Conservative force help

  • Thread starter dralion87
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  • #1
dralion87
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a bullet with a mass of .02kg and a velocity of 800m/s is fired horizontally into a tree. the tree exerts a resistive force and brings the bullet to rest after it penetrates .56m into the tree. given this information fins the constant resistive force exerted by the tree on the bullet and the power generated in bringing the bullet to a halt...
 

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  • #2
rock.freak667
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Where is your attempt?

and think energy.
 
  • #3
dralion87
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i been on this problem for the past 3 hours man/// i need help please please please
 
  • #4
rock.freak667
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In your 3 hour attempt, what methods did you try?

Did you try considering energy as I posted before?
 
  • #5
dralion87
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i don't know how to set up the problem i know that

(1/2)mv^2 + mgy + (1/2)kx^2= (1/2)mv^2 + mgy + (1/2)kx^2

so what do i do
 
  • #6
rock.freak667
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Initially the bullet has kinetic energy,[itex]\frac{1}{2}mv^2[/itex]. When it hits the tree, all that k.e. is used to cause the bullet to move a distance of 0.56m into the tree.

So Work done by tree= change in Kinetic energy of bullet.

Do you know a formula for Work done by a force,F, which moves it point of application a distance,s?
 
  • #7
dralion87
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is it (1/2)MV^2 - (1/)MV^2
 
  • #8
rock.freak667
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is it (1/2)MV^2 - (1/)MV^2

[itex]\frac{1}{2}mv^2-\frac{1}{2}mu^2[/itex] is the change in kinetic energy.


Do you know the definition of http://en.wikipedia.org/wiki/Mechanical_work" [Broken]
 
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  • #9
dralion87
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know i do after i read that thing from wikipedia... then wat do i do
 
  • #10
rock.freak667
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Fs=change in ke


you know change in ke. s=0.56. Find F
 
  • #11
dralion87
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i did (1/2)(.02)(800)^2 / (.56) =11428
 
  • #12
rock.freak667
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i did (1/2)(.02)(800)^2 / (.56) =11428

Yep.
Now how do you find power with a force and the velocity of 800m/s?
 
  • #13
dralion87
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aw man i don't know...
 
  • #14
rock.freak667
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P=Fv.


You do know these formulas right?
 
  • #15
dralion87
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yeah but i can't use those beacuse the teacher said to use the principles of energy of conservation
 
  • #16
rock.freak667
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I think you'll need to use P=Fv though.
 
  • #17
dralion87
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i did and i get 9142400 a huge number...but i won't get any credit if don't use the conservation of energy principles...
 
  • #18
dynamicsolo
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I think you'll need to use P=Fv though.

This won't help much here -- since the velocity is not constant over the path, you would have to perform an integration to find either an average power or a function for power. (The problem probably wants the average power, but in the OP's course, they probably want the students to use kinematics to find the time it takes to stop the bullet.)
 
  • #19
dynamicsolo
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i did (1/2)(.02)(800)^2 / (.56) =11428

All right, the conservation of energy equation you are using is

KE_init + W_non-conservative = KE_final ,

with the non-conservative work being given by

W_non-con = -F · (delta_x) .

Since the final KE is zero, you have

W_non-con = -KE_init and

-F · (delta_x) = -KE_init gives F = KE_init / delta_x .

You did this and found F = 11,430 Newtons.

That is the part you can do by conservation of energy principles. To find the power involved in stopping the bullet, you need to use the definition of power

P = W / delta_t , where delta_t is the time over which the force you found has acted.

You know the initial speed (800 m/sec) and final speed (0 m/sec) of the bullet. If you can find the acceleration that acted on the bullet while the trunk of the tree was stopping it, you can find the time it took to bring the bullet to rest. You also now know the force that acted on the bullet. Can you find the bullet's acceleration?

(Actually, you're getting close: the number you gave in post #17 is off by a factor of about 2...)
 
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