# Kinematics, Work, & Energy

```Bootcamp:
Kinematics, Energy, and Work
Learning
Objectives
Use kinematics equations to solve problems involving motion
Describe Newton’s 3 laws of motion
Differentiate between kinetic and potential energy
Solve problems that involve a transfer of energy
Kinematics, Energy, and Work
Use kinematics equations to solve
problems involving motion
Kinematics
Question 1
A driver decides to test the power of his new sports car. While traveling at 16.05 m/s, the driver
steps on the accelerator and reaches a speed of 45.72 m/s in just 2.5 seconds. What was the car’s
approximate acceleration, and approximately how far did it travel during this acceleration event,
respectively?
A. 1.2 m/s2, 70 m
B. 12 m/s2, 112 m
C. 1.2 m/s2, 44 m
D. 12 m/s2, 78 m
Kinematics
The Kinematic Equations
Memorize these equations:
VAT: vf = vo + at
VAX: vf2 = vo2 + 2aΔx
TAX: Δx = vot + &frac12;at2
d = vavgt
vavg = (vo + vf)/2
Kinematics
Question 1
A driver decides to test the power of their new sports car. While traveling at 16.05 m/s, the driver
steps on the accelerator and reaches a speed of 45.72 m/s in just 2.5 seconds. What was the car’s
approximate acceleration, and approximately how far did it travel during this acceleration event,
respectively?
A. 1.2 m/s2, 70 m
B. 12 m/s2, 112 m
C. 1.2 m/s2, 44 m
D. 12 m/s2, 78 m
Kinematics, Energy, and Work
Describe Newton’s 3 laws of motion
Laws of Motion
Question 2
An 80 kg man fires a 7 mm round with a mass of 10 g from a hunting rifle. The bullet experiences an
acceleration of 1,200,000 m/s2. According to Newton’s laws of motion, what is the force on the
bullet? What is the acceleration of the man after the recoil of the shot?
A. 1.2 x 104 N, 75 m/s2
B. 1.2 x 104 N, 150 m/s2
C. 1.2 x 107 N, 75 m/s2
D. 1.2 x 107 N, 150 m/s2
Laws of Motion
Newton's 1st Law
The law of inertia: an object’s velocity will remain constant unless
acted upon by a net unbalanced force.
If Fnet = zero, then anet = zero and velocity is constant.
If velocity is changing, there must be a force acting on the object.
If velocity is constant, no net force is present on the object.
Ex: car on cruise control.
Laws of Motion
Newton's 2nd Law
An object’s acceleration depends on the mass of the object and the
amount of force applied.
Heavier, more massive objects require more force to accelerate them
Laws of Motion
Newton's 3rd Law
For every action, there is an equal and opposite reaction.
When firing a rifle, the explosion exerts a forward force on the projectile, and an identical
backwards force on the rifle and the person holding it.
Laws of Motion
Question 2
An 80 kg man fires a 7 mm round with a mass of 10 g from a hunting rifle. The bullet experiences an
acceleration of 1,200,000 m/s2. According to Newton’s laws of motion, what is the force on the
bullet? What is the acceleration of the man after the recoil of the shot?
A. 1.2 x 104 N, 75 m/s2
B. 1.2 x 104 N, 150 m/s2
C. 1.2 x 107 N, 75 m/s2
D. 1.2 x 107 N, 150 m/s2
Kinematics, Energy, and Work
Differentiate between kinetic and
potential energy
Energy
Question 3
A 20 kg child at the local playground starts at rest at the top of a slide and slides down a frictionless
slide. Assuming energy is conserved, what is the height of the slide if the child comes to the bottom
with a speed of 6 m/s?
A. 0.6 m
B. 1 m
C. 1.8 m
D. 3.6 m
Energy
Defines an object’s capacity to cause change in its environment
Over time, may alter positions, temperatures, chemical bonds…
Energy* cannot be:
Created
Destroyed
Energy can be:
Transferred
Stored
Transformed
All forms of energy are quantifiable in units of joules (J)
1J = 1N⋅m =
*
1 kg⋅m2
s2
Potential Energy
Energy
Energy stored within systems as a result of specific configurations
When positions change, stored U can be released to perform work
A potential energy is associated with each conservative force (gravitational, electric, elastic)
height (h)
U = mgh
Set one endpoint to 0 to
simplify calculations!
m
g
h0
The spring constant, k, describes the
rigidity or stiffness of a given spring
U = &frac12;kx2
∆x
h
hf
k
Gravitational
Elastic
Energy
Kinetic Energy
Energy carried by massive objects in motion
1J =
1kg⋅m2
KE = &frac12;mv2
m = 2 kg
v = 1 m/s
&frac12;mv2 = ?
