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Physics 2053C – Fall 2001PowerPoint Presentation

Physics 2053C – Fall 2001

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### Physics 2053C – Fall 2001

Review for Final Exam

http://www.hep.fsu.edu/~tadams

Prof. Todd Adams,

FSU Department of Physics

Important Topics

- Kinematics
- Forces/Newton’s Laws
- Energy/Momentum Conservation
- Ideal Gases/Heat

Variable/Quantity/Units

t time s

x distance m

v velocity m/s

a acceleration m/s2

F force N = kg.m/s2

E energy J = N.m

P power W = J.s

Kinematics

- Position
- Velocity – rate of change of position
- Acceleration – rate of change of velocity
- Constant acceleration
- Constant velocity
- Constant position

a = 0

v = 0, a = 0

Equations of Motion

x = x0 + v0t + ½at2

v = v0 + at

v2 = v02 + 2a(x – x0)

x = position

x0 = initial position

v = velocity

v0 = initial velocity

a = acceleration

t = time

Forces

- Force due to gravity
- Normal force
- Force due to friction
- Tension
- Buoyancy
- External force (e.g. a push)

Gravity

- acceleration due to gravity (g = 9.80 m/s2)
- force due to gravity F = mg
- Weight = force due to gravity
- which direction???
- Also, F = G.(m1m2)/r2

Newtons

Types of Energy

- Kinetic
- Linear K = ½mv2
- Rotational

- Potential
- Gravitational U = mgh
- Spring U = ½kx2
- Internal Energy

- Heat Q = mcT
- Work W = Fdcos

Ideal Gas Law

PV = nRT

P = pressure (atm, bar, N/m2)

V = volume (m3)

n = # of moles

R = gas constant

T = temperature (K)

How to solve FORCE problems

- Read the problem.
(identify what you do and don’t know, look for “hidden” knowledge)

- Draw a free-body diagram
(identify all forces acting upon object)

- Add all forces in one direction together (x?)
F = F1 + F2 + F3 + …

(determine sum of forces, maybe Fnet = 0 or Fnet = ma)

- Add all forces in other direction together (y?)
(determine sum of forces, maybe Fnet = 0 or Fnet = ma)

- Solve for what you don’t know

T1

T2

M1

M2

Sample Force ProblemThe boxes are not moving.

- What is the value of T1?
- What is the value of T2?

M1 = 20.0 kg

M2 = 10.0 kg

= 0.3

T2

T1

M1

T1

M2

Fg

Sample Force Problem (cont)M1 = 20.0 kg

M2 = 10.0 kg

= 0.3

F = T1 – Fg = 0

T1 = Fg

T1 = M2g = (10.0 kg)(9.80 m/s2)

T1 = 98.0 N

M2

T2

T1

M1

FN

T2

T1

M2

Ffr

Fg

Sample Force Problem (cont)M1 = 20.0 kg

M2 = 10.0 kg

= 0.3

Fy = FN – Fg = 0

FN = Fg

FN = M1g = (20.0 kg)(9.80 m/s2)

FN = 196.0 N

M1

T2

T1

M1

M2

Sample Force Problem (cont)M1 = 20.0 kg

M2 = 10.0 kg

= 0.3

FN

Fx = T1 – T2 – Ffr = 0

T2 = T1 - Ffr

T2 = 98.0 N – (0.3)(196.0 N)

T2 = 39.2 N

T2

T1

M1

FN = 196.0 N

Ffr

Fg

T1

T2

M1

M2

Sample Force Problem (cont)What if the boxes are moving with constant velocity?

What if the boxes are accelerating at a = 2.2 m/s2?

What if we remove T2?

M1 = 20.0 kg

M2 = 10.0 kg

= 0.3

T1 = 98.0 N

T2 = 39.2 N

How to Solve ENERGY Problems

- Identify types of energy
Kinetic?

Gravitational Potential?

Spring Potential?

Heat?

Internal Energy?

Work?

- Identify initial and final conditions
- Find unknown quantities:
W = K + U (if W 0)

Ki + Ui = Kf + Uf (if W = 0)

Sample Energy Problem

A 25 kg block is released from rest 5.5 m up a frictionless plane inclined at 30o. The block slides down the incline and along a horizontal surface. The horizontal surface has a coefficient of static friction of 0.32.

What is the velocity of the block at the bottom of the incline?

How far along the horizontal surface will the block slide?

Sample Energy Problem (cont)

What kind of energies are present?

Kinetic energy

Gravitational potential energy

Work done by friction

What is the energy at A?

EA = KE + PE = 0 + mgh = mgdsin = (25 kg)(9.80 m/s2)(5 m)(sin 30o)

EA =612.5 J

A

B

C

Sample Energy Problem (cont)

What is the energy at B?

EB = EA = 612.5 J

What happens to the energy as the box goes from A to B?

What is the velocity at B?

EB = KE + PE = ½ mvB2 + 0

vB =7.0 m/s

A

B

C

Sample Energy Problem (cont)

What happens to the energy as the box goes from B to C?

What is the energy at C?

EC = 0.0 J

How far does the box slide?

W = KE + PE Wfr = Ffrd = FNd

W = (0.0 J – 612.5 J) + 0 J Wfr = mgd

W = Wfr .

. -612.5 J = (0.32)(25 kg)(9.8 m/s2)d = (78.4 N)d

. d = (612.5 J)/(78.4 N) = 7.81 m

A

B

C

How to Solve IDEAL GAS Problems

PV = nRT

- Identify initial and final conditions
- Take ratio
P1V1 n1RT1

P2V2 n2RT2

- Cancel anything which is constant
- Solve for what you don’t know

=

Sample Problem

V = 10 m/s

V = 18 m/s

- What are the forces on the motorcycle as it accelerates (A)?
- What are the forces on the motorcycle as it moves at constant speed (B,C)?
- How far does it travel while accelerating from rest to 30 m/s?
- What is the kinetic energy at points A, B, C?
- How much work is done by motorcycle?
- How much work is done by friction getting to A, B, C?
- What are the forces on the object as it moves upward from A to B?

M = 250 kg

30 m

F = 2500 N

25 m

A

B

C

0 to 30 m/s in 20 s

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