**$ 20.00**

# TEST BANK FOR Elementary Mechanics and Thermodynamics By Prof. John W. Norbury (Solutions Manual)

- From Engineering, Fluid Mechanics

- A-Grades
- Rating : 0
- Grade :
**No Rating** - Questions : 0
- Solutions : 275
- Blog : 0
- Earned : $35.00

SOLUTIONS MANUAL

for elementary mechanics &

thermodynamics

Professor John W. Norbury

Physics Department

University of Wisconsin-Milwaukee

P.O. Box 413

Milwaukee, WI 53201

November 20, 2000

Contents

1 MOTION ALONG A STRAIGHT LINE 5

2 VECTORS 15

3 MOTION IN 2 & 3 DIMENSIONS 19

4 FORCE & MOTION - I 35

5 FORCE & MOTION - II 37

6 KINETIC ENERGY & WORK 51

7 POTENTIAL ENERGY & CONSERVATION OF ENERGY 53

8 SYSTEMS OF PARTICLES 57

9 COLLISIONS 61

10 ROTATION 65

11 ROLLING, TORQUE & ANGULAR MOMENTUM 75

12 OSCILLATIONS 77

13 WAVES - I 85

14 WAVES - II 87

15 TEMPERATURE, HEAT & 1ST LAW OF THERMODYNAMICS

93

16 KINETIC THEORY OF GASES 99

3

Chapter 1

MOTION ALONG A

STRAIGHT LINE

5

6 CHAPTER 1. MOTION ALONG A STRAIGHT LINE

1. The following functions give the position as a function of time:

i) x = A

ii) x = Bt

iii) x = Ct2

iv) x = Dcos !t

v) x = E sin !t

where A;B;C;D;E; ! are constants.

A) What are the units for A;B;C;D;E; !?

B) Write down the velocity and acceleration equations as a function of

time. Indicate for what functions the acceleration is constant.

C) Sketch graphs of x; v; a as a function of time.

SOLUTION

A) X is always in m.

Thus we must have A in m; B in msec¡1, C in msec¡2.

!t is always an angle, µ is radius and cos µ and sin µ have no units.

Thus ! must be sec¡1 or radians sec¡1.

D and E must be m.

B) v = dx

dt and a = dv

dt. Thus

i) v = 0 ii) v = B iii) v = Ct

iv) v = ¡!D sin !t v) v = !E cos !t

and notice that the units we worked out in part A) are all consistent

with v having units of m¢ sec¡1. Similarly

i) a = 0 ii) a = 0 iii) a = C

iv) a = ¡!2Dcos !t v) a = ¡!2E sin !t

7

i) ii) iii)

x

t

v

a

x x

v v

a a

t t t

t t t

t t

C)

8 CHAPTER 1. MOTION ALONG A STRAIGHT LINE

iv) v)

0 1 2 3 4 5 6

t

-1

-0.5

0

0.5

1

x

0 1 2 3 4 5 6

t

-1

-0.5

0

0.5

1

x

0 1 2 3 4 5 6

t

-1

-0.5

0

0.5

1

v

0 1 2 3 4 5 6

t

-1

-0.5

0

0.5

1

v

0 1 2 3 4 5 6

t

-1

-0.5

0

0.5

1

a

0 1 2 3 4 5 6

t

-1

-0.5

0

0.5

1

a

9

2. The ¯gures below show position-time graphs. Sketch the corresponding

velocity-time and acceleration-time graphs.

t

x

t

x

t

x

SOLUTION

The velocity-time and acceleration-time graphs are:

t

v

t t

v

t

a

t

a

t

a

v

10 CHAPTER 1. MOTION ALONG A STRAIGHT LINE

3. If you drop an object from a height H above the ground, work out a

formula for the speed with which the object hits the ground.

SOLUTION

v2 = v2

0 + 2a(y ¡ y0)

In the vertical direction we have:

v0 = 0, a = ¡g, y0 = H, y = 0.

Thus

v2 = 0¡ 2g(0 ¡ H)

= 2gH

) v =

p

2gH

11

4. A car is travelling at constant speed v1 and passes a second car moving

at speed v2. The instant it passes, the driver of the second car decides

to try to catch up to the ¯rst car, by stepping on the gas pedal and

moving at acceleration a. Derive a formula for how long it takes to

catch up. (The ¯rst car travels at constant speed v1 and does not

accelerate.)

SOLUTION

Suppose the second car catches up in a time interval t. During that

interval, the ¯rst car (which is not accelerating) has travelled a distance

d = v1t. The second car also travels this distance d in time t, but the

second car is accelerating at a and so it's distance is given by

## [Solved] TEST BANK FOR Elementary Mechanics and Thermodynamics By Prof. John W. Norbury (Solutions Manual)

- This solution is not purchased yet.
- Submitted On 10 Feb, 2022 01:05:32

- A-Grades
- Rating : 0
- Grade :
**No Rating** - Questions : 0
- Solutions : 275
- Blog : 0
- Earned : $35.00