Exam 1 distribution
Problem 3
SWWBSB vvv ///rrr
+=
v vB/WvB/WvW/S
vB/S
36
vB/W
vW/SvB/S
WB
SW
v
v
/
/36sin =WB
SW
v
v
/
/tan =
== 36sintan xxy
Problem 13 aN
mg
maNmg =N 735scale theof reading ==N
Rv
a2
=
Rvg
Nag
Nm 2
=
=
Problem 20
( ) ( ) 31 ttFm
ta == ( )ka ta=m
=f
i
t
t
zizfz dttavv )(,,
m/s 44
2
0
42
0
3===
=
=
t
t
z
tdttv543 22
22
=+=
=+= zx vvv
Work and Kinetic Energy
General Physics IMechanics
Racquetball Striking a Wall
Kinetic Energy MechanicsPhysics 140
Mt. EtnaStewart Hall
What is Work? In physics, work is a measure of how effective a
force is in moving things Work depends on both the force and how far
something moved under the influence of that force Work is a scalar Work is a scalar For a constant force F and a straight-line
displacement s, work is their scalar product:
SI unit of work is Joule: 1 J = 1 Nm = 1 kgm2/s2
cosFssFW == rr
F
s
Work Depends on the component of the force parallel
to the displacement:
||||cos FssFFssFW ==== rr
F Fs||
If components of both force and displacement are known, can also compute work by components:
F
sF||
F
s
zFyFxFsFW zyx ++==rr
Work Depends on the component of the force parallel
to the displacement:
Force along displacement does positive work Force against displacement does negative work Force perpendicular to displacement does no
work
Uniform Circular Motion A rock of mass m is twirled on a string in a
horizontal plane. The work W done by the tension in the string on the rock is
A. W < 0A. W < 0B. W = 0C. W > 0D. More than one option above is possible
Uniform Circular Motion In uniform circular
motion a force always acts perpendicular to the motion
Fr
vr
vr
Fr
the motion It has no component
along the motion This force does no
work!
vr F
r
vr
Fr
F
0=W
A block slides along a level surface subject to the forces shown. Which force does the largest work on the block during the displacement x? The magnitude of the forces is expressed by the length of their arrows.
N
F1
F
A. F1B. F2
mg
N
x
F2B. F2C. mgD. NE. Depends on mass
Work depends only on the component of force along the motion!
Lowering a book Yuri lowers a book of mass m downward a
distance h at constant speed v. The work done by the force of gravity is
A. NegativeA. NegativeB. ZeroC. PositiveD. Impossible to tell
Lowering a book Yuri lowers a book of mass m downward a
distance h at constant speed v. The work done by the force of Yuris hand is
A. NegativeA. NegativeB. ZeroC. PositiveD. Impossible to tell
Lowering a book Yuri lowers a book of mass m downward a
distance h at constant speed v. The work done by the total force acting on the book is
A. Negative HA. NegativeB. ZeroC. PositiveD. Impossible to tell
Wtot = WG + WH = 0G
H
s
a = 0
Work in General Situations What if force is not constant and trajectory is not
a straight line? Split the path into short displacements s Find the force F acting on the body over each s Find work W over each s by computing the Find work W over each s by computing the
scalar product:
Sum the results to find the total work W of the variable force over the curved path:
cossFsFW == rr
== cossFsFWrr
Work as Integral If the force varies
over a general path, add up many little Fds pieces:
= sdFWrr
This integral means exactly the same as the sum:
= sdFW
== sdFsFsFWrrrr cos
++== dzFdyFdxFdsFW zyx cos
Problem 6.34
(c): A. 20 J B. 40 J C. 60 J D. 80 J E. 120 J
If displacement is along xaxis, only the x-component of force Fx is relevant
If Fx varies with x, sum over small displacements
L++= xFxFW
Position-Dependent Force in 1D
In the limit of each x going to zero, this sum becomes an integral
Geometrically, work is the area under the Fx(x) curve
L++= bbxaax xFxFW
=2
1
x
xx dxFW
Position-Dependent Force in 1D An object is pushed by a variable force F plotted
below as a function of position x. How much work has the force done on the object when it has moved from x = 0 to x = + 6 m?
A. 0 Force, F
2 NA. 0B. 2 JC. 4 JD. 6 JE. 12 J
x
2 N
-2 N
6 m
+ +
_
Hookes Law For a spring, the force arising as a result of
deformation is proportional to the deformation If a spring is stretched or compressed from its
equilibrium length x0 by x = x x0, then the force arising in the spring is given by the Hookes law:
xkF =
Minus sign means that the force the spring exerts is opposite to the length change
xkFx =x0
x
x
F x
Hookes Law Coefficient k is called stiffness or stiffness
coefficient or force constant or spring constant of the spring
If choose x0 = 0 and extend the spring from 0 to x, then the work done by the spring is
( ) 2kxxx
The work done by an applied external force is exactly opposite:
( )2
2
00
kxdxkxdxFWxx
x ===
2
2kxWkxFx ==
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