Potential elastic energy problem in an incline

In summary, the conversation discusses the calculation of the maximum compression and rebound distance of a 20.0 kg package released on a 50 ^\circ incline, 4.50 m from a long spring with force constant 150 N/m. The coefficient of kinetic friction between the package and the incline is \mu_k = .20 and the mass of the spring is negligible. The first part uses the equation -f_k(d+x) = (1/2)(k)(x^2) - (mg)(d+x)(sin(\Theta)) to solve for the maximum compression, which is found to be 3.69 m. The second part uses the equation K_1 + U_1 + W_n =
  • #1
erik-the-red
89
1
Question
A [tex]20.0[/tex] kg package is released on a [tex]50 ^\circ[/tex] incline, [tex]4.50 m[/tex] from a long spring with force constant [tex]150[/tex] N/m that is attached at the bottom of the incline. The block hits the spring, compresses the spring, and finally bounds back. The coefficient of kinetic friction between the package and the incline is [tex]\mu_k = .20[/tex]. The mass of the spring is negligible

1. Calculate the maximum compression of the spring by the block.

2. How far does the block rebound back along the incline?

For the first part, I used work done by nonconservative force (W_nc) = [tex]E_f - E_i[/tex].

This translates into [tex]-f_k(d+x) = (1/2)(k)(x^2) - (mg)(d+x)(sin(\Theta))[/tex].
Plugging in known values, noting that x is the unknown results in: [tex]-(.20)(20.0)(9.80)(cos(50 ^\circ)(4.50 + x)=(1/2)(150)(x^2) - (20)(9.80)(4.50+x)(sin(50 ^\circ)[/tex].

Evaluating, [tex]-25.2(4.5+x)=75x^2 - 150(4.5+x)[/tex]
[tex]0=75x^2 - 124.8(4.5+x)[/tex]
[tex]x_1 = 3.69[/tex] m.

Is this procedure correct?

2. [tex]K_1 + U_1 + W_n = K_2 + U_2[/tex]
[tex]W_n=U_2 - U_1[/tex].
[tex]-25.2x = (20.0)(9.80)(xsin(50 ^\circ) - (1/2)(150)(3.69)^2[/tex]
[tex]-25.2x=150x - 1021[/tex]
[tex]1021=175.2x[/tex]
[tex]x=5.83[/tex] m

Is this procedure correct?
 
Last edited:
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  • #2
Both seem fine to me!
 
  • #3
Thanks!

I checked today and they were correct.
 

Related to Potential elastic energy problem in an incline

1. What is potential elastic energy?

Potential elastic energy is the energy stored in a stretched or compressed elastic material. It is a form of potential energy, meaning it has the potential to do work.

2. How is potential elastic energy related to an incline?

In an incline, potential elastic energy is related to the height and angle of the incline. As an elastic material is stretched or compressed along the incline, it gains or loses potential elastic energy based on its position in relation to the ground.

3. What causes potential elastic energy to change in an incline?

The change in potential elastic energy in an incline is caused by the change in shape or position of the elastic material. As the incline angle increases, the potential elastic energy also increases.

4. How is potential elastic energy calculated in an incline?

The formula for calculating potential elastic energy in an incline is PE = mghsinθ, where PE is potential energy, m is the mass of the object, g is the acceleration due to gravity, h is the height of the incline, and θ is the angle of the incline.

5. What are some real-life examples of potential elastic energy in an incline?

Some examples of potential elastic energy in an incline include a springboard used in diving, a catapult, a rubber band being pulled back, and a bungee cord being stretched before a jump.

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