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

W=-1881.6J

Step-by-step explanation:

we have that the change in the mass is

`(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(m)/(s^2))(32m)(-0.75(8))=-1881.6J`

hope this helps!!