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 2 years 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.
 2 years 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.

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nubeer
 2 years ago
Best ResponseYou've already chosen the best response.1work done force * distance.

nubeer
 2 years ago
Best ResponseYou've already chosen the best response.1lol forgot that equal after work done.. *work done =force * perpendicular distance.

zerosniper123
 2 years ago
Best ResponseYou've already chosen the best response.0Okay well can you help me start up the equation because I'm completely lost...:O

nubeer
 2 years ago
Best ResponseYou've already chosen the best response.1it 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
 2 years ago
Best ResponseYou've already chosen the best response.2dw:1363361078018:dw

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.2which is prolly redundant information, we already know the force and the distance that it moved.

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.2.... an additional 3 inches eh we would need to recalculate the Force then right?

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.2\[750=k(3)\] \[\frac{750}{3}=k\] \[F_{new}=\frac{750}{3}(3+3)\]right?

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.2since 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
 2 years ago
Best ResponseYou've already chosen the best response.2\[W=\int Fxd\] \[W=\int_a^b(kx)x~dx\] \[W=\int_{3}^{6}\frac{750}{3}x^2~dx\] \[W=\frac{750}{9}(6^33^3)\]

zerosniper123
 2 years ago
Best ResponseYou've already chosen the best response.0Okay and that is the work done :) Thanks man! :D

zerosniper123
 2 years ago
Best ResponseYou've already chosen the best response.0Why is W= 750/9 shouldnt it be 750/3?

zerosniper123
 2 years ago
Best ResponseYou've already chosen the best response.0then times by (6^33^3)

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.2no, you are prolly not familiar with integration if you ask that question

zerosniper123
 2 years ago
Best ResponseYou've already chosen the best response.0Okay...What exactly are the units?

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.2in effect: to integrate a variable, you want to increase the exponent by 1, and divide by the new exponent \[\int~x\ dx=\frac{x^2}{2}\]

zerosniper123
 2 years ago
Best ResponseYou've already chosen the best response.0Ohhh okay thanks. :)

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.2the units will still be pounds per inches

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.2pounds per square inches that is

zerosniper123
 2 years ago
Best ResponseYou've already chosen the best response.0Okay thanks man! :D

zerosniper123
 2 years ago
Best ResponseYou've already chosen the best response.0So 15,750 pounds per square in.

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.2Force 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

zerosniper123
 2 years ago
Best ResponseYou've already chosen the best response.0Okay so how would I change it to N ewtons?

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.2conversions arent in my memory .... might have to google a conversion :)

zerosniper123
 2 years ago
Best ResponseYou've already chosen the best response.0Okay thanks anyways! :D

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.2Work is normally expressed in Newtons per meter

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.2so newtons to pounds and inches to meters would be the conversion

zerosniper123
 2 years ago
Best ResponseYou've already chosen the best response.0Yeah it gave me this 2758247.65125 per Newton Meter

zerosniper123
 2 years ago
Best ResponseYou've already chosen the best response.0http://www.convertunits.com/from/pounds+per+inch/to/newton/metre

zerosniper123
 2 years ago
Best ResponseYou've already chosen the best response.0Thats the website I used to convert.

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.2\[\frac{1575lbs*in}{1}\frac{.0254m}{1in}*\frac{4.448N}{1lbs}=177.94~Nm\]

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.2my mistake was in the lbs/in,,, its lbs x inches

zerosniper123
 2 years ago
Best ResponseYou've already chosen the best response.0Okay so its x not *

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.2more like: \[\cancel{\frac{1575~lbs}{1in}}\to1575~lbs~in\] "per" is divisiion and we didnt divide .. we multiplied

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.2lol .... and i seemed to have dropped a 0 on the end, 15750

zerosniper123
 2 years ago
Best ResponseYou've already chosen the best response.0Haha okay thanks :)

zerosniper123
 2 years ago
Best ResponseYou've already chosen the best response.0So the answer is 17,794.2Nm ;D
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