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Hwk Problem: Mechanical Energy

  1. Apr 3, 2005 #1
    A 30 g bullet, with a horizontal velocity of 516 m/s, stops 19 cm within a solid wall.
    (a) What is the change in its mechanical energy?

    (b) What is the magnitude of the average force from the wall stopping it?

    For Part a, To find the change in mechanical energy, do I use the equation: ME=(1/2)mv^2 + mgh

    For Part b, I can use this equation F=m*a
     
  2. jcsd
  3. Apr 3, 2005 #2

    quasar987

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    a. yes

    b. yes

    -------------
     
  4. Apr 4, 2005 #3
    ME = (1/2)mv^2 + mgh
    (1/2)(mass)(velocity squared) + (mass)(gravity)(height)

    F = m*a
    Force = mass * acceleration

    use this to solve your problems
     
  5. Apr 4, 2005 #4
    For a) You should use conservation of energy. i think you can assume that the height of the bullet remains the same, so there is no change in potential energy mgh. basically the clue is that you come in with velocity 516 m/s and you stop after 0.19 m.

    The change in mechanical energy is calculated by realizing that the kinetic energy goes from 1/2mv² to 0

    For b) assume all horizontal motions. You can calculate the force that makes the object stop from this equation : change in kinetic energy = travelled distance * Force

    So you'd have 1/2*30*516² = 0.19 * F...Solve for F

    marlon
     
  6. Apr 4, 2005 #5
    Thanks guys, I got the answers and they were correct!!
     
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