin(\Theta). Solving for \Delta X, I got .151 m, which is also incorrect.
When the spring is maximally compressed the force on the package will not be zero. Once again, use energy methods.
 
For part a, I did K_2+U_2=W_f.

So, (1/2)(2.00)(v_2)^2 + (2.00)(9.80)(-3.20) = -.47

v_2=7.89 m/s.

But, that is also not correct.
 
Show how you calculated the work done against friction.
 
Yeah, something was SERIOUSLY not right with that work done by friction.

I ended up getting the correct answer. I must have punched in something wrong in my calculator.

Thanks!
 
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