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Mass on incline plane sliding into spring, need to find max compression! 
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#1
Jun712, 08:49 PM

P: 15

1. The problem statement, all variables and given/known data
A 11 kg box slides 4.0 m down the frictionless ramp shown in the figure , then collides with a spring whose spring constant is 190 N/m. The angle of the ramp is 30°. What is the maximum compression of the spring? 2. Relevant equations Ei=mgh Ef=1/2kx^2 3. The attempt at a solution Ei=Ef mgh=1/2kx^2 mgLsinθ=1/2kx^2 11(9.8)(4sin30)=1/2(190)x^2 1.5m = Max compression I swear this is right but it's not, can anyone help me out? 


#2
Jun712, 08:58 PM

P: 294

Is the spring laying on the ramp, or is there a flat part before the mass hits the spring, making the spring horizontal?
Also, does the 4m refer to the distance it slid along the ramp, or the vertical displacement? 


#3
Jun712, 09:03 PM

P: 15

The spring is laying on the ramp and the 4m refers to the distance it slid along the ramp.



#4
Jun712, 09:08 PM

P: 294

Mass on incline plane sliding into spring, need to find max compression!
You made the statement that the energy of the block 4 meters above the uncompressed spring is equal to the energy of the compressed spring, but that is leaving out a portion of energy.
From the point when the block is released to the time is stopped (at the maximum compression point of the spring) what is its change in height? 


#5
Jun712, 09:16 PM

P: 15

Ohhhhhhhhh!!!
So i should do: 1/2(4+Xmax)sinθ=1/2k(Xmax)^2 ? 


#6
Jun712, 09:17 PM

P: 294

The left side should read 11(9.8) instead of (1/2), but besides that yes.



#7
Jun712, 09:27 PM

P: 15

Whoops! I guess I was so excited to have finally figured it out that I wrote the equation wrong lol.
To verify mg(4+Xmax)sinθ=1/2k(Xmax)^2 So... 1/2k(Xmax)^2  mgsinθ(Xmax)  mg4sinθ = 0 And then use quadratic equation to solve for Xmax? 


#8
Jun712, 09:29 PM

P: 294

That looks right to me.
I wonder what the nature of the roots will be of that equation? 


#9
Jun712, 09:34 PM

P: 15

Got it! Xmax is equal to 1.8m
Thanks for the help Villyer! 


#10
Jun712, 09:37 PM

P: 294

Of course



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