## zerosniper123 Group Title 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

1. nubeer Group Title

work done force * distance.

2. nubeer Group Title

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

3. zerosniper123 Group Title

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

4. nubeer Group Title

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 Group Title

|dw:1363361078018:dw|

6. amistre64 Group Title

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

7. amistre64 Group Title

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

8. amistre64 Group Title

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

9. amistre64 Group Title

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 Group Title

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

11. zerosniper123 Group Title

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

12. zerosniper123 Group Title

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

13. zerosniper123 Group Title

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

14. amistre64 Group Title

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

15. zerosniper123 Group Title

Okay...What exactly are the units?

16. amistre64 Group Title

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 Group Title

Ohhh okay thanks. :)

18. amistre64 Group Title

the units will still be pounds per inches

19. amistre64 Group Title

pounds per square inches that is

20. zerosniper123 Group Title

Okay thanks man! :D

21. zerosniper123 Group Title

So 15,750 pounds per square in.

22. amistre64 Group Title

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 Group Title

Okay so how would I change it to N ewtons?

24. amistre64 Group Title

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

25. zerosniper123 Group Title

Okay thanks anyways! :D

26. amistre64 Group Title

Work is normally expressed in Newtons per meter

27. amistre64 Group Title

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

28. zerosniper123 Group Title

Yeah it gave me this 2758247.65125 per Newton Meter

29. zerosniper123 Group Title
30. zerosniper123 Group Title

Thats the website I used to convert.

31. amistre64 Group Title

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

32. amistre64 Group Title

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

33. zerosniper123 Group Title

Okay so its x not *

34. amistre64 Group Title

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

35. amistre64 Group Title

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

36. zerosniper123 Group Title

Haha okay thanks :)

37. zerosniper123 Group Title

So the answer is 17,794.2Nm ;D