Question

A bucket of water of mass 20 kg is pulled at constant velocity up to a platform 32 meters above the ground. This takes 8 minutes, during which time 6 kg of water drips out at a steady rate through a hole in the bottom. Find the work needed to raise the bucket to the platform. Assume g

Answer 1

Answer:

W=-1881.6J

Explanation:

we have that the change in the mass is

$\frac{dm}{dt}=-c\\m(0)=60kg\\m(8)=60kg-6kg=54kg$

by solving the differential equation and applying the initial conditions we have

$\int dm=-c\int dt\\m=-ct+d\\m(0)=-c(0) + d=60 \\m(8)=-8c+d=54$

by solving for c and d

d=60

c=0.75

The work needed is

W = m(t) gh

by integrating we have

$dW_T= gh\int dm \\\\W_T=gh\int_0^8 -0.75dt\\\\W_T=(9.8\frac{m}{s^2})(32m)(-0.75(8))=-1881.6J$

hope this helps!!