finding the work needed to riaise a bucket to a platform

Question

 A bucket of water of mass 30 kg is pulled at constant velocity up to a platform 55 meters above the ground. This takes 20 minutes, during which time 7 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. (Use g=9.8m/s2.)

Work =

Answer 1

There are a few ways to approach this problem. I approached it this way, although there are other equally valid ways to solve this.

We know that:

1. W=∫F dy

with this equation we can show that the power, P, is the following

1. P=dW/dt=F*v, where v is the velocity of the object.

We can then integrate this equation with respect to time and get the following formula

1. W=∫P dt=∫F*v dt

The velocity refers to the vertical velocity, which is described as a constant v=55/20=2.75 meters/min.

The gravitational force is given by F=m(t)*g where m(t) is the mass of the bucket at some time, t, in minutes.

Since we are given that the water leaves the bucket a constant rate of 7/20=.35 kg/min we can parameterize m(t)=30-.35t

Plugging all of this in, we have

W=∫₀²⁰m(t)*g*v dt

W=g*v ∫₀²⁰m(t) dt (since g and v are constants)

W=g*v ∫₀²⁰(30-.35t)dt

solving this definite integral, we get W= 14,283.5 Joules of energy.

As a nice sanity check, it is always helpful to make sure your answer makes sense. We know that our answer for work should be less than the work needed to pull the 30kg (initial mass) bucket without leakage but more than the work needed to pull the 23kg (final mass) bucket without leakage. This is simply because you are lifting more/less mass the same distance. The equation of the work done on an object with constant mass is simply W=mgh. If you plug in these values for your upper and lower bounds you can confirm that our answer is nested comfortably within this interval.