# Homework Help: Quick work question

1. Sep 25, 2006

### 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 wouldnt be enough to move it. does that sound right?

~Amy

2. Sep 25, 2006

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.

3. Sep 25, 2006

### 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

4. Sep 25, 2006

Maybe they're included to confuse you.

5. Sep 25, 2006

### physicsgal

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

~Amy

6. Sep 25, 2006

### physicsgal

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 . 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. Sep 25, 2006

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

8. Sep 25, 2006

### physicsgal

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: Sep 25, 2006
9. Sep 25, 2006

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

10. Sep 25, 2006

### physicsgal

thanks anyways.

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

~Amy

11. Apr 2, 2008

### 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