&frac12;mv2 = 1 J
Can be converted from potential energy
s2
Energy
Conservation of Energy
The total mechanical energy (E) refers to K and U together:
When only conservative forces act:
E=U+K
ΔU = ΔK
Total E is unchanging, converting between U &amp; K:
Real-world systems lose energy to dissipative forces:
U = mgh
K = &frac12;mv2
Wnonconservative = ΔE
Energy
Conservation of energy
E=U+K
hmax
E = 90 J + 10 J
h
E = 45 J + 55 J
E = 0 J + 100 J
ho
vo
v
vx
h
v
ho
E = 45 J + 55 J
E = 0 J + 100 J
vo
Energy
Question 3
A 20 kg child at the local playground starts at rest at the top of a slide and slides down a frictionless
slide. Assuming energy is conserved, what is the height of the slide if the child comes to the bottom
with a speed of 6 m/s?
A. 0.6 m
W = ∆KE = ∆U
B. 1 m
C. 1.8 m
D. 3.6 m
&frac12; mv2 = mgh
h = v2/2g = 36/20 = 1.8 m
Kinematics, Energy, and Work
Solve problems that involve a
transfer of energy
Transfer of
Energy
Question 4
A lawnmower has a handle that is attached at an angle of 60 deg. If a man using this lawn mower to
mow 100 meters of grass uses a force of 50 N to propel the mower, how much work is the man
doing? Note: cos(60&deg;) = 0.5
A. 25 J
B. 50 J
C. 2500 J
D. 5000 J
Transfer of
Energy
Force &amp; Displacement
Energy is transferred when a force causes a displacement of a body: Work is done
W = ΔE = Fd cos θ
W∝ F
d
W∝ d
1 J = 1 N⋅m =
Wmax at θ = 0&deg;, 180&deg;
Wmin at θ = 90&deg;, 270&deg;
F
θ = 0&deg;
1 kg⋅m2
s2
Transfer of
Energy
Force &amp; Displacement
W = ΔE = Fd cos θ
If a 100 N force pushes a rock 20 m
forward, how much work is done?
If the force acts at a 30&deg; angle to the forward
path, how much work is done?
F = 100 N
100 N
20 m = d
F
θ = 0&deg;
20 m = d
87 N
θ = 30&deg;
Transfer of
Energy
Question 4
A lawnmower has a handle that is attached at an angle of 60 deg. If a man using this lawn mower to
mow 100 meters of grass uses a force of 50 N to propel the mower, how much work is the man
doing? Note: cos(60&deg;) = 0.5
A. 25 J
B. 50 J
C. 2500 J
D. 5000 J
Transfer of
Energy
Question 5
When deep under water, nitrogen gas bubbles can diffuse from the vessels into bodily tissues.
Decompression sickness results when a person experiences large, quick drops in ambient pressure,
such as when a scuba diver ascends rapidly to the surface. If a diver ascends too quickly, the
nitrogen gas bubbles in the diver’s tissue will:
A. expand and release energy.
B. expand and heat up.
C. compress and do negative work.
D. compress and gain energy.
Transfer of
Energy
Work &amp; Heat
Energy is transferred only by work (W) or heat (Q)
Work and heat describes a movement of energy
ΔUsys = Qabsorbed – Wby sys
1st Law of Thermodynamics:
ΔE = W
Conservative forces only:
Gravitational
Electric
Elastic
U0 = mgh0
Uf = mghf
W = ΔE = ΔU = mgΔh
1J=
1 kg⋅m2
s2
Transfer of
Energy
Pressure &amp; Volume
Pressure-Volume Curves
Changes to the internal energy of ideal gases alter their volume and pressure
Physical thermodynamics tracks changes in the energy of such systems
V = d3
PΔV =
P=
Fd3
d2
F
A
=
F
A
d2
d
= Fd = W
V
PΔV = W
F
P
A F
ideal gas
ΔV
Transfer of
Energy
Pressure &amp; Volume
PΔV = W
1
Isovolumetric
(Isochoric)
ΔV = 0
PΔV = 0
P
ΔP
2
2
1
W
W
Isobaric
ΔP = 0
PΔV = W
3
4
3
W
P &amp; V changing
Area beneath = W
Here: Pave ΔV = W
V
ΔV
4
Closed loop
Area within = W
P0,V0 = Pf, Vf
Here: ΔPΔV = W
Transfer of
Energy
Question 5
When deep under water, nitrogen gas bubbles can diffuse from the vessels into bodily tissues.
