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.

- anonymous

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- nubeer

work done force * distance.

- nubeer

lol forgot that equal after work done..
*work done =force * perpendicular distance.

- anonymous

Okay well can you help me start up the equation because I'm completely lost...:O

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## More answers

- 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

|dw:1363361078018:dw|

- amistre64

which is prolly redundant information, we already know the force and the distance that it moved.

- amistre64

.... an additional 3 inches eh we would need to recalculate the Force then right?

- amistre64

\[750=-k(3)\]
\[\frac{750}{3}=-k\]
\[F_{new}=\frac{750}{3}(3+3)\]right?

- 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

\[W=\int Fxd\]
\[W=\int_a^b(-kx)x~dx\]
\[W=\int_{3}^{6}\frac{750}{3}x^2~dx\]
\[W=\frac{750}{9}(6^3-3^3)\]

- anonymous

Okay and that is the work done :) Thanks man! :D

- anonymous

Why is W= 750/9 shouldnt it be 750/3?

- anonymous

then times by (6^3-3^3)

- amistre64

no, you are prolly not familiar with integration if you ask that question

- anonymous

Okay...What exactly are the units?

- 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

Ohhh okay thanks. :)

- amistre64

the units will still be pounds per inches

- amistre64

pounds per square inches that is

- anonymous

Okay thanks man! :D

- anonymous

So 15,750 pounds per square in.

- 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

Okay so how would I change it to N
ewtons?

- amistre64

conversions arent in my memory .... might have to google a conversion :)

- anonymous

Okay thanks anyways! :D

- amistre64

Work is normally expressed in Newtons per meter

- amistre64

so newtons to pounds and inches to meters would be the conversion

- anonymous

Yeah it gave me this 2758247.65125 per Newton Meter

- anonymous

http://www.convertunits.com/from/pounds+per+inch/to/newton/metre

- anonymous

Thats the website I used to convert.

- amistre64

\[\frac{1575lbs*in}{1}\frac{.0254m}{1in}*\frac{4.448N}{1lbs}=177.94~Nm\]

- amistre64

my mistake was in the lbs/in,,, its lbs x inches

- anonymous

Okay so its x not *

- amistre64

more like:
\[\cancel{\frac{1575~lbs}{1in}}\to1575~lbs~in\]
"per" is divisiion and we didnt divide .. we multiplied

- amistre64

lol .... and i seemed to have dropped a 0 on the end, 15750

- anonymous

Haha okay thanks :)

- anonymous

So the answer is 17,794.2Nm ;D

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