82 terms

# Physics Equations

MCAT physics equations
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Displacement (d)
d = Δx = final position - initial positon = net distance (plus direction)
Average velocity (v)
v = Δx/Δt = d/Δt
Average acceleration (a)
a = Δv/Δt
What assumption do we make with the BIG 5 equations? What are the 3 most frequently used equations?
That acceleration is constant (uniformly accelerated motion)

2. Missing d
3. Missing vf (v final, v)
5. Missing t
BIG 5:

1. Missing a
d = ½(v₀ + v)t
BIG 5:

2. Missing d
v = v₀ + at
BIG 5:

3. Missing v (v final)
d = v₀²t + ½at²
BIG 5:

4. Missing v₀
d = vt - ½at²
BIG 5:

5. Missing t
v² = v₀² + 2ad
For projection motion, we assume that the downward direction is...
NEGATIVE direction.

DOWNWARD = NEGATIVE DIRECTION.
PROJECTILE MOTION - Horizontal Motion
(in the x-direction)

Give the equation for:
1. Displacement
2. Velocity
3. Acceleration
1. d = v₀t (in the x-direction)

2. v₀x = vx (velocity is CONSTANT in the x-direction!!)

v₀x = v₀cosθ₀

3. ax = 0 (acceleration in the x-direction is ZERO!)
PROJECTILE MOTION - Vertical Motion

Give the equation for:
1. Displacement
2. Velocity
3. Acceleration
1. y = v₀yt + ½(-g)t² (in the y-direction)

2. vy = v₀y + (-g)t (in the y-direction)

v₀y = v₀sinθ₀

3. ay = -g (accerleration in the y-direction)
Newton's 1st Law
Fnet = 0

velocity (v) = constant!
∴a = 0 --> F = 0
Newton's 2nd Law
Fnet = m a
Newton's 3rd Law
F₁₂ = F₂₁

Action-Reaction pair
Weight
w = mg

Force due to gravity (Fg) = weight = mg
Gravitational force (Fg)
Fg = (GMm)/r²

since w = Fg,
mg = (GMm)r²
∴g = GM/r²
Kinetic friction
Ff = μkFN

μk = coefficient of KINETIC friction
FN = normal force
Static friction
Ff, max = μsFN

μs = coefficient of STATIC friction
FN = normal force
What is the relationship between μk and μs, max?
μs, max > μk

ALWAYS!
What is the force due to gravity acting PARALLEL to the inclined plane? (sliding down the plane)
F = mgsinθ

θ = angle between incline and horizontal
What is the force due to gravity acting PERPENDICULAR to the inclined plane?
F = mgcosθ

θ = angle between incline and horizontal
Center of mass xCM
xCM = (m₁x₁ + m₂x₂ + m₃x₃...)/(m₁ + m₂ + m₃...)
Center of gravity xCG
xCG = (w₁x₁ + w₂x₂ + w₃x₃...)/(w₁ + w₂ + w₃...)
In a uniform gravitational field (constant g)...
xCM = xCG
Centripetal acceleration (ac)
ac = v²/r
Centripetal force (Fc)
Fc = mac = mv²/r
Torque (τ)
τ = rFsinθ

r = distance from pivot point to F
F = force applied (Fapp)
θ = angle between r and F
When does max torque occur?
When θ =90°
τ= r F
What does it mean to be in ROTATIONAL EQUILIBRIUM?
τcw = τccw
∴τnet = 0 (net torque = 0)
clockwise torque = counterclockwise torque
τnet (net torque)
τnet = I α

I = inertia
α = rotational acceleration
*compare to Fnet = ma
Is the inertia greater for a solid sphere or a hollow sphere?
If the same net torque was applied to each sphere, which would spin faster?
Hollow sphere has GREATER inertia than solid sphere.

Solid sphere would spin faster because it has less inertia!
What does inertia depend on?
It depends on how a mass is distributed in an object.
Work (W)
W = Fdcosθ

Unit = Joules (J)

F = force applied
d = displacement
θ = angle between F and d

Wmax occurs when θ = 0
Power (P)
P = W/t

Unit: J/s = Watts (W)

W = Work
t = time
Kinetic Energy (KE)
KE = ½mv²

Unit: Joules (J)

KE always has a + value; object has to be in motion!
Work-energy theorem (Wtotal)
Wtotal = ΔKE = KEf - KE₀
Gravitational Potential Energy (PE)
PE = mgh

m = mass
g = acceleration due to gravity
h = height
Total Mechanical Energy (E)
E = KE + PE
Conservation of Energy (ideal frictionless world)
E₀ = E

KE₀ + PE₀ = KE + PE
Work Energy Theorem (including PE)
W = ΔKE = -ΔPE
Conservation of Energy (when friction is taken into account)
KE₀ + PE₀ - Wf = KE + PE

Wf = work done by FRICTION!
Momentum (p)
p = mv

*p is always in the same direction as v

Unit = kg∗m/s
Conservation of total momentum
p₀ = p
m₁v₁ = m₁v₁'

total p₀ = total pf
Impulse-momentum theorem (J)
J = Δp = FΔt

change in momentum (Δp) = impulse (J) = FΔt
During a collision (short duration of interaction between 2 objects)...
1. Total momentum (p) is ALWAYS conserved!
2. KE is NOT always conserved!
1. Elastic collision
(think of 2 pool balls)
1. p is conserved
2. KE is ALSO conserved!

