Calculating spring equilibrium length HW question HARD

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Calculating spring equilibrium length HW question.... HARD!!!!

Homework Statement


A spring is connected to the bottom of a ramp inclined at an angle of 45 degrees. In its equilibrium position, the spring is equal to the length of the ramp, L. If the spring constant is 500 N/m and I want to launch a 5 kg ball, 100 meters as measured from the base of the ramp, how long must the ramp be if my initial compression of the spring is equal to 1/2 the length of the ramp. AND. friction on the ramp cannot be ignored.... the coefficient of friction is .25


Homework Equations


I know that the spring potential equals 1/2kx^2 and the Work due to friction will equal mgcos45*L. The Total Energy at the top of the ramp will be equal to the Energy from the Spring - Work done by friction



The Attempt at a Solution



I think I understand the concepts of the problem, but I just don't know how to putit all together... You can find the Spring potential and subtract the Work done by friction to give you an energy at the top of the ramp... This energy can be converted to Kinetic energy of the ball by equaling it to 1/2mv^2. Then you can solve for a velocity. This velocity will have an x and y component which then can be used to determine the time in the y direction until it hits the ground and then this time can be used to find the range in the x direction. BUT AGAIN I can't get it all put together.
 

Answers and Replies

  • #2
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Divide it:
1)what should be the speed of the body at the top of the plane so it will reach 100m?
use kinematic when the initial Height is L*cos45.
2)find the velocity in terms of L using energy.
3) velocity in terms of L found in 2 must be equal to velocity found in 1 to reach 100 m exactly.
good luck

Edit:
1 more thing:
work of friction will be mgcos45*L/2 and not mgcos45*L.
 
  • #3
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What about this approach:
The force acting on the spring is
F = k * x
or μk* m * g = k * x
or μk* m * g = k * (L/2)
or L = (2μk* m * g/k)
L is the length of the ramp.


It looks to simple... like I am missing a key concept in this one.... Is friction accounted for correctly on the ramp?
 
  • #4
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Try to explain for yourself everything You do in your approach, try my approach, then we can discuss other approaches.
 
  • #5
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ok I thought I knew how to do the first step but i have tried many times....

To find the velocity at the top of the ramp you need to break it up into its components... First find the y velocity then the time then you can find the x velocity if you know the range (100m) and the time.

I keep getting stuck on the y velocity.
If I use the equation yFinal = yinitial + Vinitial y*t + 1/2at^2 to find the time it take to go up and come down to the same elevation of the ramp... Will not work because I have 2 unknowns.. time and initial y velocity.

If I use the equation Vfy^2 = Viy^2 +2a(change in y) I don't get an answer that works to find initial y velocity

square root of -(2(-9.81)(Lcos45)

= sqr of 18.8L

this number would be my initial y velocity and knowing that the ramp is 45 degrees the x and y components are the same so this would equal my x velocity as well. ) so the launch velocity would equal 18.8L/cos 45 or 26.6L.

Am i even close?
 
  • #6
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Yes You are very close.
Yfinal==0, he is on the ground, y initial You can found throuhh the triangle(the plane) and using trigonometry.
than You can find t in terms of Vinitial(y direction) , and substitute it in the X direction
X=Vinitial(x direction) *t
now the velocities are
in X direction: V*cos(a)
in y direction : V*sin(a)
the angle is the angle of the ramp , cause it's being shot from the edge of it
"the spring is equal to the length of the ramp, L".
and You'll find the wanted velocity of L.
and now use energies with known final velocity(final as to the ramp, it's actually the initial velocity we found).
good luck
EDIT:
Didn't see You wrote the answe of the initial velocity, let me check by myself.
 
  • #7
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Vfy^2 = Viy^2 +2a(change in y)
what did you submit here for Vfy?
btw the question isnt that easy, lots of math needed/
 
Last edited:
  • #8
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wouldn't Vfy be 0? hitting the ground right?
 
  • #9
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Assume You are right,
You can throw things at people and they will hit them with zero velocity and they won't get hurt, sounds right or wrong?
 
  • #10
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ha ha definitely wrong....
 
  • #11
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using trig the height of the ramp is Lcos45, so that would be my yinitial... but how do you find t in term of velocity?
 
  • #12
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Well, yes You're right about the initial height.
now the velocity is kinda hard to achieve.
You derived to function that looks like this:
ax^2+bx+c , right? when t is your X,so basically it's at^2 +bt +c find the roots of this function(function=0 to get roots) they will be kinda nasty...
after you've done with that, place t in the X function( Vo*t*cos45=100)
and you'll get some thing really ugly for velocity.
now the easier part IMO is the energy, but be careful there.
 
  • #13
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yea i got down to that before.. Its definatly nasty but I still have 2 unknowns.....
0= Lcos45 + Vinital*t + 4.905t^2.

I know how to do the quadradic equation, but I still have V initial and T to solve. Is there anything else I can substitute in for V initial?
 
  • #14
174
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solve for t in terms of V and L, than in the formula for the horizontal displacement :
X=Vinitial*cos45*t=100
substitute the t with the t found in terms of V, solve for V in terms of L.
this is the required initial velocity.

than take a look @ the ramp and write the energy equations
 

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