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Dec. 8, 2001 Prof. Todd Adams, FSU Department of Physics1 Physics 2053C – Fall 2001 Review for...

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Dec. 8, 2001 Prof. Todd Adams, FSU Department of Physics 1 Physics 2053C – Fall 2001 Review for Final Exam http://www.hep.fsu.edu/~tadams
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Page 1: Dec. 8, 2001 Prof. Todd Adams, FSU Department of Physics1 Physics 2053C – Fall 2001 Review for Final Exam tadams.

Dec. 8, 2001Prof. Todd Adams,

FSU Department of Physics 1

Physics 2053C – Fall 2001

Review for Final Exam

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

Page 2: Dec. 8, 2001 Prof. Todd Adams, FSU Department of Physics1 Physics 2053C – Fall 2001 Review for Final Exam tadams.

2

Important Topics

Kinematics Forces/Newton’s Laws Energy/Momentum

Conservation Ideal Gases/Heat

Page 3: Dec. 8, 2001 Prof. Todd Adams, FSU Department of Physics1 Physics 2053C – Fall 2001 Review for Final Exam tadams.

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Variable/Quantity/Units

t time sx distance mv velocity m/sa acceleration m/s2 F force N = kg.m/s2 E energy J = N.mP power W = J.s

Page 4: Dec. 8, 2001 Prof. Todd Adams, FSU Department of Physics1 Physics 2053C – Fall 2001 Review for Final Exam tadams.

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Kinematics

Position Velocity – rate of change of position Acceleration – rate of change of

velocity Constant acceleration Constant velocity Constant position

a = 0v = 0, a = 0

Page 5: Dec. 8, 2001 Prof. Todd Adams, FSU Department of Physics1 Physics 2053C – Fall 2001 Review for Final Exam tadams.

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Equations of Motion

x = x0 + v0t + ½at2

v = v0 + at

v2 = v02 + 2a(x – x0)

x = positionx0 = initial positionv = velocityv0 = initial velocitya = accelerationt = time

Page 6: Dec. 8, 2001 Prof. Todd Adams, FSU Department of Physics1 Physics 2053C – Fall 2001 Review for Final Exam tadams.

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Forces

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

Page 7: Dec. 8, 2001 Prof. Todd Adams, FSU Department of Physics1 Physics 2053C – Fall 2001 Review for Final Exam tadams.

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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

Page 8: Dec. 8, 2001 Prof. Todd Adams, FSU Department of Physics1 Physics 2053C – Fall 2001 Review for Final Exam tadams.

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Types of Energy

Kinetic Linear K = ½mv2

Rotational Potential

Gravitational U = mgh Spring U = ½kx2

Internal Energy Heat Q = mcT Work W = Fdcos

Page 9: Dec. 8, 2001 Prof. Todd Adams, FSU Department of Physics1 Physics 2053C – Fall 2001 Review for Final Exam tadams.

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Ideal Gas Law

PV = nRTP = pressure (atm, bar,

N/m2)V = volume (m3)n = # of molesR = gas constantT = temperature (K)

Page 10: Dec. 8, 2001 Prof. Todd Adams, FSU Department of Physics1 Physics 2053C – Fall 2001 Review for Final Exam tadams.

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How to solve FORCE problems

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

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

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

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

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

5) Solve for what you don’t know

Page 11: Dec. 8, 2001 Prof. Todd Adams, FSU Department of Physics1 Physics 2053C – Fall 2001 Review for Final Exam tadams.

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Sample Force Problem

The boxes are not moving.1) What is the value of T1?

2) What is the value of T2?

M1

M2

T2 T1

M1 = 20.0 kg M2 = 10.0 kg

= 0.3

Page 12: Dec. 8, 2001 Prof. Todd Adams, FSU Department of Physics1 Physics 2053C – Fall 2001 Review for Final Exam tadams.

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Sample Force Problem (cont)

F = T1 – Fg = 0

T1 = Fg

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

T1 = 98.0 N

M1

M2

T2 T1

M1 = 20.0 kg M2 = 10.0 kg

= 0.3

M2

Fg

T1

Page 13: Dec. 8, 2001 Prof. Todd Adams, FSU Department of Physics1 Physics 2053C – Fall 2001 Review for Final Exam tadams.

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Sample Force Problem (cont)

Fy = FN – Fg = 0

FN = Fg

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

FN = 196.0 N

M1

M2

T2 T1

M1 = 20.0 kg M2 = 10.0 kg

= 0.3

M1

Fg

FN

T1T2

Ffr

Page 14: Dec. 8, 2001 Prof. Todd Adams, FSU Department of Physics1 Physics 2053C – Fall 2001 Review for Final Exam tadams.

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Sample Force Problem (cont)

Fx = T1 – T2 – Ffr = 0

T2 = T1 - Ffr

T2 = 98.0 N – (0.3)(196.0 N)

T2 = 39.2 N

M1

M2

T2 T1

M1 = 20.0 kg M2 = 10.0 kg

= 0.3

M1

Fg

FN

T1T2

Ffr

FN = 196.0 N

Page 15: Dec. 8, 2001 Prof. Todd Adams, FSU Department of Physics1 Physics 2053C – Fall 2001 Review for Final Exam tadams.

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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

M2

T2 T1

M1 = 20.0 kg M2 = 10.0 kg

= 0.3T1 = 98.0 NT2 = 39.2 N

Page 16: Dec. 8, 2001 Prof. Todd Adams, FSU Department of Physics1 Physics 2053C – Fall 2001 Review for Final Exam tadams.

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How to Solve ENERGY Problems

1) Identify types of energyKinetic?Gravitational Potential?Spring Potential?Heat?Internal Energy?Work?

2) Identify initial and final conditions3) Find unknown quantities:

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

Page 17: Dec. 8, 2001 Prof. Todd Adams, FSU Department of Physics1 Physics 2053C – Fall 2001 Review for Final Exam tadams.

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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?

5 m

Page 18: Dec. 8, 2001 Prof. Todd Adams, FSU Department of Physics1 Physics 2053C – Fall 2001 Review for Final Exam tadams.

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Sample Energy Problem (cont)

5 m

What kind of energies are present?

Kinetic energy

Gravitational potential energyWork 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

AB C

Page 19: Dec. 8, 2001 Prof. Todd Adams, FSU Department of Physics1 Physics 2053C – Fall 2001 Review for Final Exam tadams.

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Sample Energy Problem (cont)

5 m

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 mvB

2 + 0

vB = 7.0 m/s

AB C

Page 20: Dec. 8, 2001 Prof. Todd Adams, FSU Department of Physics1 Physics 2053C – Fall 2001 Review for Final Exam tadams.

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Sample Energy Problem (cont)

5 m

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

EC = 0.0 JHow 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

AB C

Page 21: Dec. 8, 2001 Prof. Todd Adams, FSU Department of Physics1 Physics 2053C – Fall 2001 Review for Final Exam tadams.

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How to Solve IDEAL GAS Problems

PV = nRT1) Identify initial and final conditions2) Take ratio

P1V1 n1RT1

P2V2 n2RT2

3) Cancel anything which is constant4) Solve for what you don’t know

=

Page 22: Dec. 8, 2001 Prof. Todd Adams, FSU Department of Physics1 Physics 2053C – Fall 2001 Review for Final Exam tadams.

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Sample Problem

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

0 to 30 m/s in 20 s

25 m

30 mF = 2500 N

V = 10 m/s

V = 18 m/s

A B C


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