Answer:
a) Lim(0-inf) Work = (36 - 0.1*xi )*dx
b) Work = integral( (36 - 0.1*xi ) ).dx
c) Work = 2835 lb-ft
Step-by-step explanation:
Given:
- The weight of the bucket W = 3 lb
- The depth of the well d = 90 ft
- Rate of pull = 2.5 ft/s
- water flow out at a rate of = 0.25 lb/s
Find:
A. Show how to approximate the required work by a Riemann sum (let x be the height in feet above the bottom of the well. Enter xi∗as xi)
B. Express the Integral
C. Evaluate the integral
Solution:
A.
- At time t the bucket is xi = 2.5*t ft above its original depth of 90 ft but now it hold only (36 - 0.25*t) lb of water at an instantaneous time t.
- In terms of distance the bucket holds:
( 36 - 0.25*(xi/2.5)) = (36 - 0.1*xi )
- Moving this constant amount of water through distance dx, we have:
Work = (36 - 0.1*xi )*dx
B.
The integral for the work done is:
Work = integral( (36 - 0.1*xi ) ).dx
Where the limits are 0 < x < 90.
C.
- Evaluate the integral as follows:
Work = (36xi - 0.05*xi^2 )
- Evaluate limits:
Work = (36*90 - 0.05*90^2 )
Work = 2835 lb-ft
