Transcript Work, Power, Energy

Work
The work done on an object by a constant force
is the product of the force that is parallel to the
displacement (θ = 0°) times the displacement.
Work is computed by
W = F × Δx × cosθ
where F is measured in N, Δx is the
displacement measured in m and θ is the angle
between the force and the displacement.
 The unit of work is
1N•1m = 1J
where J is the work done measured in
joules.
Do not confuse a N•m which is also used to
measure torque.
Work is a scalar quantity meaning it has
only direction.
When is there no work done in a “physics sense”?
 When Δx = 0, W = 0 J in all cases.
 No matter how long you hold a heavy
package, if your displacement is zero, so is
the work done.
F
•
Fw
You are applying a force but
Δx = 0 which makes W = 0.
 What happens when you carry this heavy
object a displacement of 2.0 m?
F
•
Δx
Fw
You are exerting a force, you do have a
displacement, but the angle between the
applied force and the displacement is
90°, therefore the W = 0.
 What about an object undergoing uniform
horizontal circular motion?
Fc
Δx
You are exerting a force, you do have a
displacement, but the angle is 90°, and
the cos 90° = 0, therefore the W = 0.
Work Problems
How much work would you do if you pushed a
140 kg crate 12.7 m across a floor at a constant
velocity if µ = 0.60?
m = 140 kg
FN
Ff
F
Fw
Δx = 12.7 m
µ = 0.60
Fy = 0
Fx = 0
FN = Fw
F = Ff
µ = Ff/FN
Ff = µFN = µFw = µmg
Ff = 0.60 × 140 kg × 9.80 m/s2 = 820 N
W = F × Δx × cosθ
W = 820 N × 12.7 m × 1 = 1.0 x 104 J
A 60. kg crate is pulled 50. m along a horizontal
surface by a constant force of 110. N which acts
at an angle of 40° with the horizontal. The
frictional force is 50. N.
(a) What is the work done by each force acting
on the crate.
FN
FV
Ff
Fw
F
θ
FH
m = 60. kg
F = 110. N
Δx = 50. m
θ = 40°
Ff = 50. N
W(Fw) = Fw × Δx cos 90° = 0 J
W(Fv) = FV × Δx cos 90° = 0 J
W(FN) = FN × Δx cos 90° = 0 J
W(FH) = FH × Δx cos 40°
W(FH) = 110. N × 50. m × cos 40° = 4.2 x 103 J
W(Ff) = Ff × Δx cos 180°
W(Ff) = 50. N × 50. m × cos 180° = -2.5 x 103 J
(b) What is the net work done on the crate?
WV = 0 J in a vertical direction.
WT = FH + Ff + WV
WT = 4.2 x 103 J – 2.5 x 103 J + 0
WT = 1.7 x 103 J
Work-Energy Principle
Energy is the ability to do work.
Kinetic energy (translational energy) is the
energy resulting from the motion of an object
and is given by
 KE = ½mv2 = 1 kg•1 m2/s2 = N•m = J
 This is no coincidence that work and
energy are measured in the same units.
The work-energy principle states that the net
work done on or by an object is equal to its
change in kinetic energy.
 W = ΔKE = ½mv2
 When ΔKE > 0, work is done on the object
and its kinetic energy increases.
 When ΔKE < 0, work is done by the object
and its kinetic energy decreases.
Gravitational potential energy is the stored
energy an object has because of its position
relative to the earth.
The work-energy principle states that the net
work done on or by an object is equal to its
change in potential energy.
 W = F × Δx × cosθ
 W = ΔPEg = mgΔh = kg•m/s2•m = J
Δh is the height measured from a convenient
reference point.
 The reference point can be anywhere
because you are only interested in the
change in energy, not the actual energy.
 When ΔPEg > 0, work is done on the
object and its potential energy increases.
 When ΔPEg < 0, work is done by the
object and its potential energy decreases.
Work-Energy Principle Problem
How much work is required to accelerate a
1250 kg car from 30. m/s to 40. m/s?
m = 1250 kg
vi = 30. m/s
W = ΔKE = ½mvf2 - ½mvi2
W = ½ × 1250 kg × (40. m/s)2 –
½ × 1250 kg × (30. m/s)2
vf = 40. m/s
W = 4.4 x 105 J
Because W > 0, work is done on the car
increasing its kinetic energy.
