Potential elastic energy problem in an incline

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SUMMARY

The discussion centers on calculating the maximum compression of a spring and the rebound distance of a 20.0 kg package on a 50° incline, utilizing a spring with a force constant of 150 N/m. The calculations incorporate the coefficient of kinetic friction (μ_k = 0.20) and apply the work-energy principle. The maximum compression of the spring is determined to be 3.69 m, while the rebound distance along the incline is calculated to be 5.83 m. Both procedures and results presented by the participants are confirmed as correct.

PREREQUISITES
  • Understanding of the work-energy principle in physics
  • Familiarity with spring mechanics and Hooke's Law
  • Knowledge of frictional forces and their calculations
  • Basic trigonometry for resolving forces on an incline
NEXT STEPS
  • Explore advanced applications of Hooke's Law in dynamic systems
  • Learn about energy conservation in systems with friction
  • Investigate the effects of varying spring constants on compression and rebound
  • Study the impact of different incline angles on motion dynamics
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Students and educators in physics, mechanical engineers, and anyone interested in understanding the dynamics of spring systems and inclined planes.

erik-the-red
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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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Both seem fine to me!
 
Thanks!

I checked today and they were correct.
 

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