I'm stcuk, sort of, not really, I need some resurance is all

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Homework Help Overview

The discussion revolves around a physics problem involving a dog jumping into a boat and the resulting kinetic energy dissipated. The subject area includes concepts of kinetic energy and conservation of momentum.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the calculation of kinetic energy before and after the dog jumps into the boat. Questions arise regarding the inclusion of the boat's mass and the application of conservation of momentum. Some participants express confusion about the correct approach to calculating the kinetic energy of the system.

Discussion Status

Several participants have provided guidance on using conservation of momentum and the need to consider both the dog and the boat in the kinetic energy calculations. There is an ongoing exploration of how to correctly compute the kinetic energy dissipated, with some participants expressing uncertainty about the calculations.

Contextual Notes

Participants discuss the implications of the mass of the boat and the final velocity of the combined system, indicating a need for clarity on these assumptions. There is also a mention of potential confusion regarding the subtraction of kinetic energies leading to negative values.

cooper43
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Homework Statement



Ive got another question that I think I might have failed at.

A dog weighing 250 kg jumps into an empty boat by the dock at a speed of 5m/s causing the boat to move away form the shore at 1.75m/s. Find the amount of kinetic energy dissipated.

Homework Equations



KE=1/2*m*v^2

The Attempt at a Solution



Before

KE=1/2*m*v^2
KE=(.5)(250kg)(5.0m/s)^2
KE(1)=3125

and after

KE=1/2*m*v^2
KE=(.5)(250)(1.75m/s)^2
KE(2)=382.813

KE1-KE2=2742.19

KE dissipated = 2742

So am I right?
 
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Welcome to PF!

cooper43 said:
A dog weighing 250 kg jumps into an empty boat by the dock at a speed of 5m/s causing the boat to move away form the shore at 1.75m/s. Find the amount of kinetic energy dissipated.

Hi cooper43! Welcome to PF! :smile:

You need to include the KE of the boat, so put the mass of the boat = M, and use conservation of momentum. :wink:
 
Is conservation of momentum P=mv?

And hat do you mean Mass of the boat. Should I leave it as M or do I replace it with the mass of the dog
 
cooper43 said:
Is conservation of momentum P=mv?

Nooo … conservation of momentum is mvi + MVi = mvf + MVf

ie total momentum before = total momentum after …

it is always true for a collision. :smile:
And what do you mean Mass of the boat …

Now you're just being ridiculous.

Stop asking questions and start answering the problem.
 
tiny-tim said:
Nooo … conservation of momentum is mvi + MVi = mvf + MVf

ie total momentum before = total momentum after …

it is always true for a collision. :smile:


Now you're just being ridiculous.

Stop asking questions and start answering the problem.

LMAO! Well said!

As Tiny-Tim said, Final Momentum = Initial Momentum

Therefore,

mv = mv and you can work out your mass from there if necessary...
 
Okay, so i did the m1v1i + M2V2i = mvf + MVf
and M2=464.286

The KE of the dog is 3125

but I am not sure how to gt the other KE

should it be the dog+boat and the final velocity?
 
yes, the final value of the kinetic energy is found by considering the mass of the boat and the dog and the final velocity. They act as a single object (theoretically)
 
So its KE1=3125 and KE2=1093.75

what do i subtract from what?

wouldnt it normaly be KE1-KE2,,,but it would end up negative
 
cooper43 said:
So its KE1=3125 and KE2=1093.75

what do i subtract from what?

wouldnt it normaly be KE1-KE2,,,but it would end up negative

never mind, that made me look like a dumb ***

I figured it out

The amount of kinetic energy dissipated is KE=2031.25
 

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