Physics Learning Cheat Sheet
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1The Physics Learning Resource Center Reference sheet
Vectors
Some Basic Considerations
The Basic Kinematic Equations
Forces Mediated Through Fields
Potential Energy
For a particle experiencing constant
If F is a conservative force and
is any path
The sum, A + B, of two vectors, A and B,
Draw a picture to help you visualize the
The force exerted on a particle with mass,
acceleration
from an arbitrary location, called ground, to
is the vector created by placing the tail of
problem. Label it with the algebraic
m, from a gravitational field, g (r), at a
the point determined by r, then the potential
B on the head of A and then drawing a
variables for the quantities given to you.
position, r, is F = mg (r).
2
1
r(t) =
at
+ v
t + r
.
energy, U (r), from F is
0
0
line segment from A’s tail to B’s head:
2
Think of what relationships exist between
The force exerted on a particle with
2
2
v
= v
+ 2a ∆r.
0
what you are given and what is needed.
charge, q, with a velocity, v, from an
v = at + v
.
U (r) =
F (r) dr =
W
.
0
Ground to
electric field, E (r), and a magnetic field,
Work the problem with the algebraic
v + v
0
v
=
.
B (r), at a position, r, is
ave
variables for as long as possible. Only
The potential energy at ground is always
2
insert numbers at the end.
zero.
F = q [E (r) + v
B (r)] .
To determine the difference between two
Use units when inserting the numbers and
With ground infinitely far away,
vectors, rewrite the equation as a sum:
Fundamental Dynamic Quantities
This is the Lorentz force equation.
make sure they match correctly.
– The potential energy from gravity is
A can be rewritten as
B = C
m is the mass of the particle; both its
Use significant digits correctly while doing
A + B = C, so B in the above illustration
mM
tendency to resist acceleration and its
calculations.
U (r) =
G
.
is the difference between C and A.
Classical Fields
interaction strength with gravitational
r
A and B are components of C because they
fields.
A USEFUL APPROXIMATION At an
The gravitational field, g (r), at a position,
add up to C. Components that are
Fundamental Kinematic Quantities
altitude of less than 1100 meters, the
q is the charge of the particle; its
r, created by a point source mass, M , at
perpendicular to each other are
potential energy at a height, h, is
the origin is
interaction strength with electric fields.
t is the instant of time that we are looking
particularly useful. We commonly use i,
at the system.
which points in the positive x-direction, j,
M
U (h) = mgh.
g (r) =
G
ˆ r .
is the instant that the initial conditions of
which points in the positive y-direction,
t
r
2
Newton’s Laws
0
– The potential energy from an electric
and k, which points in the positive
the system were set; often has a value of
A USEFUL APPROXIMATION At an
1. If there is no net force on a particle, it
field is
zero.
z-direction. All three vectors have a length
altitude of less than 1100 meters, g is
1
qQ
experiences no acceleration. The particle is
of one.
U (r) =
.
r(t)ˆ r (t) is the
downward with a constant value of
r(t)
x(t)i + y(t)j + z(t)k
4π
r
in equilibrium.
0
The dot product, A B, between two
position vector. It indicates the particle’s
g = 9.81 m/s
2
.
The potential energy from a magnetic field
2. Let
F be the net force on the
location and it is time-dependent.
vectors, A and B, is the real number
The electric field, E (r), at a position, r,
is zero because magnetic forces never do
particle—the vector sum of all the forces
) is the initial position of the
created by a point source charge, Q, at the
r
r (t
work.
0
0
A B = AB cos θ = A
B
+ A
B
+ A
B
.
acting on it. This net force produces an
x
x
y
y
z
z
particle.
origin is
The potential energy of a spring is
acceleration on the particle determined by
1
Q
Some useful dot products are
is the displacement vector.
∆r = r
r
E (r) =
ˆ r .
0
F = ma.
1
2
4π
r
A CALCULUS QUANTITY dr is the
0
2
U (x) =
kx
.
3. If Particle 1 exerts a force on Particle 2,
i i = j j = k k = 1.
2
The magnetic field, B (r), at a position, r,
infinitesimal displacement.
Particle 2 exerts a force equal in magnitude
created by a point source charge, Q, at the
All other possible dot products between i,
A CALCULUS QUANTITY
and opposite in direction on Particle 1.
origin and moving with a velocity, v, is
j, and k are zero.
Potential
dr
The cross product, A
B, between two
v =
µ
Qv
ˆ r
When potential energy comes from a field,
0
dt
B (r) =
.
vectors, A and B, is a vector whose length
Contact Forces
2
4π
r
there exists a quantity known as potential
is the instantaneous velocity of the
is A
B = AB sin θ.
(not to be confused with potential energy).
When a particle exerts a force on a surface,
particle. Its magnitude, v, is the
You find the direction of the cross product
The electric potential, V (r), is
the surface feels the component of that
instantaneous speed of the particle.
Angular Kinematics and Dynamics
using the right-hand rule: point the fingers
force that is perpendicular, or normal, to
) is the initial velocity of the
v
v (t
1
Q
Everything that happens in a straight line
of your right hand along the direction of A
0
0
the surface. By Newton’s third law, the
V (r) =
.
particle.
4π
r
and hold your hand so that you can turn it
has its equivalent when rotating:
0
surface exerts a force, N , equal and
Linear Quantity Angular Quantity
towards B. Your thumb is pointing in the
Note that, of F , E, U , and V , the potential is
opposite to this normal component.
∆r
∆r
direction of the cross product.
=
t
t
v
the easiest quantity to calculate. The four
ave
The frictional force, f , has maximum
t
t
∆t
0
x
θ
quantities are related by
magnitude µN .
is the average velocity of the particle over
v
ω
the period of time ∆t. Its magnitude, v
,
The fluid resistance, f has a
ave
=q
a
α
F
E
is the average speed of the particle.
velocity-dependent magnitude, so the
m
I
=
U
=
V
basic kinematic equations do not apply.
The vector area of a surface is a vector
A CALCULUS QUANTITY
q
q
U =qV
whose length is equal to the area of the
U
V
F
τ
dv
low velocities,
kv,
surface and that points “out” at ninety
The linear and angular quantities are
a =
f =
dt
. high velocities.
degrees from the surface.
2
Dv
connected by the relationship s = rθ when
Gauss’s Law
is the instantaneous acceleration of the
the rotational motion is circular.
A CALCULUS CONCEPT The path
The buoyant force, B, is upward with a
particle.
integral of a vector function of position,
Q
inside
magnitude equal to the weight of the fluid
Φ
E (r) da =
.
f (r), along a path, , defined by the curve
0
displaced.
Work Energy Theorem
∆v
(x(λ), y(λ), z(λ)), is
THE TRICK In situations where the charge
a
=
ave
∆t
A spring “stretched” a distance, x, (a
distribution reduces the spatial dependency
is the average acceleration of the particle
dx
dy
dz
negative value means the spring is
of E, draw a Gaussian surface that keeps E
f (r) dr =
f
+ f
+ f
dλ.
1
2
1
2
over the the period of time ∆t.
W
F (r) dr = ∆K
m
v
m
v
.
x
y
z
squished) exerts a force, F =
f
i
dλ
dλ
dλ
kxi.
f
i
2
2
constant and pull it outside the integral.
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