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 .

BuyFind

Fundamentals of Chemical Engineeri...

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

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 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.

ΔP.E=ΔK.EΔP.E+ΔK.E=0MgΔz+MΔv22=0(v22v12)2=g(z2z1)v22=v12+[2g(z2z1)]

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 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.

The law of energy conservation for the thrown ball

ΔP.E+ΔK.E=0MgΔz+MΔv22=0(v22v12)2=g(z2z1)v2=2g(z1z2)+v12

C)

Interpretation Introduction

Interpretation:

Examples for the assumption made in part A and B have to be determined.

Concept introduction:

Air resistance is depending on the ratio of surface area to the mass, if the mass is less and the surface area is more means air resistance is more.

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