Explanation Work is defined as the amount of energy needed to perform any physical task. If constant force is applied, then the work is simply given by the following equation W = F × d Where W = W o r k d o n e F = C o n s t a n t F o r c e d = d i s t a n c e t h r o u g h w h i c h t h e f o r c e a c t s If the force applied is not constant and changes over time, then integral equation is used to calculate the work done. W = ∫ x = a x = b f ( x ) d x Where, W = W o r k d o n e f ( x ) = f o r c e a s a f u n c t i o n o f d i s t a n c e x = d i s t a n c e t h r o u g h w h i c h t h e f o r c e a c t s Given: Given a horizontal cylinder tank having radius of the base is r m e t e r . Also, the density of water is 1000 k g / m 3 Formula Used: We know that force is given as F = m a or F = m g , where is acceleration due to gravity. Also, Mass, m = V D , where V is the volume of the object and d is the density. Thus, force is given as F = D V g Calculation: Let us place the origin along the axis of the cylinder. At the location y , the layer of water is a rectangular slab of length l , width 2 r 2 − y 2 and thickness Δ y . Thus, the volume of one layer of water is given as V o l u m e = 2 l r 2 − y 2 Δ y m 3 Hence, the force required to lift each layer of water is V o l u m e = ( 1000 ) ( 2 l r 2 − y 2 Δ y ) ( 9...

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 6.5, Problem 21E

To determine

The work done (in Joules) required for pumping all of the water out of the full tank if the density of the water is $1000 k g / m^{3}$ . Given that tank has the horizontal cylinder shape and water exits from the small hole at the top.

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Chapter 6.5, Problem 20E

Chapter 6.5, Problem 22E

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Calculus - Standalone book

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The work done (in Joules) required for pumping all of the water out of the full tank if the density of the water is 1000 k g / m 3 . Given that tank has the horizontal cylinder shape and water exits from the small hole at the top.

Solution Summary: The author explains the work required for pumping all of the water out of a full tank from the small hole at the top. If the force applied is not constant and changes over time, then integral equation is used.

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The work done (in Joules) required for pumping all of the water out of the full tank if the density of the water is $1000 k g / m^{3}$ . Given that tank has the horizontal cylinder shape and water exits from the small hole at the top.

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Chapter 6 Solutions

The work done (in Joules) required for pumping all of the water out of the full tank if the density of the water is 1000 k g / m 3 . Given that tank has the horizontal cylinder shape and water exits from the small hole at the top.

Solution Summary: The author explains the work required for pumping all of the water out of a full tank from the small hole at the top. If the force applied is not constant and changes over time, then integral equation is used.

Videos

The work done (in Joules) required for pumping all of the water out of the full tank if the density of the water is 1000 kg/m3. Given that tank has the horizontal cylinder shape and water exits from the small hole at the top.

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Chapter 6 Solutions

The work done (in Joules) required for pumping all of the water out of the full tank if the density of the water is $1000 k g / m^{3}$ . Given that tank has the horizontal cylinder shape and water exits from the small hole at the top.