Decompression sickness results when a person experiences large, quick drops in ambient pressure,
such as when a scuba diver ascends rapidly to the surface. If a diver ascended too quickly, the
nitrogen gas bubbles in the diver’s tissue will:
A. expand and release energy.
B. expand and heat up.
C. compress and do negative work.
D. compress and gain energy.
Transfer of
Energy
Question 6
If an 1800 kg car accelerates from 0–20 m/s in 3 seconds, how much power is this car capable of
producing?
A. 60 kW
Variables given:
Variables needed:
B. 120 kW
vo = 0 m/s
Power, P = ?
C. 240 kW
v = 20 m/s
D. 360 kW
m = 1800 kg
t=3s
Transfer of
Energy
Power
Denotes the rate at which work is performed
Measured in units of watts (W)
Calculated from energy change between two points in time
W
P=
1W=
t
J
s
=
Also used for rate of energy dissipation by circuit elements
Expressed via current (I), voltage (ΔV), and resistance (R)
Uf
U0
P = ΔU/ Δt = mgΔh/Δt
P = IΔV
P = I2R
=
ΔE
Δt
1 kg⋅m2
s3
Transfer of
Energy
Question 6
If an 1800 kg car accelerates from 0–20 m/s in 3 seconds, how much power is this car capable of
producing?
A. 60 kW
B. 120 kW
C. 240 kW
D. 360 kW
P = ∆E
t
∆E = ∆W = ∆KE = &frac12;mv2
P = &frac12; (1800 kg) (20 m/s)2
3s
P = (900 kg) (400 m2/s2)
3s
P = 120,000 W = 120 kW
Work-Energy
Theorem
Question 7
A 1 kg block slides down a ramp from rest. After sliding for a brief period, the block has lost 3
meters of height and is traveling at 6 m/s. How much energy has been lost to friction?
A. 0 J
B. 12 J
C. 18 J
D. 30 J
Work-Energy
Theorem
Calculating Energy Loss
Together, work by all the forces acting on an object equals to the change to its kinetic energy
No appeal to magnitude, displacement, or time over which forces act
Can track net energy loss due to dissipative forces
Wnet = ΔK
m = 1 kg
v0 = 0 m/s
Δh = 3 m
f
N
vf = 6 m/s
mg
How much energy is dissipated by friction?
ΔK = Kf
ΔK = &frac12;mv2
Wnet
ΔK = &frac12;(1)(36) = 18 J
The block has
18 J of KE
ΔE = ΔU + ΔK = Wfriction
ΔU = mgΔh
ΔU = (1)(–10)(3) = –30 J
Wfriction = – 30 J + 18 J
Wfriction = –12 J
The system
started with 30 J
12 J was lost to
friction
Work-Energy
Theorem
Question 7
A 1 kg block slides down a ramp from rest. After sliding for a brief period, the block has lost 3
meters of height and is traveling at 6 m/s. How much energy has been lost to friction?
A. 0 J
B. 12 J
C. 18 J
D. 30 J
Kinematics, Energy, and Work
Takeaways
Takeaways
Kinematic Equations
vf = vo + at
vf2 = vo2 + 2aΔx
Δx = vot + &frac12;at2
d = vavgt
vavg = (vo + vf)/2
Takeaways
First law
m
Second law
Fnet = 0
m
a
F = ma
F
a
m
m
Third law
FAB = –FBA
Engine
F (Identical force)
F = ma
Takeaways
When only conservative forces act, straightforward calculations track changes in mechanical energy
Energy is a transferrable,
transformable, storable,
property of objects
It measures the ability to
change the environment in
specific ways
Total mechanical energy is
comprised of kinetic and
potential energies
Conservative forces do not
alter the total amount of
useful energy in a system
KE = &frac12;mv2
Ug = mgh
Uspring = &frac12;kx2
E=U+K
Wnonconservative = ΔE
Takeaways
Work is the major process by which energy is transferred between objects
W = Fd cos θ
Work:
Is typically encountered when
forces cause displacements in the
directions they act
Measures an amount of energy
transferred or transformed by a
force
Accompanies changes to the
volume of pressurized gases
The work-energy theorem
accounts for all forces acting
W = ΔE
d
1st Law:
θ
F cos θ
F
ΔUsys = Qabsorbed – Wby sys
P
PΔV = W
ΔP
W
V
ΔV
Wnet = ΔK
Bootcamp:
Kinematics, Energy, and Work
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