Work Problem (easy but weird)

  • Thread starter fomenkoa
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  • #1
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This seems easy but I'm not getting the right answer!

The question: A 0.5 kg ball is thrown into the air. At a height of 20m above the ground, it is travelling at 15 m/s.

a)How much work was done by someone at ground level throwing the ball up into the air


Ok, for a), Work=Force x Distance
so,

W = (0.5 kg x 9.81 N/kg) x 20 m
= 98 J

But that's not the answer in my textbook! I realize my substitution for "Force" is wrong but I dont know why

Anton
 

Answers and Replies

  • #2
dextercioby
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That work (which is asked) is nothing but the initial KE...Find it.

Daniel.
 
  • #3
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What is the answer your book got?

Wouldn't your force be [tex]0.5kg[/tex] x [tex]15m/s^2=7.5N[/tex]

[tex]7.5N[/tex] x [tex]20m=150J[/tex]

*Note, just a guess. I probably don't know the right answer either. I really should have taken a physics class this year.

Just want to try to be helpful.
 
  • #4
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Another thing... remember that once the person's hands leave the ball, the only force acting on it is gravity (and air resistance, but that's negligable). Remember the nice rule that [tex]\bigtriangleup{K.E.} = W[/tex]
 
  • #5
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Lucretius said:
What is the answer your book got?

Wouldn't your force be [tex]0.5kg[/tex] x [tex]15m/s^2=7.5N[/tex]

[tex]7.5N[/tex] x [tex]20m=150J[/tex]

*Note, just a guess. I probably don't know the right answer either. I really should have taken a physics class this year.

Just want to try to be helpful.

Yes! The answer is 150 J! However, Force=mass x accel
You said that the accel is 15 m/s/s.....why?? Isnt that the speed? Why is the speed and accel the same?
 
  • #6
dextercioby
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Shouldn't it be [itex] 156.25 \ \mbox{J} [/itex]...?

Daniel.
 
  • #7
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fomenkoa said:
Yes! The answer is 150 J! However, Force=mass x accel
You said that the accel is 15 m/s/s.....why?? Isnt that the speed? Why is the speed and accel the same?

I got something right? Wow! Confidence boost.

[tex]15m/s[/tex] would be the speed. However, [tex]15m/s^2[/tex] would be acceleration.

Speed can be meters per second, but acceleration the change of the the rate of speed, which would be [tex]15m/s^2[/tex].
 
  • #8
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dextercioby said:
Shouldn't it be [itex] 156.25 \ \mbox{J} [/itex]...?

Daniel.
My book says 1.5e2 J...I guess thats to 2 significant digits...but can anyone explain to me why the 15 m/s became a 15 m/s/s?
 
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  • #9
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Lucretius said:
I got something right? Wow! Confidence boost.

[tex]15m/s[/tex] would be the speed. However, [tex]15m/s^2[/tex] would be acceleration.

Speed can be meters per second, but acceleration the change of the the rate of speed, which would be [tex]15m/s^2[/tex].

But....to know accel, wouldnt you need the initial velocity which we dont know? I still dont understand :redface:
 
  • #10
dextercioby
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If you mean [tex] 1.5 \cdot 10^{\mbox{2}} [/tex] ,then i believe you.

You have to realize that u're being asked to find the work done to imprime a certain velocity to the ball.And that work is exactly the initial KE of the ball.U'll have to use the theorem of varation of KE (or the law of the total mechanical energy) to find it.

It's not difficult at all.

Daniel.
 
  • #11
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dextercioby said:
If you mean [tex] 1.5 \cdot 10^{\mbox{2}} [/tex] ,then i believe you.

You have to realize that u're being asked to find the work done to imprime a certain velocity to the ball.And that work is exactly the initial KE of the ball.U'll have to use the theorem of varation of KE (or the law of the total mechanical energy) to find it.

It's not difficult at all.

Daniel.
Oups yeah I meant 1.5e2 ....so

Ek = 1/2 x m x v^2
I know the mass, but not the initial velocity!
 
  • #12
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fomenkoa said:
But....to know accel, wouldnt you need the initial velocity which we dont know? I still dont understand :redface:

[tex]A=\frac{\delta v}{\delta t}[/tex].

So your book says [tex]15m/s[/tex], and not [tex]15m/s^2?[/tex]
 
  • #13
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Are you sure thye answer is supposed to be 150J? I get alittle over 154J.
 
  • #14
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Lucretius said:
[tex]A=\frac{\delta v}{\delta t}[/tex].

So your book says [tex]15m/s[/tex], and not [tex]15m/s^2?[/tex]

Right. It says that at 20m, it is travelling at 15 m/s
 
  • #15
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jdavel said:
Are you sure thye answer is supposed to be 150J? I get alittle over 154J.

It's to 2 significant digits (rounding)
Oh boy I still dont know how to solve this problem :confused:
 
  • #16
dextercioby
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Lucretius,please.You're iincorrect.As to the OP,i already said what u should be doing...

Daniel.
 
  • #17
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dextercioby said:
Lucretius,please.You're iincorrect.As to the OP,i already said what u should be doing...

Daniel.

:frown: what'd I do wrong?
 
  • #18
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Lucretius said:
:frown: what'd I do wrong?
That's what I wanna know too,,I still dont get it
 
  • #19
dextercioby
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There's no indication of any force acting on the ball during flight,therefore,it's natural to assume that the only force doing work on the ball while in motion is the force of earth's gravitational attraction.

Daniel.
 
  • #20
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fomenkoa,

What Lucretius did to solve this makes no sense; he lucked out getting close to the right answer (no offense Lucretius!).

Dextercioby is steering you the right way, pay attention to him.
 
  • #21
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I guess I shouldn't attempt to help people anymore, heh.

Alright, bye.
 

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