anonymous
  • anonymous
I'm completely lost...can someone show me the steps on how to solve these... A force of 750 pounds compresses a spring 3 inches from its natural length of 15 inches. Find the work done in compressing the spring an additional 3 inches.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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nubeer
  • nubeer
work done force * distance.
nubeer
  • nubeer
lol forgot that equal after work done.. *work done =force * perpendicular distance.
anonymous
  • anonymous
Okay well can you help me start up the equation because I'm completely lost...:O

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nubeer
  • nubeer
it has told u 750 is the force which is not going to cahnge .. so now u just have to find the distance the spring is going to compressed.
amistre64
  • amistre64
|dw:1363361078018:dw|
amistre64
  • amistre64
which is prolly redundant information, we already know the force and the distance that it moved.
amistre64
  • amistre64
.... an additional 3 inches eh we would need to recalculate the Force then right?
amistre64
  • amistre64
\[750=-k(3)\] \[\frac{750}{3}=-k\] \[F_{new}=\frac{750}{3}(3+3)\]right?
amistre64
  • amistre64
since the force to move the spring varies across its distance, we could use integration to determine the total amount of work done .... if you believe that the question is asking for a more complicated result
amistre64
  • amistre64
\[W=\int Fxd\] \[W=\int_a^b(-kx)x~dx\] \[W=\int_{3}^{6}\frac{750}{3}x^2~dx\] \[W=\frac{750}{9}(6^3-3^3)\]
anonymous
  • anonymous
Okay and that is the work done :) Thanks man! :D
anonymous
  • anonymous
Why is W= 750/9 shouldnt it be 750/3?
anonymous
  • anonymous
then times by (6^3-3^3)
amistre64
  • amistre64
no, you are prolly not familiar with integration if you ask that question
anonymous
  • anonymous
Okay...What exactly are the units?
amistre64
  • amistre64
in effect: to integrate a variable, you want to increase the exponent by 1, and divide by the new exponent \[\int~x\ dx=\frac{x^2}{2}\]
anonymous
  • anonymous
Ohhh okay thanks. :)
amistre64
  • amistre64
the units will still be pounds per inches
amistre64
  • amistre64
pounds per square inches that is
anonymous
  • anonymous
Okay thanks man! :D
anonymous
  • anonymous
So 15,750 pounds per square in.
amistre64
  • amistre64
Force over a Distance .... nah, its just pounds per inch you might want to make it more presentable tho and convert it to Newtons of something
anonymous
  • anonymous
Okay so how would I change it to N ewtons?
amistre64
  • amistre64
conversions arent in my memory .... might have to google a conversion :)
anonymous
  • anonymous
Okay thanks anyways! :D
amistre64
  • amistre64
Work is normally expressed in Newtons per meter
amistre64
  • amistre64
so newtons to pounds and inches to meters would be the conversion
anonymous
  • anonymous
Yeah it gave me this 2758247.65125 per Newton Meter
anonymous
  • anonymous
http://www.convertunits.com/from/pounds+per+inch/to/newton/metre
anonymous
  • anonymous
Thats the website I used to convert.
amistre64
  • amistre64
\[\frac{1575lbs*in}{1}\frac{.0254m}{1in}*\frac{4.448N}{1lbs}=177.94~Nm\]
amistre64
  • amistre64
my mistake was in the lbs/in,,, its lbs x inches
anonymous
  • anonymous
Okay so its x not *
amistre64
  • amistre64
more like: \[\cancel{\frac{1575~lbs}{1in}}\to1575~lbs~in\] "per" is divisiion and we didnt divide .. we multiplied
amistre64
  • amistre64
lol .... and i seemed to have dropped a 0 on the end, 15750
anonymous
  • anonymous
Haha okay thanks :)
anonymous
  • anonymous
So the answer is 17,794.2Nm ;D

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