What is Bob's maximum power output as he throws the rock?

AI Thread Summary
Bob throws a 500 g rock at a speed of 30 m/s, moving his hand forward 1.0 m. He calculates the acceleration to be 450 m/s², resulting in a force of 225 N applied to the rock. The work done by Bob on the rock is determined to be 225 J. His maximum power output is calculated as 6750 W. The discussion also explores alternative methods for solving the problem, particularly using work-energy principles.
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Homework Statement


Bob can throw a 500 g rock with a speed of 30 m/s. He moves his hand forward 1.0 m while doing so.
a. How much work does Bob do on the rock?
b. How much force, assumed to be constant, does Bob apply to the rock?
c. What is Bob's maximum power output as he throws the rock?


Homework Equations


vf^2= vi^2 + 2ad
F=ma
W= F.d

P= F.v

The Attempt at a Solution



Find the acceration provided by the applied force.

vf^2= vi^2 + 2ad
30^2= 2a (1)
a= 450 m/s^2

Then find the force using Newton's second law:
F=ma
F= 450 (0.5)
F= 225 N

Now calculate work using the force applied.

W= F.d
W= 225 *1= 225 J

Power= F.v= 225*30= 6750 W.

Is this the correct way to do it? my textbook doesn't have solutions to even-numbered problems and I'm not sure if my way of calculating acceleration and then using Newton's second law is correct. Any help would be greatly appreciated!
 
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That looks just fine. You could have done it perhaps a bit quicker using work-energy methods (if you are familiar with this method), but there's always more than one way to attack a problem.
 
PhanthomJay said:
That looks just fine. You could have done it perhaps a bit quicker using work-energy methods (if you are familiar with this method), but there's always more than one way to attack a problem.

We apparently covered this method but the only work-energy equation that I know is: Ki+ Ui + Wext= Kf+ Uf+ delta E (thermal). Can you please guide me on how to set it up the work-energy way because I've seen other questions in the textbooks like that but I do them different.Thanks for your help =).
 
habibclan said:
We apparently covered this method but the only work-energy equation that I know is: Ki+ Ui + Wext= Kf+ Uf+ delta E (thermal). Can you please guide me on how to set it up the work-energy way because I've seen other questions in the textbooks like that but I do them different.Thanks for your help =).

Hi habibclan! :smile:

Just put delta E = 0.

So in this case the equation is KE = work times distance.

Try it! :smile:
 
tiny-tim said:
Hi habibclan! :smile:

Just put delta E = 0.

So in this case the equation is KE = work times distance.

Try it! :smile:

But I don't know work. I still don't get it :(.
 
habibclan said:
We apparently covered this method but the only work-energy equation that I know is: Ki+ Ui + Wext= Kf+ Uf+ delta E (thermal). Can you please guide me on how to set it up the work-energy way because I've seen other questions in the textbooks like that but I do them different.Thanks for your help =).
As tiny-tim noted, delta E =0. So using your equation, 0 + 0 + Wext = 1/2mv^2 + 0 + 0. You know v=30, and m= .5, solve for Wext (the work done by Bob on the rock). Then since Wext = Fd, solve for F.
 
Thanks a lot tiny-tim and PhantomJay! The help is much appreciated!
 
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