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

In summary, the package rebounds back up the incline and gets very close to its original position. The speed before it reaches the spring is 8.50 m/s, and the maximum compression of the spring is .151 m.
  • #1
erik-the-red
89
1
A 2.00-kg package is released on a [tex]53.1 ^\circ[/tex] 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[tex] \mu_{s} \;=\; 0.40[/tex] and [tex]\mu_{k} \;=\; 0.20.[/tex] 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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  • #2
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 [tex]K_1+U_1+W_f=K_2+U_2[/tex]. But, [tex]U_1=K_1=0[/tex]. 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 [tex]K\Delta X=f_k+mgsin(\Theta)[/tex]. Solving for [tex]\Delta X[/tex], I got .151 m, which is also incorrect.

I don't know how to do part C.
 
  • #3
erik-the-red said:
I started off by breaking the force of weight into x and y components. I used the equation [tex]K_1+U_1+W_f=K_2+U_2[/tex]. But, [tex]U_1=K_1=0[/tex]. 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 [tex]K\Delta X=f_k+mgsin(\Theta)[/tex]. Solving for [tex]\Delta X[/tex], 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.
 
  • #4
For part a, I did [tex]K_2+U_2=W_f[/tex].

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

[tex]v_2=7.89[/tex] m/s.

But, that is also not correct.
 
  • #5
Show how you calculated the work done against friction.
 
  • #6
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!
 

1. What is the difference between final speed and compression in a kinetic-potential energy problem?

Final speed refers to the velocity of an object at the end of its motion, while compression refers to the decrease in volume or size of an object due to external forces. In a kinetic-potential energy problem, the final speed is determined by the conversion of potential energy to kinetic energy, while compression is a result of the transfer of energy between different forms.

2. How is final speed and compression calculated in a kinetic-potential energy problem?

The final speed can be calculated using the equation v = √(2gh), where v is the final speed, g is the acceleration due to gravity, and h is the height of the object. Compression can be calculated using the equation ΔV = V - V0, where ΔV is the change in volume, V is the final volume, and V0 is the initial volume.

3. Can the final speed and compression values be negative in a kinetic-potential energy problem?

Yes, the final speed and compression values can be negative in a kinetic-potential energy problem. A negative final speed indicates that the object is moving in the opposite direction of its initial motion, while a negative compression value indicates that the object has expanded rather than compressed.

4. How does the mass of an object affect the final speed and compression in a kinetic-potential energy problem?

The mass of an object affects the final speed and compression by influencing the amount of kinetic and potential energy that the object possesses. Heavier objects will have a greater potential energy and therefore a higher final speed when converted to kinetic energy. The mass of an object also affects compression, as a heavier object will resist changes in volume more than a lighter object.

5. Are there any real-life applications of final speed and compression in kinetic-potential energy problems?

Yes, there are many real-life applications of final speed and compression in kinetic-potential energy problems. Some examples include calculating the speed and compression of a roller coaster at different points in its track, determining the speed and compression of a projectile launched from a slingshot, and understanding the changes in speed and compression of a compressed spring when released.

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