More Work-Energy Problems
A 75 kg crate is pushed up a rough inclined
plane by a force of 470 N which is parallel to the
plane. The plane is 20. m long and makes an
angle of 25° with the ground.
(a) How much work is done by pushing the
crate from the bottom to the top of the
plane?
m = 75 kg
θ = 25°
FT = 470 N
l = 20. m
FN
.
Fp
Ff
FT
•
θ
θ F
w
FN
W = F × Δx × cosθ = 470 N × 20. m × 1 = 9.4 x 103 J
(b) What is the change in potential energy of the
crate?
ΔPEg = mgh = 75 kg × 9.80 m/s2 × 20. m ×
sin 30°
ΔPEg = 7.4 x 103 J
(c) What is the work done against friction?
Fp = Fwsinθ
Fp = 75 kg × 9.80 m/s2 × .500 = 370 N
FT = Fp + Ff
Ff = 470 N – 370 N = 1.0 x 102 N
W = Ff × Δx × cosθ
W = 1.0 x 102 N × 20. m × -1 = -2.0 x 103 J
It is important to realize that something is
remaining constant in the problem.
 That something is the sum of the ΔPEg
(gravitational potential energy) and Wf
(work done against friction) equals the
work done in pushing the crate the
entire length of the plane.
A 70. g arrow is fired from a bow whose string
exerts a force of 85 N on the arrow over a
distance of 90. cm. What is the speed of the
arrow as it leaves the bow?
m = 70. g
Δx = 90. cm
W = ΔKE
F × Δx × cosθ = ½mv2
F = 85 N
F × Δx × cosθ = ½mv2
v = (2 × F × Δx × cosθ/m)½
v = ((2 × 85 N × 90. cm × 1 m/100 cm × 1)/
70. g × 1 kg/103 g))½
v = 47 m/s
A 65 kg boy scout starts at an elevation of
1200. m and climbs to an elevation of 2700. m.
(a) What is the boy scout’s change in potential
energy?
m = 65 kg
hi = 1200. m
g = 9.80 m/s2
hf = 2700. m
ΔPEg = mgΔh = 65 kg × 9.80 m/s2 ×
(2700. m – 1200. m) = 9.6 x 105 J
(b) What is the minimum work?
W = ΔPEg = 9.6 x 105 J
(c) Is it possible for the work required to be
greater than the minimum amount of work?
Yes, if friction was present, WT = Wf + ΔPEg
Conservation of Energy
Gravity is considered to be a conservative
force.
 When computing ΔPEg, the value does
not depend on the path but only on the
initial and final positions.
 Sometimes gravity is referred to as a
state function because it is independent
of path.
Friction is considered to be a non-conservative
force or a dissipative force.
 Because friction is dependent on path, it
is sometimes referred to as a path
function.
 Whenever friction is present, there will
always be a certain amount of
mechanical energy that gets “wasted” or
turned into heat.
 The amount of energy “wasted” due to
friction does depend on the path taken.
The conservation of mechanical energy holds
true only if there are conservative forces acting
on a system.
 ΔE = 0
ΔKE + ΔPE = 0
A Pendulum Problem
A 0.57 kg pendulum bob at the end of a 0.90 m
cord is displaced 0.10 m and then released.
Using only energy considerations, determine
the maximum speed of the bob. The center of
gravity was raised 0.0050 m.
m = 0.57 kg
Δh = 0.0050 m
l = 0.90 m
g = 9.80 m/s2
Δx = 0.10 m
.
Δh
ΔE = 0
(KE + PEg)amp = (KE + PEg)eq pos
0
0
.
(PEg)amp = (KE)eq pos
mgΔh = ½mv2
v = 2gΔh
v = (2 × 9.80 m/s2 × 0.0050 m)½
v = 0.31 m/s
Power
Power is the rate at which work is done or the
rate at which energy is transformed into another
form of mechanical energy.
P=
W
Δt
=
J
s
= W
where P is the power measured in watts, W, W
is the work in joules, J, and Δt is the time in
seconds, s.
Power is sometimes expressed in horsepower,
1 hp = 746 W.
