12/12 do now. An 8. kg ball is fired horizontally from a 2.0 x 10 3 kg cannon initially at rest. After having been fired, the momentum of the ball is 2.40 x 10 3 kg·m/s east (neglect friction.]. Calculate the magnitude of the cannon’s velocity after the ball is fired. [show all work].
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Lesson 1: Basic Terminology and Concepts
Work
Definition and Mathematics of Work
Calculating the Amount of Work Done by Forces
Potential Energy
Kinetic Energy
Mechanical Energy
Power
Lesson 2: The Work-Energy Theorem
Internal vs. External Forces
The Work-Energy Connection
Analysis of Situations Involving External Forces
Analysis of Situations in Which Mechanical Energy is Conserved
Application and Practice Questions
Bar Chart Illustrations
Definition and Mathematics of Work
force
cause
displacement
cause
no work
work
no work
work
F
θ
d
W = F∙d∙cosθ
scalar
Work done – positive, negative or zero work
Positive work
negative work - force acts in the direction opposite the objects motion in order to slow it down.
no work
To Do Work, Forces Must Cause Displacements
W = F∙d∙cosθ = 0
Only the horizontal component of the force (Fcosθ) causes a horizontal displacement.
F & d are in the same direction,
θ is 0o.
d
F
Each path up to the seat top requires the same amount of work. The amount of work done by a force on any object is given by the equation W = F∙d∙cosθ
where F is the force, d is the displacement and θ is the angle between the force and the displacement vector. In all three cases, θ equals to 0o
Given: F = 20.0 N; d = 5.00 meters; m = 2.00 kg; θ = 0
Unknown: W = ?
W = F∙dcosθ
W = (20.0 N)(5.00 m)cos0o
W = 100. J
20.0 N
Given: d = h = 1.2 meters; m = 5.0 kg; θ = 0
Unknown: W = ?
W = F∙dcosθ
F = mg = (5.0 kg)(9.81 m/s2) cos0o = 49 N
W = F∙d = (49 N) (1.2 m) = 59 J
m = 2.3 kg
d = 2.0 m
θ= 30o
5.0 N
30o
2.3 kg
Fy = 6.0 N
dx = 3.0 m
6.0 N
W = Fxdx
W = (8.0 N)(3.0 m) = 24 J
8.0 N
3.0 m
Wyou = Fd
Wneighbor = (½ F)(4d) = 2 Fd = 2 Wyou
The neighbor, twice as much
Example: a block is pulled along a table with 10. N over a distance of 1.0 m.
W = Fd = (10. N)(1.0 m) = 10. J
work
Force (N)
Displacement (m)
height
base
area
Potential energy
vertical position (height).
gravitational
gravitational potential energy
PEgrav = m∙g∙h
Joules
Work and energy has the same unit
zero
ground
Solve:
∆PE = mg∆h
∆PE = (2.00 kg)(9.81m/s2)(0.92 m) = 18 J
The weight of the object is the slope of the line.
Weight = 25 J/1.0 m = 25 N
m = weight / g = 2.5 kg
elastic
F = kx
Spring force = spring constant x displacement
force
elongation
The slope represents spring constant: k = F / x
elongation
force
The slope represents the inverse of spring constant:
Slope = 1/k = x / F
PEs = Favg∙d = Favg∙x = (½ k∙x)∙x = ½ kx2
Note: F is the average force
PEs = ½ k∙x2
Elastic potential energy
elongation
Given:
k = 120 N/m
x = 0.20 m
Unknown:
PEs = ? J
Solve:
PEs = ½ kx2
PEs = ½ (120 N/m)(0.20 m)2
PEs = 2.4 J
PEs = ½ kx2
PEs = ½ k(0.20 m)2
PEs = (0.020 k) J
Solve:
PEs = ½ kx2
To find x, use Fs = kx,
(2.50 N) = (25.0 N/m)(x)
x = 0.100 m
PEs = ½ (25.0 N/m)(0.100 m)2
PEs = 0.125 J
Given:
Fs = 2.50 N
k = 25.0 N/m
Unknown:
PEs = ? J
PE = ½ kx2
15 J = ½ k (0.50 m)2
k = 120 N/m
Given:
F = 10. N
x = 0.20 m
Unknown: PEs = ? J
PEs = ½ kx2 = ½ (kx)(x)
PEs = ½ Fx
PEs = ½ (10. N)(.20 m)
PEs = 1.0 J
Given:
F = 0.2 N
x = 0.02 m
Unknown: PEs = ? J
PEs = ½ kx2 = ½ (kx)(x)
PEs = ½ Fx
PEs = ½ (0.2 N)(0.02 m)
PEs = 0.002 J
KE = ½ mv2
motion
__________________
Kinetic energy
speed
Note: Kinetic energy (along with every other type of energy) is a scalar, not a vector!
