Some error here I think.
Extending from 0 to 0.5 work done= integral of kx from 0 to 0.5=k(0.5)^2/2 So 0.25k/2=100 k=800
Hello Mathhelpforum!
I have two questions. One with an answer I wanted to double check, and the second needing help on.
The first one is if it takes 100J of work to stretch a spring 0.5m from it's equilibrium position, how much work is needed to stretch an additional .75m?
So I set my limits at x = .5 and x = 1.25.
100 = F(.5)
F = 200.
kx = F
k(.5) = 200
k = 100.
So the spring constant is 100. Then the integral of 100x dx from .5 to 1.25 = 50x² -> 5(1.25)² - (50(.5)²) = 65.625J
And this problem I'm stuck on.
A swimming pool has a shape of a box with a base of 25m by 15m and a depth of 2.5m. How much work is required to pump the water out of the pool?
Would it be Integral from 0 to 2.5m (pg)dy * integral from 0 to 2.5m of (2.5-y)dy?