# Math Help - Physics Spring problem. Work is shown!

1. ## Physics Spring problem. Work is shown!

Suppose that 2 Joules of work is needed to stretch a spring from its natural length of 35 cm to a length of 54 cm. How much work is needed to stretch it from 50 cm to 55 cm?

First change cm to m.
W = the integral from Xi to Xf of kx dx

so W=(1/2)k(Xf^2) - (1/2)k(Xi^2)
So solve for k
2=(1/2)(k(.54^2) - k(.35^2))
4=k(.1691)
k=23.6546

them solve for W with your new k value but with a different distance
W=(23.6546/2)(.55^2 - .50^2)
W=.620975 J

2. Not sure if you simply made a typo, but I get W=0.6209 instead of 0.06209. Also do you know what is the right answer, or just that yours is wrong?

Not sure if you simply made a typo, but I get W=0.6209 instead of 0.06209. Also do you know what is the right answer, or just that yours is wrong?

Hi, I made a typo on typing the answer. I did also get .6209 which is the answer I put and it was wrong...

I have no idea what I did wrong. It is in Joules right? I am pretty sure it is..... so frustrating!

4. $W = \dfrac{1}{2} kx^2$

where $x$ is the length of stretch or compression in meters measured from the spring's natural length.

if the spring's natural length is 35 cm and the spring is stretched to a length of 54 cm , then ...

$x = .54 - .35 = .19$

... use this value of $x$ to determine the spring constant, $k$ , then determine the work required to stretch the spring from 50 to 55 cm