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Work and Energy
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turn your lab
notebooks into
the bin next to
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Energy
Objects (or systems) can have energy and
transfer or transform it.
When Energy is transferred, we call it “doing
work”.
Since energy is conserved, we state that:
W = ΔE
Work
Remember: Work energy required to make something move.
Angle b/t F and d
Force (N)
Work = F d cos (θ)
Work Energy (J)
Distance (m)
Use only the magnitude of F and d in the equation. The
angle will determine if work is positive or negative.
Work is measured in:
Newton meters (N *m)
Joules (J)
d
F
WORK DONE
F
d
NO WORK
A shopper pushes a shopping cart on a rough surface with a force of 8.9 N
at an angle of 60° to the left of the negative y- axis. While the cart moves a
horizontal distance of 10.0 m, what is the work done by the shopper on the
shopping cart?
Fa = 8.9 N at an angle of 60° to
the left of the negative y- axis
Y-axis
8.9 N
Θ=60°
10 meters
A 40-N force pulls a 4-kg block a horizontal distance of 8 m.
The rope makes an angle of 350 with the floor and μk = 0.2.
What is the work done by each force acting on the block AND
what is the NET Work (total work)?
Fa = 40 N
x
= 8m
q = 35°
4kg
μk= 0.2
Find the Work done by each force
Work = F d cos (θ)
WFa = F d cos(θ)
Fa =40N
35°
FF
WFa = (40) (8) cos(35)
WFa = 262.13 J
WFn = 0 J
WFg = 0 J
FN
Fg = mg
Each of these forces are perpendicular to
the distance (90°), so that the works are
zero. (cos 900=0):
d = 8m
WFf = FF d cos(θ)
WFf = 3.25 (8) cos(180)
WFf = -26 J
We need Ff to find the work done by friction.
Remember Ff =μFn
Fn does not equal Fg because the Fa is at an angle. (Fa is “helping” Fn). To find Fn:
Fn + (Fa sin(35)) = Fg
y-component of Fa
Ff =μFn
Fn = mg - (Fa sin(35))
Ff =(0.2)(16.26)
Fn = 16.26 N
Ff =3.25 N
Work done by each force and the total
work
WFa = 262.13 J
WFn = 0 J
WFg = 0 J
WFf = -26 J
Total work = Wfa + WFn + WFg + WFf
Total work = 262.13 + 0 + 0 + (-26)
Total work = 236.13 J
When Fnet = 0, All forces (and components) are equal
(constant velocity / no acceleration)
Fy
FN
Fa
θ°
FF
Remember, the applied
force is at an angle. There
is an x and y component.
Fg = mg
Fn + Fa sinθ = Fg
Fx
Fx = Fa cos θ
Fy = Fa sin θ
Ff = Fa cos θ
Work as the area under the graph
Force vs Position graph
Force
F
Positive
work done
Position x
Negative work done
Graph of Force vs. Displacement
The area under the
curve is equal to the
work done.
Force, F
F
Work = F(x2 - x1)
Area
x1
x2
Displacement, x
Work  F x
What work is done by a constant force of 40 N
moving a block from x = 1 m to x = 4 m?
Work  Fx
Force, F
40 N
Work = F(x2 - x1)
Area
1m
Work = (40 N)(4 m - 1 m)
4m
Displacement, x
Work = 120 J
Mechanical Energy
ME is the energy related to an objects motion or position
Potential Energy
(PE or Ug)
Kinetic Energy
(KE or K)
Stored energy
due to position
Energy in motion
No PE when
object is at h=0
PE =mgh
No KE if object is
at rest
KE= ½mv2
Elastic Potential
Energy
(ePE or Us)
Energy stored in
springs due to
compression or
stretching
ePE= ½kx2
Kinetic and Potential Energy are EQUAL!!
Total PE = Total KE
There is
height
but no
velocity
There is
height
and
velocity
There is
velocity
Height is
0
P.E. = max
K.E. = 0
P.E. = K.E.
K.E. = max
P.E. = 0
Energy is converted but not lost!
Total Energy in this system = 100 J
P.E. = 100 J
P.E. = 50 J
K.E. = 100 J
K.E. = 0
K.E. = 50 J
P.E. = 0
Elastic Potential Energy
ePE is Potential energy stored as a result of
deformation of an elastic object, such as the
stretching of a spring.
Since we know that energy is conserved,
we can make the assumption that:
KE = PE = ePE = W
½ mv2 = mgh = ½ kx2 = Fd
The Work-Energy Theorem
Work is equal to the
change in Kinetic
Energy
W = ΔKE
Work  mv  mv
1
2
2
f
1
2
The Work-Energy Theorem: The work
done by a resultant force is equal to the
change in kinetic energy that it produces.
2
0
Example 1: A 20-g projectile strikes a mud bank, going
through a distance of 6 cm before stopping. Find the
stopping force F if the entrance velocity is 80 m/s.
0
Work = ½ mvf2 - ½ mvo2
80 m/s 6 cm
x
F=?
F d = - ½ mvo2
F (0.06 m) = - ½ (0.02 kg)(80 m/s)2
F (0.06 m) = -64 J
F = 1067 N
Work to stop bullet = change in K.E. for bullet
Example 2: A bus slams on brakes to avoid an accident.
The tread marks of the tires are 25 m long. If m = 0.7, what
was the speed before applying brakes?
Work = K
Work = F d
FF = mFN = m mg
Work = -m mg d
-½ mvo2 = -mk mg d
25 m
FF
0
K = ½ mvf2 - ½ mvo2
vo  2mgd
vo  2(0.7)(9.8m / s)(25m)  18.52m / s
Example 3: A 4-kg block slides from rest to the
bottom of the 300 inclined plane. Find velocity at
bottom. (h = 20 m)
x
FN
h
mg
300
Work  PE  KE
mgh  mv  v  2 gh
1
2
2
v  2 gh  2(9.8m / s 2 )( 20m)  19.80m / s
Power
Power is defined as the rate at which work
is done:
Work (J)
Force (N) Distance (m)
P = ΔE = W = Fd cosθ = Fvcosθ
t
Power (watts)
t
t
Time (sec)
Power is measured in Watts
Power Example: What power is
required to lift a 900-kg elevator
at a constant speed of 4 m/s?
v = 4 m/s
P = F v cos θ= mg v cos θ
P = (900 kg)(9.8 m/s2)(4 m/s) cos (0°)
P = 35300 W
Hooke’s Law
When a spring is stretched, there is a restoring
force that is proportional to the displacement.
The larger the displacement, the larger the
restoring force
F  kx
x
m
F
The spring constant k is a property
of the spring given by:
F
k
x
Example: A 4-kg mass suspended from a spring
produces a displacement of 20 cm. What is the
spring constant?
The stretching force is the weight
(W = mg) of the 4-kg mass:
20 cm
F = (4 kg)(9.8 m/s2) = 39.2 N
F
m
Now, from Hooke’s law, the force
constant k of the spring is:
k=
F
x
=
39.2 N
0.2 m
k = 196 N/m