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Potential elastic energy problem in an incline

 
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Oct19-05, 08:04 PM   #1
 

Potential elastic energy problem in an incline


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?
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Oct19-05, 09:58 PM   #2
 
Both seem fine to me!
Oct20-05, 08:14 PM   #3
 
Thanks!

I checked today and they were correct.
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