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Projectile Motion and Spring Question

  1. Nov 7, 2004 #1
    These are two questions from an assignment which I just don't get. I'm in grade 12. I just want to be pointed in the direction, not given the answer. I wanna make sure I know this for the test, my teachers hard :(

    1) A 1.00kg mass and a 2.00kg mass are set gently on a platform mounted on an ideal spring of force constant 40.0N/m. The 2.00kg mass is suddenly removed. How high above this starting position does the 1.00kg reach?

    I was thinking of determining the energy at equilibrium, because Ee = 0 and Ek = max at that point...but it would become zero, would it not? or, could i equal Ek1 + Ee1 = Ee2 + Ek2? I dunno...

    2) A ball of mass 0.25kg is thrown vertically upward from the roof of a building 18m high with a speed of 16m/s, and just misses the building on the way down
    a) To what vertical height above Earth does the ball rise?
    b) With what vertical velocity does the ball hit the ground?

    dy = v1t + -.5at^2? But I don't know time, or the vertical distance...I dunno where to begin, I used to be able to do this...but slipped my mind completly.

    Thanks for any help :D
     
  2. jcsd
  3. Nov 7, 2004 #2
    Law of Conservation of Energy : Q#2

    Gravitational potential energy calculated here is relative to the Earth's surface.
    Initially, the ball has kinetic energy and gravitational potential energy.
    [tex]E_{initial}=\frac{1}{2}mu^2+mgh[/tex]
    When the ball rises to its max height, the velocity of the ball is zero, so it doesn't have any kinetic energy but only gravitational potential energy, so
    [tex] E_{max height}=mgH[/tex]
    Law of Conservation of Energy requires
    [tex]E_{initial}=E_{max height}[/tex]
     
  4. Nov 7, 2004 #3
    wow...damn...that makes a lot of sense...thanks :D :D :D

    Anychance for #1?

    Thanks alot though :D
     
  5. Nov 7, 2004 #4
    Law of Conservation of Energy Q#1

    Potential energy calculated here is relative to the horizontal line where the center or mass of block 1.00 kg lies. Assume that the 1.00-kg block(m1) is beneath the 2.00-kg block(m2). Define our system to be consisted of the spring and the 1.00-kg block only.
    The initial energy of the system defined consisted of the spring potential energy only. No kinectic energy because m1 is not in motion and the gravitational potential energy of m1 is zero because the reference we have defined earlier.
    [tex]E_{intial}=\frac{1}{2}kx^2[/tex]
    When m1 rises to its max height, its velocity is zero and there is no spring potential energy because the spring is relax.
    [tex]E_{max height}=m_{1}gh[/tex]
    Law of conservation of energy requires
    [tex]E_{initial}=E_{max height}[/tex]
    To find x, you will have to analyze the system of m1,m2 and the spring in equilibrium.
     
  6. Nov 7, 2004 #5
    Before you're reply I tried doing two systems...A and B, A has the two masses, and B is the change and only the one mass. In both cases, Fnet = 0 because they both reach a new equilibrium, therefore Fg = Fx for both A and B. x = mg/k. Solve for both and subtract the two to find the change.

    Does that make any sense?
     
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