A bucket of water of mass 30 kg is pulled at a constant velocity up to a platform 30 meters above...

Question:

A bucket of water of mass 30 kg is pulled at a constant velocity up to a platform 30 meters above the ground. This takes 16 minutes, during which time 3 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 = 9.8 \hbox{eq}{m/s}^2. {/eq})

Work Done:

If the mass of an object being raised varies with position, then so does the force required to just balance the weight of the object. In this case, the total work required to raise the object by a specific length equals the integral of the small work required to raise the object a small length.

Answer and Explanation:

The constant speed at which the bucket is pulled up equals {eq}\displaystyle \frac {\text{Distance}}{\text{time}} = \frac {30\mathrm{\,\,m}}{16*60\mathrm{\,\,s}} = \frac 1{32}\mathrm{\,\,m\,s^{-1}} {/eq}. So, at a general time {eq}t {/eq} s, the bucket is {eq}\displaystyle y= \frac 1{32}t {/eq} m above the floor of the well.


The rate at which water drips out of the bucket equals {eq}\displaystyle \frac {3\mathrm{\,\,kg}}{16*60\mathrm{\,\,s}} = \frac 1{320}\mathrm{\,\,kg\,s^{-1}} {/eq}

So, at a general time {eq}\displaystyle t,\,\,\,\,\frac 1{320}t {/eq} kg of water has leaked out of the bucket, and the mass of the bucket plus water remaining in the bucket equals {eq}\displaystyle 30 - \frac 1{320}t = 30-\frac 1{320}*\frac y{\frac 1{32}} =(30-0.1y) {/eq} kg.


A force (say {eq}F(y) {/eq}) equal to the sum of weights of the bucket and the water remaining must be applied upward to raise the bucket.

So {eq}F(y) = (30-0.1y)*9.8=(294-0.98y) {/eq} N.

The work done in raising the bucket plus water a small length {eq}\mathrm{d}y {/eq} equals {eq}F(y)\mathrm{d}y = (294-0.98y)\mathrm{d}y {/eq} and the work done in raising the bucket to the top of the platform equals

{eq}\begin{align*} \int_{y=0}^{30} (294-0.98y)\mathrm{d}y &=294\bigg[y \bigg]_{y=0}^{30}-0.98\left[ \frac {y^{2}}{2}\right]_{y=0}^{30} \\ &= 294*30-0.98\frac {30^{2}}{2}\\[2ex] &=\color{Blue}{8379\mathrm{\,\,J}\hspace{0.8cm}\mathrm{(Answer)}} \end{align*} {/eq}


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Work as an Integral

from Physics: High School

Chapter 7 / Lesson 10
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