∴p₀ = pf
∴KE₀ = KEf
2. Inelastic collision
(think of a car crash)
1. p is conserved
2. KE is NOT conserved: one object becomes deformed; loss in energy

m₁v₀₁ + m₂v₀₂ = m₁v₁' + m₂v₂'
3. Perfectly inelastic collision
(think of 2 objects becoming stuck together)
1. p is conserved
2. objects get stuck together and move with the same v' (final velocity!)

m₁v₀₁ + m₂v₀₂ = (m₁ + m₂)v'
During the collision, there is a short duration of interaction between the 2 objects so we can apply...
Newton's 3rd Law: action-reaction pair!
F₁₂ = F₂₁
Angular momentum (L)
L = Iω

I = moment of inetia
ω = angular velocity

*compare to p = mv
Conservation of total angular momentum
L₀ = L'
Density (ρ)
ρ = m/V

m = mass
V = volume
Specific gravity (sp gr)
sp gr = ρ/ρH₂O

ρ = density of object or fluid
ρH₂O = density of water
What is the ρ of water?
ρH₂O = 1000 kg/m³ = 1 g/cm³ = 1 kg/L
Pressure (P)
F/A

F = perpendicularly applied force
A = area

Unit = N/m₂ = Pascal (Pa)
1atm = 100 kPa
Gauge Pressure (Pg)
Pressure felt at a particular depth = hydrostatic pressure
Pg = ρ g D

ρ = density of fluid
g = accl due to gravity
D = depth
Total hydrostatic pressure (Ptotal)
Ptotal = Patm + Pg

Patm = pressure at the surface
Pg = gauge pressure
Buoyant Force (Fb)
*Archimedes Principle
Fb = ρ V g

ρ = density of fluid
V = Vsub = volume of object submerged = volume of fluid displaced
g = accl due to gravity
Archimedes Principle
There is an upward acting force on an object that is rising, sinking, floating, or completely submerged in a fluid. This force is called the buoyant force (Fb) and it equals the weight of the fluid displaced.
Case 1: Object is sinking
ρ object > ρ fluid
Case 2: Object is completely submerged and reaches the bottom of the tank
Fnet = Fn + Fb - Fg = 0
∴Fn = Fg - Fb

ρ object > ρ fluid

*Fn = normal force: weight of the object inside the fluid = apparent weight
Case 3: Object is rising up in the fluid
Fnet = Fb - Fg = ma

ρ fluid > ρ object
Case 4: Object is floating on the surface
Fnet = Fb - Fg = 0
∴Fb = Fg (= weight of object)
ρ V g = m g
ρ V = m = ρ₀V₀ (object)
ρ fluid > ρ object

V/V₀ = ρ₀/ρ = specific gravity
Remember, ice is always
90% submerged! Think of the Titanic...
What are the 4 conditions to have an IDEAL FLUID?
1. No or negligible viscosity.
2. Laminar flow (continuous, streamline flow).
3. Fluid must be incompressible.
4. Flow must be continuous.
Flow rate (f)
f = A v
A: cross-sectional area
v: velocity of fluid
Unit: m³/s
Continuity equation (f₁ = f₂)
A₁v₁ = A₂v₂
Bernoulli's equation (ideal fluid)
describes ideal fluid flow through a pipe
*looks similar to conservation of energy equation

P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂

*pgh is NOT gauge pressure, but the PE of the fluid at a certain height, h.
Apply Bernoulli's equation to a HORIZONTAL PIPE.
Since h₁ = h₂, the equation looks like this:
P₁ + ½ρv₁² = P₂ + ½ρv₂²
For a horizontal pipe, describe the relationships between pressure and velocity.
1. If v₁ > v₂, then P₁ < P₂.
2. If v₂ > v₁, then P₂ < P₁.
Objects will move from _______ pressure to _______ pressure. Think of an example.
High P --> Low P

Think of air planes and the lift force!
Toricelli's result (velocity of fluid coming out of a hole in a tank)
*v efflux
v = √(2gD)
Stress
stress = F/A
Strain
strain = ΔL/L₀

ΔL = change in length
L₀ = initial length
Hooke's Law
stress = modulus ∗ strain
Tension or compression (ΔL)
ΔL = FL₀/EA

*FLEA
Shear (X)
X = FL₀/AG

*FLAG
What is the elementary charge of an e-?
1.6 ∗ 10⁻¹⁹ C
Coulomb's law (Electric force: force on a charged particle)
F = kQq/r²

k = Coulomb's constant = 9 ∗10⁹ Nm²/C²
Q = charge that applies force
q = test charged being acted on by Q
r = distance between 2 charges

Positive F = repelling force
Negative F = attractive force
1 microcoulomb equals how many Coulomb?
1μC = 10⁻⁶ C
1C = 10⁶μC
State the principal of superposition.
The net electric force on a charge (q) due to a collection of other charges (Q's) is equal to the sum of the individual forces that each of the Q's alone exerts on q