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Chapter 7

Notes

Common student difficulties and concerns (Keep these in mind while reviewing thewritten problems)

1)

Students tend to think of gravitational potential energy as belonging to a

single object rather than being a property of the system including both the

object and Earth (or whatever the source of the gravitational force on the

object is). In general, students often dont realize that an interaction must

take place between two objects as opposed to just happening to a single

object.

2)

As discussed in Chapter 4, students have difficulty choosing and defining a

system and recognizing whether a system is isolated or is closed or open.

3) Students have trouble maintaining a choice of system through an entire

problem. This produces energy accounting difficulties.4) Students often have a difficult time observing or identifying an interaction.

An interaction between two carts, which occurs over a short period of time, is

easily identified by students. However, friction between a cart and its track,

which gradually decreases the carts kinetic energy over a long time, is not as

easily identified as an interaction.

5)

Students have difficulties distinguishing reversible processes from

irreversible processes and distinguishing between closed and open systems.

6)

The idea of a field is difficult for many students to grasp.

Suggested Additional Problems

Section 7.1

5

Section 7.2

9, 10

Section 7.3

13

Section 7.4

22

Section 7.5

24

Section 7.6

29, 30

Section 7.7

33, 35, 39

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before it hits the ground? (c) Suppose you throw the ball directly downward at the

speed calculated in part a. What is its kinetic energy just before it hits the ground?

Suggested Solution: (a) Using kinematics we can say vy,i2

= vy,f2

+ 2ay y so

that if the ball stops exactly as it reaches your friend

vy,i =

vy,f2

2ay

y=

0

2

9.8m/s

2

( ) 11m( )=

15m/s (b) The initial energywhen the ball is thrown must be the same as the final energy the instant

before it strikes the ground. Thus Ef =Ki +Ui = 12mvi

2+mghi = 27J (c) Since

the kinetic energy only depends on the speed, not the direction of motion,

nothing about the above expression changes, and we see that the energy is

still equal to 27 J.

Problem 68

A 0.70-kg basketball dropped on a hardwood floor rises back up to 65% of its

original height. (a) If the basketball is dropped from a height of 1.5 m, how much

energy is dissipated in the first bounce? (b) How much energy is dissipated in the

fourth bounce? (c) To which type of incoherent energy is the dissipated energy

converted?

Suggested Solution: (a) If we call the moment of release the initial time, and

the moment at which the ball reaches its peak after bouncing the final time,

then the energy lost can be written entirely in terms of the difference in

gravitational potential energies (because the speed is zero in both cases).

Then E=Uf Ui =mg 0.65hi hi( ) =3.6J (b) Assuming the percent ofenergy lost in each bounce is roughly independent of the initial height, the

maximum height between the third and fourth bounce will be 0.65( )3hi and

the maximum height between the fourth and fifth bounce will be 0.65( )4hi .

Hence the energy change in the fourth bounce will be

E=Uf Ui =mghi 0.65( )4

0.65( )3

= 0.99J . Hence 0.99 J are lost in the

fourth bounce. (c) Most energy is converted to heat; some is converted to

sound.

Problem 81

Not looking where you are going, you and your bike collide at 12 m/s into the back

of a car stopped at a red light. The car does not have its brakes applied and so isjolted forward. The driver immediately leaps out crying, Whiplash! Facing a day in

court, you have to determine the acceleration of the car as a result of the collision.

You note that you and your bike (combined inertia 80 kg) came to a complete halt in

the collision and that the rim of your front wheel was pushed all the way to the

center hub. The diameter of the bike wheel before the crash was 0.75 m, and a

reference book tells you that the inertia of the car is 1800 kg.

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