Answer:
Work done in stretching the spring = 7.56 lb-ft.
Step-by-step explanation:
Normal length of the spring = 8 in or
ft
=
ft
If the spring has been stretched to 11 inch then the stretched length of the spring is = 11 in
=
ft
Force applied to stretch the spring = 12 lb
By Hook's law,
F = kx [where k is the spring constant and x = length by which the spring is stretched]
12 = k(
)
k = 18
Work done (W) to stretch the spring by
ft will be
W = 
= 
= ![18[\frac{x^{2}}{2}]^{\frac{11}{12}}_0](https://tex.z-dn.net/?f=18%5B%5Cfrac%7Bx%5E%7B2%7D%7D%7B2%7D%5D%5E%7B%5Cfrac%7B11%7D%7B12%7D%7D_0)
= 9(
)²
= 7.56 lb-ft
5.1 i think. is it mutipul question
Answer:
I just had this problem on a test! I hope this helps, i will attach a picture of my messy work, but hopefully that will give you an idea on how do do it!
How to Solve!-
Repeat this distance formula for each side
You just need to plug those expressions inside the formula: it doesn't matter if they're expressions involving a variable instead of plain numbers: the formula becomes

If you want, you can simplify it by expanding the square and then multiply the two parenthesis:
