Question

Hooke's law

Answer 1

need help with the last part

Answer 2

need help setting up F(x) for work integral

Answer 3

Sorry can't help you on dis one :/

Answer 4

well thanks for looking

Answer 5

have you learnt calculus, integration

Answer 6

yeah i know calc but i need help setting up F(x)

Answer 7

I have not learnt it (the hooke's dynamics), but probably i could do it by calculus but it would be really hairy and there are some big formulae which you might not know , so sry!

Answer 8

Maybe @ganeshie8 or @Vincent-Lyon.Fr or @hartnn could help you

Answer 9

hooke's dynamics is linear it's just F=kx lol

Answer 10

F is the restorative force

Answer 11

yea i know f=kx lol but i did not not do that formally so i do not know how to apply that

Answer 12

in school i mean to say

Answer 13

K is the constant holding the spring in equilibrium and x is distance travelled on x axis

Answer 14

\[W=\int\limits_{0}^{x}kx\ dx=\frac{1}{2}kx ^{2}\]
this is the work done in stretching a spring from 0 to x.

Does that help?

Answer 15

It's a lot of stuff to type out, so here's a good site that explains it with examples.

http://tutorial.math.lamar.edu/Classes/CalcI/Work.aspx

Answer 16

@kropot72
you shouldn't use x as a limit of integration and the variable of integration ;)

Answer 17

x is 1/5 i believe

Answer 18

the formula helped forsure

Answer 19

In spite of the reservation by @Zarkon the definite integral is valid and can be further validated graphically by finding the area under the force curve, which is the work done in stretching the spring a distance x.

Answer 20

\[W=\int\limits_{0}^{d}kx\ dx=\frac{1}{2}kd ^{2}\]

this would be valid...but you can't use the same variable in the limit and as the variable of integration

Answer 21

@drbgonzal
Yes, the limit of integration is x=1/5m, as you have suggested in the first screen shot.
In any case, W=(1/2)kx^2 is used routinely in physics as a formula. Your x replaces the constant d in @Zarkon's formula. We're all talking about the same result. Zarkon is trying to explain to @kropot72 about mathematical conventions and he's correct.

Answer 22

The area under the force-displacement curve and the x-axis from x = 0 to x = x is the area of the hatched triangle with base x and height kx. The area is therefore:
\[\frac{1}{2}\times x \times kx=\frac{1}{2}kx ^{2}\]