Goes over... Chapter 6: Work and Kinetic Energy Chapter 7: Potential Energy, Energy Conservation Chapter 8: Momentum, Impulse, Collision
|
m = mass
|
g = 9.8 m/s
|
|
F = Weight (N)
|
F = m ⋅ g
|
|
s = distance
|
K
E = Kinetic Energy |
|
W = Workdone
|
|
|
Power = P (Watts)
|
x = cos
y = sin |
|
1 km = 1000m
|
1kg = 1000g
|
|
ΔK = K
f - Ki |
Friction = always negative
|
|
g = -9.8 (decreasing)
|
g = 9.8 (normal)
|
|
a = g (gravitational acceleration)
|
Θ = Angle between F and s
|
|
|| = Component of F parallel to dl
|
v = velocity
|
|
W = (∫P
2 to P1) F⋅dl |
W = F
|| dl |
|
W = (∫P
2 to P1) F⋅cosΘ⋅dl |
W = F ⋅ s (Joules)
|
|
P
av = ΔW / Δt |
P = lim Δt > 0 (ΔW / Δt) = dW / dt
|
|
V
f = Vi2 + 2 ⋅ a ⋅ s |
P = (W/t)
|
|
Constant Speed : (a = 0)
|
F = force
P = F ⋅ v |
|
Friction (opposite) = cos(180o)
|
|
|
W
x = F (cosΘ)⋅s || Wy = F (sinΘ)⋅s |
|
|
a = (V
f2 - Vi2) / (2⋅ s) |
|
|
F
s = (1/2)⋅m⋅Vf2 - (1/2)⋅m⋅Vi2 |
|
|
W
grav = m⋅g⋅h |
P = (W/t)
|
|
K
E = (1/2)⋅m⋅V2 |
P
E = m⋅g⋅h |
|
Potential Energy = U, P
E |
ΔK = -ΔU
grav |
|
K = Kinetic Energy
|
R = Radius
|
|
s = y
f - yi |
U
grav = m ⋅ g ⋅ y |
|
Δs = Δxî + Δyĵ
|
cm = circular motion |
|
k = constant of spring
|
|
|
P
E = (1/2)⋅k⋅x2 |
|
|
W
grav = w-vector ⋅ Δs-vector |
Diameter = 2 ⋅ Radius
|
|
W
f = Work Done by Friction |
if elastic...
K E = PE |
|
W
grav = F × s |
W
grav = m ⋅ g ⋅ yi - m ⋅ g ⋅ yf |
|
K
i + Ui = Kf + Uf |
(1/2)⋅m⋅V
i2 + m⋅g⋅yi = (1/2)⋅m⋅Vf2 + m⋅g⋅yf |
|
if gravity does work....
E = K + U grav |
W
total = Kf - K i W total = Wgrav + Wel + Wother |
|
W
other + Ui - Uf = Kf - Ki arrange to.... K i + Ui + Wother = Kf + Uf |
Work done on a spring
W = (1/2)K E⋅Xf2 - (1/2)KE⋅Xi2 Work done by a spring W = (1/2)K E⋅Xi2 - (1/2)KE⋅Xf2 |
|
U
cm = m ⋅ g ⋅ R |
Elastic Potential Energy
U el = (1/2)⋅KE⋅x2 Work Done by Elastic Force W el = (1/2)⋅KE⋅xi2 - (1/2)⋅KE⋅xf2 |
|
if elastic force does work, and mechanical energy is conserved
K i + Uel, i = Kf + Uel, f |
|
|
Work Done by Friction:
W f = -WfricW f = -(-fk⋅ s) W f = μk⋅m⋅g⋅s |
Law of Conservation of Energy
ΔK + ΔU + ΔU int = 0 |
|
F = F
x + Fy + FzF x(x) = -m⋅g F y(y) = -m⋅g F z(z) = -m⋅g |
F
x = (1/2)⋅K⋅x2 |
|
p = momentum
|
J = Impulse
|
|
m = mass
|
v = velocity
|
|
P = m ⋅ v (kg ⋅ m/s)
|
|
|
F = d
p / dt J y = (∫tf to ti) ΣFy dt J y= (Fav)y (tf - ti) J y = Pfy - Piy J y = (m⋅Vfy) - (m⋅Viy) |
J = ΣF (t
f - ti) J = ΣF⋅Δt J = (∫t f to ti) ΣF dt J x = (∫tf to ti) ΣFx dt J x= (Fav)x (tf - ti) J x = Pfx - Pix J x = (m⋅Vfx) - (m⋅Vix) |
|
ΣF = (P
f - Pi) / (tf - ti) |
J = F
av(tf- ti) |
|
J = (P
f - Pi) = {F} : Change in Momentum |
|
|
P = P
A + PB = |PA + PB| |
|
|
Assuming m
1 and m2 don't change |
|
|
m
1⋅v1+m2⋅v2 = constant |
(P
1+P2)i = (P1+P2)f |
|
P
1+P2 = constant |
P
i = Pf |
|
V
f = (m1⋅v1+m2⋅v2) / (m1++m2) |
|
No comments yet.
Cheatography n. The art of mapping complex subjects into clear, instantly navigable reference guides.