KE= ½ mv2
KE = 0.5 (55) (3) 2 = 247.5 J
Known:
Unknown:
Solve:
KE = ½ mv2
KE = ½ (2.7 kg)(1.5 m/s)2
KE = 3.0 J
Solve:
KE = ½ mv2
To find mass, we can use
Weight = mg
m = weight / g = 27 N / 9.81 m/s2
m = 2.8 kg
KE = ½ (2.8 kg)(1.5 m/s)2
KE = 3.1 J
Known:
Unknown:
Known:
Unknown:
Solve:
KE = ½ mv2
450 J = ½ (m)(25 m/s)2
m = 1.4 kg
a
b
c
d
force
weight
elongation
mass
elongation
A
B
force
Gravitational potential energy
C
D
height
TME = PE + KE
TME = PEgrav + PEspring + KE
TME =
TME =
TME =
TME =
TME =
Force vs. elongation
Force (N)
Elongation (m)
TME = PE + KE
TME = PEgrav + PEspring + KE
TME =
TME =
TME =
TME =
TME =
It can be assumed that Ben must apply an (80 kg x 9.81 m/s2) -Newton downward force upon the stairs to elevate his body.
ability
scalar
Wnet = ∆KE = KE2 – KE1
Fd = KE2 – KE1
Wnet = ∆KE = KEf - KEi
(-8000N) • d = -312 500 0 J
d = 39.1 m
Wnet = ∆KE = KEf - KEi
TMEi + Wext = TMEf
KEi + Wext = 0 J
½ •m•vi2 + F•d•cos(180o) = 0 J
F•d = ½ •m•vi2
d vi2
Stopping distance is dependent upon the square of the velocity.
4 m x 22 = 16 m
4 m x 32 = 36 m
4 m x 42 = 64 m
4 m x 52 = 100 m
TMEi + Wext = TMEf
or
KEi + PEi + Wext = KEf + PEf
The equation states that the initial amount of total mechanical energy (TMEi) plus the work done by external forces (Wext) is equal to the final amount of total mechanical energy (TMEf). Note the work can be positive and negative.
KEi + PEi + Wext = KEf + PEf
The external forces are applied force and friction force
0 + 0 + Wapp + Wf = 0 + (15 N)(0.20 m)
5.0 J + Wf = 3.0J
Wf = -2.0 J
2.0 J of work is done to overcome friction
example
Initially:PE = (0.25 kg)(9.81 m/s2)(2 m) = 4.9 J
KE = 0 J (the peach can is at rest)
Finally: PE = 0 J (the can's height is zero)
KE = 0 J (the peach can is at rest)
Wext = (500 N) • (d) • cos 180 = -(500 N)•d
KEi + PEi + Wext = KEf + PEf4.9 J + (-500 N)∙d = 0 J
d = 0.0098 m
In the diagram below, 450. joules of work is done raising a 72-newton weight a vertical distance of 5.0 meters. How much work is done to overcome friction as the weight is raised?
