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Answer Check: A golf ball is hit and hits a bee. find the work done on the bee

  1. Nov 11, 2012 #1
    1. The problem statement, all variables and given/known data

    pbd3e.png

    2. Relevant equations

    F=ma
    W= Fdcosθ

    3. The attempt at a solution

    I first found the acceleration:
    t= d/v
    t= 37m/28m/s = 1.32s
    a = v/t
    a = 28m/s / 1.32s
    a= 21.21 m/ss

    Second, the force:
    Fclub - Fgsinθ = ma
    Fclub = a +gsinθ
    Fclub = 21.21m/ss + 9.8m/ss sin84
    Fclub = 30.96N

    Then I plugged in:
    W=(30.96N)(37m)cos84
    W=119.75J

    I'm pretty sure this is correct. only thing I'm worried about really is the acceleration.
     
  2. jcsd
  3. Nov 11, 2012 #2
    You have to use potential and kinetic energy conservation to solve this problem.
     
  4. Nov 11, 2012 #3

    haruspex

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    That calculation is completely wrong. 28m/s is the initial speed of the ball, and it is at some angle to the vertical. The flight time to reach the bee would be 1.32s if the ball were travelling straight up at 28m/s and not slowed by gravity along the way.
    What acceleration do you think this is calculating?
    It is impossible to calculate the force from the club since you do not know how long the impact lasted. You can calculate the impulse (momentum), but that doesn't help with this question.
     
  5. Nov 12, 2012 #4
    So, since we can ignore resistance, the energy is conserved and the potential energy at the beginning equals the kinetic energy at the end.

    PEi = KEf
    mgy = 1/2mv2
    v = sqrt(2gy)
    v = sqrt(2 * 9.8m/s2 * 37m)
    v= 26.93m/s
     
  6. Nov 12, 2012 #5

    haruspex

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    This isn't a stone being dropped from rest. Total energy is conserved here means PEi + KEi= PEf + KEf
    What is KEi?
     
  7. Nov 12, 2012 #6
    KEi = 0 because the initial velocity is 0. And would make 1/2mv2 = 0

    And PEf = 0 because y is the vertical distance to move the object. we dont need to move the object because its already at its finaly position of 37m. If wrong on this and I should use 37m for y and solve for v, I end up with 0
     
  8. Nov 12, 2012 #7
    The initial kinetic energy is not zero. The ball has some initial speed, namely 28 m/s.

    It's usually more convenient to have the starting, not final position have position and potential energy 0. This means that the final position is at y=37m and a potential energy of mgy.

    You know the initial potential and kinetic energies, and the final potential energy. The only thing you don't know is the final kinetic energy. Solve for this and then determine the speed.
     
  9. Nov 12, 2012 #8
    Ok, I understand this and I've come up with:
    1/2mvi2 = mgyf + 1/2mvf2
    vf = sqrt [ vi2 - 2(gyf) ]
    vf = 7.67m/s
     
  10. Nov 12, 2012 #9
    There you go. You've got it.
     
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