Solutions manual for mechanics and thermodynamics
.pdfChapter 6
KINETIC ENERGY & WORK
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CHAPTER 6. KINETIC ENERGY & WORK |
Chapter 7
POTENTIAL ENERGY & CONSERVATION OF ENERGY
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54CHAPTER 7. POTENTIAL ENERGY & CONSERVATION OF ENERGY
1.A block of mass m slides down a rough incline of height H and angle µ to the horizontal. Calculate the speed of the block when it reaches the bottom of the incline, assuming the coe±cient of kinetic friction
is k.
SOLUTION
The situation is shown in the Øgure.
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The work-energy theorem is
¢U + ¢K = WNC
= Uf ° Ui + Kf ° Ki
but Uf = 0 and Ki = 0 giving
KF = Ui + WNC
Obviously WNC must be negative so that Kf < Ui
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where we have used Fk = kN. To get N use Newton's law
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56CHAPTER 7. POTENTIAL ENERGY & CONSERVATION OF ENERGY
Chapter 8
SYSTEMS OF PARTICLES
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CHAPTER 8. SYSTEMS OF PARTICLES |
1.A particle of mass m is located on the x axis at the position x = 1 and a particle of mass 2m is located on the y axis at position y = 1 and
a third particle of mass m is located oÆ-axis at the position (x; y) = (1; 1). What is the location of the center of mass?
SOLUTION |
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The position of the center of mass is |
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mi~ri |
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xcm = M Xi |
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m + 2m + m |
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Thus the coordinates of the center of mass are
µ1 3 (xcm; ycm) = 2; 4
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2.Consider a square at table-top. Prove that the center of mass lies at the center of the table-top, assuming a constant mass density.
SOLUTION
Let the length of the table be L and locate it on the x{y axis so that one corner is at the origin and the x and y axes lie along the sides of the table. Assuming the table has a constant area mass density æ, locate the position of the center of mass.
xcm = M |
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A 2 0 0A
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L2 £ L = |
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CHAPTER 8. SYSTEMS OF PARTICLES |
3.A child of mass mc is riding a sled of mass ms moving freely along an icy frictionless surface at speed v0. If the child falls oÆ the sled, derive a formula for the change in speed of the sled. (Note: energy is not conserved !) WRONG WRONG WRONG ??????????????
speed of sled remains same - person keeps moving when fall oÆ ???????
SOLUTION
Conservation of momentum in the x direction is
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pix = pfx
(mc + ms)v0 = msv where v is the new Ønal speed of the sled, or
v = |
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the change in speed is
v ° v0 = mc v0 ms
which will be large for small ms or large mc.