The velocity of dropped object when hits the ground Concept introduction: Law of conservation of energy: The energy can neither be created nor be destroyed but it changes from one form to another. Accordingly, the energy entering into the open system will be equivalent to the energy leaving out of the same system, thereby conserving energy. Δ P .E= Δ K .E Δ P .E + Δ K .E = 0 M g Δ z + M Δ v 2 2 = 0 ( v 2 2 − v 1 2 ) 2 = − g ( z 2 − z 1 ) v 2 2 = v 1 2 + [ − 2 g ( z 2 − z 1 ) ] Here, gravity is g , mass is M , velocity is v , initial height is z 1 , final height is z 2 , and difference in height is Δ z . Fundamentals of Chemical Engineeri...

1st Edition
Kevin D. Dahm + 1 other
Publisher: Cengage Learning
ISBN: 9781111580704 Fundamentals of Chemical Engineeri...

1st Edition
Kevin D. Dahm + 1 other
Publisher: Cengage Learning
ISBN: 9781111580704

Solutions

Chapter 1.7, Problem 22P

A)

Interpretation Introduction

Interpretation:The velocity of dropped object when hits the groundConcept introduction:Law of conservation of energy:  The energy can neither be created nor be destroyed but it changes from one form to another. Accordingly, the energy entering into the open system will be equivalent to the energy leaving out of the same system, thereby conserving energy.ΔP.E=ΔK.EΔP.E+ΔK.E=0MgΔz+MΔv22=0(v22−v12)2=−g(z2−z1)v22=v12+[−2g(z2−z1)]Here, gravity is g, mass is M, velocity is v, initial height is z1, final height is z2, and difference in height is Δz.

B)

Interpretation Introduction

Interpretation:The velocity of thrown down object when hits the groundConcept introduction:Law of conservation of energy:  The energy can neither be created nor be destroyed but it changes from one form to another. Accordingly, the energy entering into the open system will be equivalent to the energy leaving out of the same system, thereby conserving energy.The law of energy conservation for the thrown ballΔP.E+ΔK.E=0MgΔz+MΔv22=0(v22−v12)2=−g(z2−z1)v2=2g(z1−z2)+v12

C)

Interpretation Introduction

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