Integrals in Physics

1. Oct 1, 2006

lemurs

Ok This is a Calculus assignment question but more of af physics thing..

First were using integrals so there is the first part.

I understand Intergration and all it jusst how to get a equation to apply it to is the hard part.

a 4 lb bucket is lifted 80 Ft from a bottom of a well full of water.
the water in bucket weighs 40lb but is leaking out of a hole at a rate of .2 lb/s. It rate of accent is 2 ft/s.

As far as I can tell there will be a constant in the equation since a 4lb bucket will rise 80 ft. so that is constant thye whole time.. but since the water is leaking how do i use that rate of change in with the speed.

Plus I am figure this out in the British system.. Lb/ft not metric newton joules.

so a far as i got i would have 4*80 + (40-.2x)/2

As far as I can figure i would have to use that formula.. but it just doesn't look right or sound right..

2. Oct 1, 2006

HallsofIvy

.2 lb/s is the rate at which the water leaks out. During those two seconds, the bucket rises 2 ft/s so the water leaks out at .1 lb/ft. The weight of the water in the bucket after rising x feet is 40- .1x. The "40" is not divided by 2. Now you can do either of two things: argue that the weight of bucket and water, after rising x feet is 4+ 40- .1x= 44- .1x and integrate that as x goes from 0 to 80 or argue that it requires 4*80 ft-lbs of work to raise the 4 pound bucket so integrate 40- .1x to find the work necessary to raise the water and then add 4*80 to that.

3. Oct 1, 2006

lemurs

Thanks now that you got it making complete sense I can get this stuff done..