Physics: Principles with Applications
Physics: Principles with Applications
6th Edition
ISBN: 9780130606204
Author: Douglas C. Giancoli
Publisher: Prentice Hall
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Chapter 6, Problem 52P

Part (a)

To determine

Fraction of its initial energy is lost during the bounce.

Part (a)

Expert Solution
Check Mark

Answer to Problem 52P

Solution: 20%

Explanation of Solution

Given:

Initial height of ball =2.0m

Maximum height attained by ball after bounce =1.6m

Formula Used: P.E of a body at height from a point (reference point) =mgh

Conservation of mechanical energy.

  KE2+PE2=KE1+PE1

When only conservative force is acting.

Calculation:

  Physics: Principles with Applications, Chapter 6, Problem 52P , additional homework tip  1

Initial height of ball h1=2.0m

  P.E1=mgh1

Maximum height attained by ball is h2

  PE2=mgh2

Loss in P.E (P.E2)=P.E1-P.E2

  =mg h1-mg h2=mg(h1-h2).

Fraction of its potential energy losses.

  =mg( h 1 - h 2 )mg h1×100=h1-h2h1×100=2-1.62×100=0.42×100=20%

Conclusion:The fraction of energy lost during the bounce =20%

Part (b)

To determine

The ball’s speed just before and just after the bounce.

Part (b)

Expert Solution
Check Mark

Answer to Problem 52P

Solution: 6.3 m/s (just before bounce), 5.6 m/s (just after bounce)

Explanation of Solution

Given: Initial height of ball =2.0m

Maximum height attained by ball after bounce =1.6m

Formula Used: P.E of a body at height from a point (reference point) =mgh

Conservation of mechanical energy.

  KE2+PE2=KE1+PE1

  (when only conservative force is acting)

Calculation:

  Physics: Principles with Applications, Chapter 6, Problem 52P , additional homework tip  2

Before and after bounce only gravitational force is acting on ball (conservative force) so apply conservation of energy before bounce and after bounces.

Initial mechanical energy of ball (Ei)

  =P.E1+K.E1=mg h1+12m(0)2=mg h1.

Mechanical energy of ball just before collision (Ef)

  =P.E2+K.E2=mg(0)+12m( v 2)2

Reference position for potential energy is taken ground surface.

  Ei=Efmg h1=12m( v 2)2v2=2gh1   =2×9.80×2   =6.26 m/s6.3 m/s

Speed of ball just before bounce =6.26m/s2 after bounce

Initial mechanical energy of ball (Ei)

  =P.E1+K.E1=mg(0)+12mv12=12mv12

Final mechanical energy of ball (Ef)

  =P.E2+K.E2=mg h2+12m(0)2  (at the heighest point velocity=0)=mg h2

  12mv22=mgh2            (conservation of energy)v2=2gh2    =2×9.80×1.60    =5.6 m/s.

Conclusion:

Speed of ball just after bounce =5.6 m/s

Part (c)

To determine

Where did the energy go

Part (c)

Expert Solution
Check Mark

Answer to Problem 52P

Solution:Into heat and sound energy

Explanation of Solution

Loss in mechanical energy is converted into heat and sound energy during collision with Earth’s surface.

Conclusion:

Loss in mechanical energy is converted into sound and heat energy.

Chapter 6 Solutions

Physics: Principles with Applications

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