Lecture 14
Work: Varying forces and
curved trajectories
Power
ACT: Zero net work
ACT: Zero net work
The system of pulleys shown below is used to lift a bag of mass
M
at constant speed a distance
h
from the floor. What is the work
done by the person?
M
h
A.
Mgh
B. ½
Mgh
C.
2
Mgh
The force by the person is
Mg
/2 (see lecture 10), but he needs to
pull on a length of rope of 2
h.
net
0 (constant speed)
W
KE
= ∆
=
(
)
g
T
W
W
Mgh
= −
= − −
The story so far:
net
W
K
= ∆
Kinetic energy
2
1
2
K
mv
=
r
F
r
F
r
F
W
∆
=
∆
=
∆
⋅
=
cos
//
θ
G
G
Work by a constant force,
along a straight path:
Work/Kinetic energy theorem
The journey is divided up into a series of segments over which the force
is constant.
Work
by non-constant force, with
straight line trajectory
An object moves along the x-axis from point x
1
to x
2
. A non-constant
force
is applied on the object. What is the work done by this force?
x
1
F
A
∆
x
A
F
B
∆
x
B
F
C
∆
x
C
F
D
∆
x
D
F
E
∆
x
E
E
Ex
D
Dx
C
Cx
B
Bx
A
Ax
x
F
x
F
x
F
x
F
x
F
W
∆
+
∆
+
∆
+
∆
+
∆
=
The total work is the sum of the works for each of the intervals:
x
2

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F
x
x
∆
x
A
∆
x
B
∆
x
C
∆
x
D
∆
x
E
Work=area
F
A
x
F
B
x
F
C
x
F
D
x
F
E
x
E
Ex
D
Dx
C
Cx
B
Bx
A
Ax
x
F
x
F
x
F
x
F
x
F
W
∆
+
∆
+
∆
+
∆
+
∆
=

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