Ask your own question, for FREE!
Mathematics 19 Online
OpenStudy (anonymous):

A spring requires a force of 25 N to produce an extension of 2 cm in it. What is the work done in further extending it by 3 cm.

OpenStudy (zion):

its physics, i guess,.

OpenStudy (amistre64):

its application of integration; work section

OpenStudy (anonymous):

25.21 / 4

OpenStudy (amistre64):

\[\lim\sum F(x)\triangle x\]

OpenStudy (anonymous):

c'mon amistre.. he can use formulae as well. k=25/2 and the work done he needs is \[1\div2 \times K \times (a ^{2} - b ^{2})\] a is 5 & b is 2. or else, worry about the integration of k*x .

OpenStudy (amistre64):

F(x) = 25x \[\int_{2}^{3}(25x)dx\] maybe?

OpenStudy (amistre64):

they can yes; but wheres the fun in that :)

OpenStudy (anonymous):

for the record it is a riemann sum as △x approaches zero. thats like the rectangles becoming small enuff that their edges approximate the curve really well. the curve here is a straight line.

OpenStudy (amistre64):

25(3^2)/2-25(2^2)/2

OpenStudy (anonymous):

the problem isnt fun anyway, its in other word the area under a line of given slope b/w tweo points. or the area of a trapezium. :)

OpenStudy (amistre64):

25(9-5)/2 = 25(2) = 50 perhaps?

OpenStudy (amistre64):

messed it up lol

OpenStudy (amistre64):

25(9-4)/2 lol 25(5)/2 = 125/2 might be better

OpenStudy (anonymous):

LOL! All the possible answers are between the numbers 1 and 3 and are in J, joules.

OpenStudy (amistre64):

i dont know the conversions right off hand :)

OpenStudy (amistre64):

50+12+.5 = 62.5 is the answer I get; but thats prolly in newton-centimeter

OpenStudy (amistre64):

a Joule is: 1 newton is displaced through a distance of 1 metre

OpenStudy (amistre64):

.03^2 = .0009; ... 25(.0005)/2 = .00625 http://www.wolframalpha.com/input/?i=int%2825x%29dx+from+.02+to+.03

OpenStudy (amistre64):

CAN YOU POST YOUR OPTIONS?

OpenStudy (anonymous):

http://www.onlineconversion.com/force.htm

OpenStudy (amistre64):

IM GONA BREAK OFF THE CAPS KEY LOL

OpenStudy (anonymous):

Every time I post my options, my questions get deleted

OpenStudy (amistre64):

.625 still aint between 1 and 3

OpenStudy (anonymous):

exactly.

OpenStudy (amistre64):

if you get it wrong, will it tell you the right answer?

OpenStudy (anonymous):

No :/

OpenStudy (anonymous):

These are my options 1.3125J 1.876J 2.7121 2.85J

OpenStudy (amistre64):

close your eyes, use the force, scream at the top of your lungs while clicking your best guess ....

OpenStudy (anonymous):

Dont delete my post now :P

OpenStudy (anonymous):

lmao. thats not going to help me much lol. The force isnt always the answer

OpenStudy (amistre64):

im thinking i missed a part; something to do with x-someting

OpenStudy (anonymous):

Oh the horror. Where's einstein when you need him?! Oh wait its amistre :)

OpenStudy (amistre64):

25x(.02+x) maybe? gotta re read up some stuff

OpenStudy (anonymous):

ohh.. okay.. Dont take too long :)

OpenStudy (amistre64):

25 = 2x F(x) = 25x/2 might be my undoing :)

OpenStudy (amistre64):

to get it into newtons tho lets adjust it as this 25x^2 ------ [.02,.03] .04 25(.03^2)/.04 - 25(.02^2)/.04 = .3125 .... doh!! re reading ...

OpenStudy (amistre64):

read thru the problem and make sure you havent left out anything or typoed it please

OpenStudy (amistre64):

the best i can come up with the given information is .3125

OpenStudy (amistre64):

the measurements in the stated problem are simply to small to come up with anything substantive to me

OpenStudy (anonymous):

Ohhh I see what you're saying. I guess I'll just go with that answer.

OpenStudy (anonymous):

thanks :)

OpenStudy (amistre64):

im gonna see if i can interpret this; "What is the work done in further extending it by 3 cm." to mean the over all work done is .5625 J so it cant go higher than that ... srry

OpenStudy (anonymous):

Curiousmath, I posted on this in the physics forum, let me know if you still have questions :)

OpenStudy (amistre64):

i still have questions :)

OpenStudy (anonymous):

I posted it in the physics section too, but that section isnt as popular as this one :)

OpenStudy (anonymous):

what's up amistre?

OpenStudy (amistre64):

i just looked at your post over there; and its the same information i keep coming up with; so im gonna assume the question is just bad :)

OpenStudy (anonymous):

I don't think the answers given are correct, if that's what you are referring to. :)

OpenStudy (amistre64):

yep ;)

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!