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Third Law Problem

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  • #1
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Homework Statement
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?
 

Answers and Replies

  • #2
D H
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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.
 
  • #3
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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.
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:
  • #4
D H
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You went from 382.5 N to 385.5 N. That's a mistake you carried through to your final answer.
 
  • #5
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You went from 382.5 N to 385.5 N. That's a mistake you carried through to your final answer.
Thanks, fixed my solution. But if the diver was not travelling horizontally and the direction west was not stated. Then would gravitational force apply?
 
  • #6
D H
Staff Emeritus
Science Advisor
Insights Author
15,393
683
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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