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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.
 one year ago
 one year ago
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.
 one year ago
 one year ago

This Question is Closed

nubeerBest ResponseYou've already chosen the best response.1
work done force * distance.
 one year ago

nubeerBest ResponseYou've already chosen the best response.1
lol forgot that equal after work done.. *work done =force * perpendicular distance.
 one year ago

zerosniper123Best ResponseYou've already chosen the best response.0
Okay well can you help me start up the equation because I'm completely lost...:O
 one year ago

nubeerBest ResponseYou've already chosen the best response.1
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.
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
dw:1363361078018:dw
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
which is prolly redundant information, we already know the force and the distance that it moved.
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
.... an additional 3 inches eh we would need to recalculate the Force then right?
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
\[750=k(3)\] \[\frac{750}{3}=k\] \[F_{new}=\frac{750}{3}(3+3)\]right?
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
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
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
\[W=\int Fxd\] \[W=\int_a^b(kx)x~dx\] \[W=\int_{3}^{6}\frac{750}{3}x^2~dx\] \[W=\frac{750}{9}(6^33^3)\]
 one year ago

zerosniper123Best ResponseYou've already chosen the best response.0
Okay and that is the work done :) Thanks man! :D
 one year ago

zerosniper123Best ResponseYou've already chosen the best response.0
Why is W= 750/9 shouldnt it be 750/3?
 one year ago

zerosniper123Best ResponseYou've already chosen the best response.0
then times by (6^33^3)
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
no, you are prolly not familiar with integration if you ask that question
 one year ago

zerosniper123Best ResponseYou've already chosen the best response.0
Okay...What exactly are the units?
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
in effect: to integrate a variable, you want to increase the exponent by 1, and divide by the new exponent \[\int~x\ dx=\frac{x^2}{2}\]
 one year ago

zerosniper123Best ResponseYou've already chosen the best response.0
Ohhh okay thanks. :)
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
the units will still be pounds per inches
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
pounds per square inches that is
 one year ago

zerosniper123Best ResponseYou've already chosen the best response.0
Okay thanks man! :D
 one year ago

zerosniper123Best ResponseYou've already chosen the best response.0
So 15,750 pounds per square in.
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
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
 one year ago

zerosniper123Best ResponseYou've already chosen the best response.0
Okay so how would I change it to N ewtons?
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
conversions arent in my memory .... might have to google a conversion :)
 one year ago

zerosniper123Best ResponseYou've already chosen the best response.0
Okay thanks anyways! :D
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
Work is normally expressed in Newtons per meter
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
so newtons to pounds and inches to meters would be the conversion
 one year ago

zerosniper123Best ResponseYou've already chosen the best response.0
Yeah it gave me this 2758247.65125 per Newton Meter
 one year ago

zerosniper123Best ResponseYou've already chosen the best response.0
http://www.convertunits.com/from/pounds+per+inch/to/newton/metre
 one year ago

zerosniper123Best ResponseYou've already chosen the best response.0
Thats the website I used to convert.
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
\[\frac{1575lbs*in}{1}\frac{.0254m}{1in}*\frac{4.448N}{1lbs}=177.94~Nm\]
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
my mistake was in the lbs/in,,, its lbs x inches
 one year ago

zerosniper123Best ResponseYou've already chosen the best response.0
Okay so its x not *
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
more like: \[\cancel{\frac{1575~lbs}{1in}}\to1575~lbs~in\] "per" is divisiion and we didnt divide .. we multiplied
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
lol .... and i seemed to have dropped a 0 on the end, 15750
 one year ago

zerosniper123Best ResponseYou've already chosen the best response.0
Haha okay thanks :)
 one year ago

zerosniper123Best ResponseYou've already chosen the best response.0
So the answer is 17,794.2Nm ;D
 one year ago
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