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 Foo Off-Topic chit chat with no general subject.

 12-15-06, 04:08 PM #1 phantomcow2 la vache fantôme Thread Starter     Join Date: Aug 2004 Location: NH Bikes: Posts: 6,266 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 1 Post(s) Work done on the book Okay I had a test question today, was unsure what the question meant. Basically here it is: A person applies 35N at a 25degree angle on a book, sliding it for 40m. What is the total work done on the book? One possible answer was simply 1400joules, or 35*40. But because it is being applied at a 25 degree angle to the horizontal, that to me means not all of the force the person exerts is going toward moving the book, some is just pushing downward on the earth. Some of it is In fact, 31.172N is being applied and actually moving the book forward. This means the total work done is 1268.83Joules Was I correct? __________________ C://dos C://dos.run run.dos.run
12-15-06, 04:20 PM   #2
PCS2
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 Originally Posted by phantomcow2 But because it is being applied at a 25 degree angle to the horizontal, that to me means not all of the force the person exerts is going toward moving the book, some is just pushing downward on the earth.
yes, some of the force is just pushing the book down. Just break up the force into it's x and y components. Fx is the only one that does work.

Sounds like you did, but I didn't check the results.

 12-15-06, 04:20 PM #3 phantomcow2 la vache fantôme Thread Starter     Join Date: Aug 2004 Location: NH Bikes: Posts: 6,266 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 1 Post(s) Thats what I thought. Thanks for teh assurance . __________________ C://dos C://dos.run run.dos.run
12-15-06, 04:23 PM   #4
Tom Stormcrowe
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 Originally Posted by PCS2 yes, some of the force is just pushing the book down. Just break up the force into it's x and y components. Fx is the only one that does work. Sounds like you did, but I didn't check the results.
Sounds suspiciously like a LaGuerre polynomial in for! Factor, young lad!
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 12-15-06, 04:28 PM #5 skiahh Senior Member     Join Date: Aug 2003 Location: CO Springs, CO Bikes: 08 Stumpjumper FSR Expert, 02 Litespeed Tuscany, 04 Specialized S-Works Epic Posts: 1,035 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) I'm definately not a physicist, but if you're applying the force directly to the book, wouldn't the first answer be correct? I mean, I understand the logic behind it taking the 31.172N to move the book the 40m, but you're still applying the total force (35N) the entire time, right? Sounds like a trick question to me.
12-15-06, 04:28 PM   #6
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 Originally Posted by Tom Stormcrowe Sounds suspiciously like a LaGuerre polynomial in for! Factor, young lad!
eh?

much faster to use components......infact, should be able to get a rough estimate w/o a calc in seconds.....

ie w=Fx(D)

Fx=35 cos 25 ~~ 35 * 0.9 ~~ 31.5

31.5 N * 40m ~ 1260 J.

No prob

12-15-06, 04:30 PM   #7
phantomcow2
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 Originally Posted by skiahh I'm definately not a physicist, but if you're applying the force directly to the book, wouldn't the first answer be correct? I mean, I understand the logic behind it taking the 31.172N to move the book the 40m, but you're still applying the total force (35N) the entire time, right? Sounds like a trick question to me.
Well I see what you mean about applying the total force for 50 meters. But, the question specifically asked "Work done on the book".
I think that this means only the force that actually moves the book counts. If they had said, the total work done by the person applying the force, then it would be 1400
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12-15-06, 04:30 PM   #8
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 Originally Posted by skiahh I'm definately not a physicist, but if you're applying the force directly to the book, wouldn't the first answer be correct? I mean, I understand the logic behind it taking the 31.172N to move the book the 40m, but you're still applying the total force (35N) the entire time, right? Sounds like a trick question to me.
Yeah, I see what you are saying, but go back to the definition of work, Work = force times distance, so if you break up the force into it's two components, Fx and Fy, the x force is the only one that provides movement. If you apply 10000000 N of force to an object straight down into the ground and the object doesn't move, the total work you have done (by defn) is zero.

Does that help?

12-15-06, 04:32 PM   #9
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 Originally Posted by phantomcow2 If they had said, the total work done by the person applying the force, then it would be 1400
No, this would be incorrect, remember, w = F x D, no distance moved, no work, again, by the definition. If you pushed on a brick wall with 100 N (provided the wall didn't move of course), your total work is zero.

12-15-06, 04:43 PM   #10
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 Originally Posted by phantomcow2 Okay I had a test question today, was unsure what the question meant. Basically here it is: A person applies 35N at a 25degree angle on a book, sliding it for 40m. What is the total work done on the book? One possible answer was simply 1400joules, or 35*40. But because it is being applied at a 25 degree angle to the horizontal, that to me means not all of the force the person exerts is going toward moving the book, some is just pushing downward on the earth. Some of it is In fact, 31.172N is being applied and actually moving the book forward. This means the total work done is 1268.83Joules Was I correct?
Well can't help ya there not one of my strong points.