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Work and Energy
Conservation Law
Work and Energy
 Work is a force applied to an object that
causes the point of application of the
force to move through some distance
 Energy is the capacity of an object to do
work
 Unit: Joules
 Energy and work are scalar quantities
Work
 Is work a dot product or a cross
product?
W  Fd cos
Example problem
 A man cleaning a floor pulls a vacuum
cleaner with a force of magnitude F=
50.0 N at an angle of 30.0º with the
horizontal. Calculate the work done by
the force on the vacuum cleaner as the
vacuum cleaner is displaced 3.00m to
the right.
Answer
 W=Fd cos θ
 W= 50.0 N (3.00m) cos 30º
 W = 130 J
Work a Scalar Product
 Work is the dot product of Force and
Displacement
W  F d
Work Done by a Constant
Force
 A particle moving in the xy plane undergoes
a displacement as shown below in a constant
force as given below acts on the particle. A)
Calculate the magnitude of the displacement
and that of the force. B) Calculate the work
done on the object.

d  (2.0i  3.0 j )m

F  (5.0i  2.0 j ) N
Answer
d  x2  y2
d  (2.0) 2  (3.0) 2
d  3.6m
F  Fx2  Fy2
F  (5.0) 2  (2.0) 2
F  5.4 N
W  F d
W  (5.0i  2.0 j )  (2.0i  3.0 j )
W  (5.0i  2.0i)  (5.0i  3.0 j )  (2.0 j  2.0i)  (2.0 j  3.0 j )
W  10  0  0  6  16 J
Work Done by a Varying
Force
 Imagine a particle undergoes a very
small displacement Δx
 The Fx is approximately constant over
this small displacement.
 Work done can be expressed as
W  Fx x
Graph of a Varying Force
Force
Force vs. displacement
6
5
4
3
2
1
0
Series1
1
2
3
4
Displacement
5
6
Work Done by a Varying
Force
W   Fx x
xf
W  lim  Fx x
x  0
xf
W
xi
 F dx
x
xi
Work
Integration
Work Video
Power
 Power is the rate of doing work or of
transferring energy.
dw
P
dt
Power Units
 Power is measured
in joules per second.
 1 J/s = Watt
 1 hp = 746 W
 1 kWh = 3.6 x 106W
Power and Force
 If force F causes a
particle to undergo a
displacement ds, the
work done is
dW=F•ds.
 Since ds/dt is v the
power is given by
the following
formula:
dW
P
dt
F  ds
P
dt
P  F v
Problem
 Consider a car traveling at a steady
speed of 60 km/h (16.7 m/s). It
encounters a frictional force (rolling and
air drag) of 520 N. At what power level
does the engine deliver energy to the
wheels?
Answer
 P = Fv
 8.68 kW
 11.6 hp