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Expert-verified Found in: Page 247 ### Physics For Scientists & Engineers

Book edition 9th Edition
Author(s) Raymond A. Serway, John W. Jewett
Pages 1624 pages
ISBN 9781133947271 # Water falls without splashing at a rate of 0.250 L/s from a height of 2.60 m into a bucket on a scale. If the bucket is originally empty, what does the scale read in Newton’s 3.00 s after water starts to accumulate in it?

The scale read in Newton’s 3.00 s after water starts to accumulate in it is 16.5 N.

See the step by step solution

## Step 1: Concept

The kinetic energy to particle is:

$\mathrm{K}=\frac{1}{2}{\mathrm{mv}}^{2}$

The potential energy to particle is:

U = mgy

## Step 2: Reading of scale

After 3.00 s of pouring, the bucket contains $\left(3.00\text{s}\right)\left(0.250\text{L/s}\right)=0.750\text{L}$ of water with mass,$\left(0.750\mathrm{L}\right)\left(1\mathrm{kg}/1\mathrm{L}\right)=0.750\mathrm{kg}$ and feeling gravitational force $\left(0.750\mathrm{kg}\right)\left(9.80\mathrm{m}/{\mathrm{s}}^{2}\right)=7.35\mathrm{N}$.

Water is entering the bucket at the speed that can be calculated as,

${\mathrm{v}}_{\mathrm{impact}}=\sqrt{2{\mathrm{gy}}_{\mathrm{top}}}$

Substitute the values, and we get,

${\mathrm{v}}_{\mathrm{imapact}}=\sqrt{2\left(9.8{\text{m/s}}^{\text{2}}\right)\left(2.600\text{m}\right)}\phantom{\rule{0ex}{0ex}}{\mathrm{v}}_{\mathrm{imapact}}=7.14\text{m/s}\phantom{\rule{0ex}{0ex}}$

The rate of change of momentum is the force itself:

$\frac{\mathrm{dm}}{\mathrm{dt}}{\mathrm{v}}_{\mathrm{impact}}+{\mathrm{F}}_{\mathrm{extra}}\frac{\mathrm{dt}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}{\mathrm{mv}}_{\mathrm{f}}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{dm}}{\mathrm{dt}}{\mathrm{v}}_{\mathrm{impact}}+{\mathrm{F}}_{\mathrm{extra}}=0\phantom{\rule{0ex}{0ex}}$

Substitute the values, and we get,

${\mathrm{F}}_{\mathrm{extra}}=-\frac{\mathrm{dm}}{\mathrm{dt}}{\mathrm{v}}_{\mathrm{impact}}\phantom{\rule{0ex}{0ex}}{\mathrm{F}}_{\mathrm{extra}}=\left(-0.250\text{kg/s}\right)\left(-7.14\text{m/s}\right)\phantom{\rule{0ex}{0ex}}{\mathrm{F}}_{\mathrm{extra}}=1.78\text{N}\phantom{\rule{0ex}{0ex}}$

Altogether the scale must exert $=7.35\mathrm{N}+7.35\mathrm{N}+1.78\mathrm{N}=16.5\mathrm{N}$.

Thus, the scale read in Newton’s 3.00 s after water starts to accumulate in it is 16.5 N. ### Want to see more solutions like these? 