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Third Law Problem

  1. Oct 9, 2013 #1
    The problem statement, all variables and given/known data
    A diver with a weight of 833 N dives from a boat with a mass of 375kg. If he leaves the boat with a velocity of 2.25m/s [W] after accelerating for 0.50s, what will be the final velocity of the boat?

    The attempt at a solution
    ad=Vfd-Vid/t
    =2.25-0/0.5
    ad=4.5m/s^2 [W]

    Fg=mdg
    833=m(9.8)
    md=85kg

    Fnet=mdad
    Fg+Fd=mdad
    833+Fd=-450.5

    Fb+Fd=0
    Fb= -Fd
    Fb= 450.5N

    Fb=mbab
    450.5=(375)(ab)
    ab=1.26m/s^2

    ab=Vfb-Vib/t
    1.26=Vf-0/0.50
    Vf= 0.63m/s

    Is this correct?
     
  2. jcsd
  3. Oct 9, 2013 #2

    D H

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    Staff Emeritus
    Science Advisor

    No, that's not correct. You should ignore gravitational force here. Assume the diver leaves the boat with a horizontal velocity of 2.25 m/s. The only thing for which you need gravitational acceleration is the relationship between weight and mass.
     
  4. Oct 9, 2013 #3
    Fd=mdad
    =(85)(4.5)
    Fd=382.5 N

    Fb= -382.5 N

    Fb=mb(-ab)
    -382.5=(375)(-ab)
    (ab)= 1.02 m/s^2

    ab= Vf-Vi/t
    1.02=Vf-0/0.50
    Vfb= 0.51m/s

    So I would do this then?

    Also, if the direction of the velocity was not given to be [W] then would I use the gravitational force?
     
    Last edited: Oct 9, 2013
  5. Oct 9, 2013 #4

    D H

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    Staff Emeritus
    Science Advisor

    You went from 382.5 N to 385.5 N. That's a mistake you carried through to your final answer.
     
  6. Oct 9, 2013 #5
    Thanks, fixed my solution. But if the diver was not travelling horizontally and the direction west was not stated. Then would gravitational force apply?
     
  7. Oct 9, 2013 #6

    D H

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    Staff Emeritus
    Science Advisor

    You have to know the direction. Velocity, acceleration, and force are vectors. Moreover, this is a 3rd law problem. The diver's interaction with the boat and with the Earth are two different interactions. You shouldn't mix them up.
     
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