Explanation Given: The height and square basea box: h = 3 m b = 2 m Density at height y is f ( y ) = 1000 − 100 y k g / m 3 Calculation: It is known that that the work done is W = ∫ y 1 y 2 F ( y ) d y . Here the equation is V = ( b , b ) ∂ h Mass is the product of volume and density. then m = p ⋅ V p = f ( y ) = ( 1000 − 100 ) kg/m 3 And the force is F = m g Now mass and force of the structure is m = ( 1000 − 100 y ) × 4 × ∂ h F = ( 1000 − 100 y ) × 4 × ∂ h × ( 9.81 ) The work done is W = ∫ 0 h ( F ( x ) y ) y d y W = ∫ 0 h ( 1000 − 100 y ) 4 ( 9

4th Edition

ISBN: 9781319279813

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 6.5, Problem 16E

To determine

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the work against gravity required to build a box.

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Chapter 6.5, Problem 15E

Chapter 6.5, Problem 17E

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Chapter 6 Solutions

CALCULUS:EARLY TRANS.-SAPLINGPLUS

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the work against gravity required to build a box.

Solution Summary: The author calculates the work against gravity required to build a box. The equation is V=(b,b)partial h

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To calculate: the work against gravity required to build a box.

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Chapter 6 Solutions

To calculate:

the work against gravity required to build a box.