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Physics force and work problem

  1. Mar 15, 2016 #1
    1. The problem statement, all variables and given/known data
    A 65-kg worker at a bakery loses his balance and falls 4.0 m before hitting the surface of a large vat of cake batter. He continues to travel downwards an additional 2.0 m before the cake batter finally brings him to rest. Calculate the work done on the worker by the cake batter.

    2. Relevant equations
    F = ma
    ΔKE = Wnet
    W = Fscosθ
    PEgravity = mgh

    3. The attempt at a solution
    ΔKE = KEfinal - KEinitial = 0 - KEinitial ⇒ KEintitial = -(1/2)mv^2
    PEgravity = mgh = (65 kg)(9.80 m/s^2)(4.0 m) = 2548 J
     
  2. jcsd
  3. Mar 15, 2016 #2

    gneill

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    Staff: Mentor

    Consider that the worker is still falling (moving vertically) through the cake batter until he comes to rest.
     
  4. Mar 15, 2016 #3
    So,
    F = ma = (65 kg)(9.80 m/s^2) = 637 N
    W = Fscosθ = (637 N)(6.0 m)cos180° = -3822 J (6.0 as the displacement considering his whole movement?)
     
  5. Mar 15, 2016 #4

    gneill

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    Staff: Mentor

    That will work. The basic premise is that the work falls a total of 6 m through the gravitational field and so gains energy Mgh from the change in gravitational potential energy. That total amount of energy is "lost" to the cake batter over distance that he travels through the batter.
     
  6. Mar 15, 2016 #5
    Was my answer right or do I still have to factor in the energy he lost while going through the batter?
     
  7. Mar 15, 2016 #6

    gneill

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    Staff: Mentor

    Your answer was fine. The energy lost to the batter is the energy gained from falling. That's why the worker comes to rest in the batter. That energy represents the work done by the batter. The only way you could improve your result would be to make sure that its stated to the correct number of significant figures.
     
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