Power can also be expressed as
P=
W
Δt
=
F × Δx × cosθ
Δt
= F × v × cosθ
A Power Problem
A 60. kg boy runs up a flight of stairs in 3.7 s.
The vertical height of the stairs is 4.2 m.
(a) What is the boy’s output power both in watts
and horsepower?
m = 60. kg
Δt = 3.7 s Δh = 4.2 m
ΔPEg
W
mgΔh
P=
=
=
Δt
Δt
Δt
.
P =
60. kg × 9.80 m/s2 × 4.2 m
3.7 s
P = 670 W ×
1 hp
746 W
= 0.90 hp
(b) How much energy is involved?
P=
E
Δt
E = 670 W × 3.7 s = 2.5 x 103 J
Elastic Potential Energy
To stretch or compress a spring an amount s
from its equilibrium position requires a force F.
Hooke’s Law is given by
Fr = -ks
where Fr is the restoring force in N, k is the
spring constant in N/m, and s is the
displacement of the spring from the equilibrium
position.
 The negative sign indicates the restoring
force of the spring.
 The restoring force of the spring is always
trying to restore the spring to its
equilibrium position.
 The greater the value of k, the stiffer the
spring and the more difficult it is to stretch
or compress it.
Because Fr α s, the more a spring is
compressed or stretched, the greater the
restoring force Fr.
 W = ½•F•s•cosθ
 The cos θ always equals 1 and Fr = ks
which gives
W = ΔPEe = ½ks2
Applying the Conservation of Mechanical
Energy to a spring gives
 ΔE = 0
 (PEe + KE)amp = (PEe + KE)eq pos
0
PEe = KE
½ks2 = ½mv2
0
Spring Problem
A spring stretches 0.150 m when a 0.300 kg
mass is suspended from it. The spring is
stretched an additional 0.100 m from this
equilibrium position and released.
0.300 kg
0.150 m
0.100 m
0.300 kg
(a) Determine the spring constant.
m = 0.300 kg
Δx = 0.100 m
s = 0.150 m
Fr = ks
k=
k=
Fr
s
Fw
=
s
mg
=
s
0.300 kg × 9.80 m/s2
0.150 m
= 19.6 N/m
(b) What is the amplitude?
The amplitude is 0.100 m which is the
maximum displacement from the equilibrium
position.
(c) Determine the maximum velocity.
(PEe + KE)eq pos = (PEe + KE)amp
0
0
½mv2 = ½kΔx2
v = (kΔx2/m)½
v = (19.6 N/m × (0.100 m)2/0.300 kg)½
v = 0.808 m/s in the direction of either
amplitude
(d) Determine the velocity when the mass is
0.050 m from the equilibrium position.
(PEe + KE)amp = (PEe + KE)disp
0
½kΔx2 = ½ks2 + ½mv2
19.6 N/m × (0.100 m)2 =
19.6 N/m × (0.050 m)2 + 0.300 kg × v2
v = 0.700 m/s in the direction of either
amplitude
(e) Fnet = ma
a = Fnet/m = Fr/m = kΔx/m
19.6 N/m × 0.100 m
a =
= 6.53 m/s2
0.300 kg
Wrap Up Questions
A mass that is attached to the free end of a
spring undergoes simple harmonic motion.
Does the total energy change if the mass is
doubled and the amplitude remains the same?
The total energy is given by PEe = ½kxmax2,
therefore changing the mass has no effect on
the total energy.
Does the kinetic energy depend on mass?
Yes, because kinetic energy, KE, is given by
KE = ½mv2.
If the net work done on an object is equal to
zero, what is true about the object’s velocity?
The velocity remains the same because
W = ΔKE = 0 meaning that vf = vi.
Why is the work done by a frictional force, Ff, is
always negative?
The frictional force always opposes the
displacement of the object making the angle
between the force and displacement 180°, and
the cos 180° = -1.
Can the average power ever equal the
instantaneous power?
Yes, if the power remains constant throughout a
time interval.
Using the work-energy principle, show why
friction always reduces the kinetic object of an
object.
Because the angle between the displacement
and the frictional force is always 180°,
W = ΔKE = ½mvf2 - ½mvi2 < 0 because vf < vi.