Calculate Work Done with 104 N Force on 26 kg Mass

[physicsgal]

"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 wouldn't be enough to move it. does that sound right?

~Amy

 

[radou]

"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 wouldn't 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.

 

[physicsgal]

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

 

[radou]

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.

 

[physicsgal]

k, looks like youre right. thanks for the help!

~Amy

 

[physicsgal]

another question. there's a force displacement graph. (quick vent) i asked the school that I am signed up with and they have no idea about the below question . 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". I am not sure what "possible outcomes" mean.

~Amy

 

[radou]

Could you make a sketch of the diagram and post it on a link?

 

[physicsgal]

i don't know 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

 

[radou]

Hm, the work should equal the area under the graph, but I don't understand the 'two possible outcomes' part, neither.

 

[physicsgal]

thanks anyways.

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

~Amy

 

[Spazolio]

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