KEi + PEi + Wext = KEf + PEf
There are two external forces: applied force and friction force
The applied force did 450 J of work:
0 + 0 + 450 J + Wf = 0 + (72 N)(5.0m)
450 J + Wf = (72 N)(5.0 m)
Wf = -90 J
90 J of work is done to overcome friction
B
0.50 m
KEi + PEi + Wext = KEf + PEf
0+mg(0.80m)+Wf=0+ mg(0.50m)
Wf= -0.12 J
0.12 J of work is done to overcome friction
Wf = Ff∙d = 0.12 J
Ff∙(5.00m) = 0.12 J
Ff = 2.4 x 10-2 N
A
0.80 m
KEi + PEi + Wext = KEf + PEf
The external force is friction force only
0 + (40.N)(8.0m) + Wf = KEf + 0
320 J – 50 J = KEf
KEf = 270 J
KEi + PEi + Wext = KEf + PEf
The external forces are applied force and friction force
0 + 0 + Wapp + Wf = 0 + (15 N)(0.20 m)
4.0 J + Wf = 3.0J
Wf = -1.0 J
1.0 J of work is done to overcome friction
KEi + PEi + Wext = KEf + PEf
If onlyinternal forces are doing work (no work done by external forces), then there is no change in the total amount of mechanical energy. The total mechanical energy is said to be conserved.
Wext = 0
KEi + PEi = KEf + PEf
KEi + PEi = KEf + PE
KEi + PEi = KEf + PEf
KEi - KEf = PEf - PEi
Decrease in KE = Increase in PE
mechanical energy
constant
0.153
0
0.306
0.306
1.73
2.45
use the heights and the speeds given in the table below to fill in the remaining cells at the various locations in a 0.200-kg bob's trajectory.
3.920
0
3.920
3.162
2.920
1.000
4.134
2.211
1.709
3.920
1.748
2.172
3.920
4.660
5.675
0.700
3.220
3.920
PE
pos
conserved
conserved.
increases
decreases
The graph shows as a ball is dropped, how its energy is transformed.
constant
KEi + PEi = KEf + PEf
0 + mg(3.0m) = KEf + mg(1.0m)
KEf = mg∆h = (3.0 kg)(9.81 m/s2)(3.0 m – 1.0 m) = 59 J
(KE gained at B is potential lost from A to B
KEi + PEi = KEf + PEf
KEf - KEi = PEi - PEf
The KE gained is PE lost
∆KE = ∆PE = mg∆h
KE = (0.20 kg)(9.81 m/s2)(1.00 m) = 2.0 J
KEi + PEi = KEf + PEf
KEf - KEi = PEi - PEf
∆KE (gained) = GPE (lost)
∆KE = (1.0 kg)(9.81 m/s2)(20.m-0) = 2.0x102 J
20. m/s
KEi + PEi + Wext = KEf + PEf
Wext = KEf + PEf – (KEi + PEi)
Wext = 100 J - 0
Wext = 100 J
10. m
In this situation, the force doing the work is gravity, which is an internal force.
KEi + PEi = KEf + PEf
KEi = 0J;
PEi = mgh=(55.0kg)(9.81m/s2)(3.00m) = 1619 J
PEf = mgh = (55.0kg)(9.81m/s2)(1.00m) = 540. J
KEf = KEi + PEi - PEf = 1080 J
KEi + PEi + Wext = KEf + PEf
0 + 0 + 100 J = KEf + 0
KEf = 100 J
KE = ½ mv2
100 J = ½ (0.5 kg)v2 v = 20 m/s
KEi + PEi = KEf + PEf
0 + mgh = 0 + ½ kx2
mgh = ½ kx2
(0.1 kg)(9.81m/s2)(2m) = ½∙k (0.1m)2
k = 400 N/m
KEi + PEi = KEf + PEf
0 + ½ kx2 = ½ mv2 + 0
½ kx2 = ½ mv2
k = mv2/x2
16 m/s