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Physics For Scientists & Engineers
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Short Answer

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 by Step Solution

Step 1: Concept

The kinetic energy to particle is:

K=12mv2

The potential energy to particle is:

U = mgy

Step 2: Reading of scale

After 3.00 s of pouring, the bucket contains 3.00 s0.250 L/s=0.750 L of water with mass,0.750 L1 kg/1 L=0.750 kg and feeling gravitational force 0.750 kg9.80 m/s2=7.35 N.

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

vimpact=2gytop

Substitute the values, and we get,

vimapact=29.8 m/s22.600 mvimapact=7.14 m/s

The rate of change of momentum is the force itself:

dmdtvimpact+Fextradtdt=ddtmvfdmdtvimpact+Fextra=0

Substitute the values, and we get,

Fextra=-dmdtvimpactFextra=-0.250 kg/s-7.14 m/sFextra=1.78 N

Altogether the scale must exert =7.35 N+7.35 N+1.78 N=16.5 N.

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

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