Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

zerosniper123

  • 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.

  • This Question is Closed
  1. nubeer
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    work done force * distance.

  2. nubeer
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  3. zerosniper123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  4. nubeer
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 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.

  5. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1363361078018:dw|

  6. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  7. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  8. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  9. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 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

  10. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  11. zerosniper123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  12. zerosniper123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  13. zerosniper123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  14. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  15. zerosniper123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Okay...What exactly are the units?

  16. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    in effect: to integrate a variable, you want to increase the exponent by 1, and divide by the new exponent \[\int~x\ dx=\frac{x^2}{2}\]

  17. zerosniper123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Ohhh okay thanks. :)

  18. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    the units will still be pounds per inches

  19. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    pounds per square inches that is

  20. zerosniper123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Okay thanks man! :D

  21. zerosniper123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So 15,750 pounds per square in.

  22. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 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

  23. zerosniper123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Okay so how would I change it to N ewtons?

  24. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  25. zerosniper123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Okay thanks anyways! :D

  26. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Work is normally expressed in Newtons per meter

  27. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  28. zerosniper123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yeah it gave me this 2758247.65125 per Newton Meter

  29. zerosniper123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  30. zerosniper123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thats the website I used to convert.

  31. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  32. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  33. zerosniper123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Okay so its x not *

  34. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  35. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  36. zerosniper123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Haha okay thanks :)

  37. zerosniper123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So the answer is 17,794.2Nm ;D

  38. Not the answer you are looking for?
    Search for more explanations.

    Search OpenStudy
    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.