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by: Justin Kidd

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# Equation List with description PHYS 140

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Justin Kidd

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Equations for up to Exam 2.
COURSE
College Physics
PROF.
Dr. Mattson
TYPE
Class Notes
PAGES
2
WORDS
CONCEPTS
Algebra, Physics, kinematics, rotational kinematics, Universal Gravitation, Friction, conservative force, nonconservative forces
KARMA
25 ?

## Popular in Physics and Astronomy

This 2 page Class Notes was uploaded by Justin Kidd on Thursday October 13, 2016. The Class Notes belongs to PHYS 140 at James Madison University taught by Dr. Mattson in Fall 2016. Since its upload, it has received 4 views. For similar materials see College Physics in Physics and Astronomy at James Madison University.

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Date Created: 10/13/16
2 Kinematics x f x +iv t +xia t x v xf v  xia t x vxf= v  xi2a (x –xx)f i 2 y f y +iv t +yia t y v  = v  + a t 2 2 yf yi y vyf= v  yi2a (y –yy)f i 2  max height = v / 2gyi Newton’s Second Law F = ma  or  a = ΣF / m   Weight and Force of Gravity w = F  =gmg Normal force (a = 0, flat surface) FN = mg Normal force on elevator upward acceleration F N= mg + ma  downward acceleration F N= mg ­ ma Tension when supporting weight of object at rest T = mg Other Tension equations Fnet T −  w = 0  T =  w =  mg        ma = T ­ mg        T = mg + ma Static Friction and Kinetic Friction fs =  μ s N             kk N Hooke’s Law  (k = proportionality constant; ΔL=deformation) F =  kΔL or F = kx Rotational kinematics same as kinematics, but note: remember thumb rule for angular velocity direction replace x with θ,        v with  ω,        a with α  Rotation angle  (Δs = arc length) Δθ = Δs / r Centripetal acceleration (uniform circular motion) a   =  v /r  =  rω   2 c Centripetal force Fc =  ma  c =  mv /r   =  mrω2 2 Centripetal acceleration about a banked path a = gtanθ = v /r) Max velocity to take banked curve  (disregarding friction) tan θ = v /rg      //      θ = tan (v /gr) Relating tangential and angular variables xT = s or speed = rθ    //    v  =Trω    //    a  = Tα 2 2 Net acceleration of circular motion |anet=  (a c+ a ) T ­11 3 ­1 ­2 2   2 2 Universal Gravitational Force  (G = 6.67408 × 10  m  kg  s ) F =  Gm M 1/ r2(G = 6.674Å~10–11 N ⋅ m /kg  ) Calculate gravitational field near surface of large object g = GM / r 2 Keplar’s Law  (ratio of two planetary orbits around sun) T 1/ T  2 r  /1r   23 2 2 3 Keplar’s Law  (relation of satellite orbit about a larger body) T  = 4π  r / GM r  / T  =  GM / 4π 2 Centripetal application to gravitational force  F G= F c  (velocity of satellite, mass of larger object) v =  (GM / 1) Relation of weight of a mass at the surface of the Earth to its  R /(R  + h) 2 E E weight at an orbit that is an altitude h above the surface 2 Escape velocity ½mv  + e(GM m  / r1 =20 + 0    v  = (2GM e r)  2 2 Rotating cylinder g = v /r = rw Work W =  Fd cos θ 2 2 W =  ΔKE = ½mv  ­ ½mv f i  W  = (ΣFcos θ ) Δx = F Δx cos θ total i  i net net Derivation of kinetic energy in relation to work W = FΔx = ma(vt + ½ati) 2 Change in potential energy due to gravity U G (uniform gravitational field)= GPE = ΔPE   =  mgh Gravitational field starts to drop (0 is max value) U = ­GM m  / r G  1 2 Object descends without friction ΔKE = −ΔPE g 2 Potential energy of spring PE s  = ½kx    Mechanical energy for conservative force KE + PE KE i + PE i = KE f + PE f Nonconservative force W nc = ΔKE + ΔPE KE i + PE i + W nc  = KE f + PE f Conservation of Energy  (OE = all other forms of energy) KE i + PE i + W nc  + OE i = KE f + PE f  + OE f Energy E = KE + GPE + Elastic PE Work done by other forces ΔKE + ΔPE Work done by gravity    W G F ΔG cos0 = mg Δh = ­GPE   (change in y is negative)  Work done by any particular conservative force W con PE Power P =  W /  t          P = ΔE / t

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