Quick work question

  • Thread starter physicsgal
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  • #1
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"a force of 104 N is applied to a 26 kg mass at an angle of 32 degrees to the horoizontal. calculate the work done when the mass is moved through a horizontal distance of 5 m."

my answer is zero because 26kg times 9.8 m/s^2 = 255 N, so a 104N force wouldnt be enough to move it. does that sound right?

~Amy
 

Answers and Replies

  • #2
radou
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physicsgal said:
"a force of 104 N is applied to a 26 kg mass at an angle of 32 degrees to the horoizontal. calculate the work done when the mass is moved through a horizontal distance of 5 m."

my answer is zero because 26kg times 9.8 m/s^2 = 255 N, so a 104N force wouldnt be enough to move it. does that sound right?

~Amy

Since there is no friction mentioned, the force should move the mass. It is not the same as lifting the mass. In order to lift it, you should have a vertical force which is greater than the weight of the mass. So, the force F = 104 * cos32 is doing work along a distance of 5 m. Unless I'm missing something big here. :biggrin:
 
  • #3
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that was my first answer F = 104 * cos32 = 441 J.

i dunno. it just seems like the 26 kg should be included somehow.

can anyone else shed some insight?

~Amy
 
  • #4
radou
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physicsgal said:
that was my first answer F = 104 * cos32 = 441 J.

i dunno. it just seems like the 26 kg should be included somehow.

can anyone else shed some insight?

~Amy

Maybe they're included to confuse you. :smile:
 
  • #5
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k, looks like youre right. thanks for the help!

~Amy
 
  • #6
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another question. there's a force displacement graph. (quick vent) i asked the school that im signed up with and they have no idea about the below question :eek:. they have these chat rooms that take for ever to get any help and more than 50% of the time with my questions they are no help at all. today i was lucky and only wasted a half hour waiting (usually it takes an hour or more). but if you send an email they'll give you a good answer, but it takes 5 business days. grrrr...

part a starts at 40N, 0 displacement in m
then it goes to 40N at 4 m
=160J

then from 40N it slopes at 60N by 8 m
= 200J

so the total = 360J

and the question asks "state two possible outcomes of the work (above) being done on the object". im not sure what "possible outcomes" mean.

~Amy
 
  • #7
radou
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Could you make a sketch of the diagram and post it on a link?
 
  • #8
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i dunno how links work. but i'll try to sketch it here just using typing.

N
60...................... /
..................../
40------- /

20

0......2.....4.... 6....8 (displacement in m)

imagine the /s are a straight normal slope..

k, this is the best drawing i can do.

~Amy
 
Last edited:
  • #9
radou
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Hm, the work should equal the area under the graph, but I don't understand the 'two possible outcomes' part, neither.
 
  • #10
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thanks anyways. :smile:

if anyone else can help, it'd be appreciated!

~Amy
 
  • #11
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Its because the graph starts from 10N of force instead of 0. which means if you were to count the squares, instead of using the formula W=F*D , you will notice that you are not able to count from 0-10 N because its not included on the graph, thus changing the outcome
 

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