deck_2435790-2 Flashcards
Center of mass
Cmass= (r1m1+r2m2+r3m3…)/mtotal
Newton’s 2nd law: Force
F=ma
Avg velocity
Vavg=(V1+V2)/2
Distance traveled
Distance=rate*time
Range
Range=Vx*time
Object falling-its distance traveled
X=1/2 at²
Final velocity given drop height
V=√(2gh)
Round trip times or time in the air
tair=2V/g*V=must be the vertical component of initial velocity
At terminal velocity
mg=F air
Force DUE to gravity or “inverse square law”* “G”*NOT gravity itself
F=Gm1m2/r²
“gravity” or “the strength of the gravitational field” or “acceleration due to gravity” * “g”
g=Gm/r²
Near earth, force due to gravity
F=mg
Gravitational Potential energy*near earth
PE=mgh
Gravitational Potential energy*in space, or near earth if NOT assuming g=10m/s²
PE= -Gm1m2/r
Friction formulas
Ff= µ(static)Fnormal or Ff=µ(static)mgcosθFf= µ(kinetic)Fnormal or Ff=µ(kinetic)mgcosθsliding=kineticno sliding=static
Force down an inclined plane, parallel to the surface
F = mgsinθ
Normal force on an inclined plane (always perpendicular to surface)
Fn=mgcosθ
Velocity of a particle at the base of an inclined plane
Vf=√(2gh)
Hooke’s Law (springs)
F = k∆x*(where ∆x is the displacement ofthe spring from its equilibrium point, NOT thelength of the spring)
Elastic Potential energy
PE = 1/2k∆x²*likely to be used in conservation of energy [KE–>PE]. This can tell us how far the spring will compress when something hits the spring1/2mv²= 1/2k∆x²
Period of a Spring*(time needed for one complete cycle)
T = 2π√(m/k) [mass on a spring]*solving for frequency: just invert it T=1/f f=1/T
Period of a Pendulum
T = 2π√(L/g) [pendulum]*solving for frequency: just invert it T=1/f f=1/T
Torque
T=Fℓ*ℓ=lever arm
Centripetal Force
Fc=mv²/rvelocity of satellite: mv²/r=Gm1m2/r²centriFUgal= the force “going away” from center of circle
How many radians in 1 circle
6 radians* 2π radians in 1 circle (360°) so… ≈ 6 radians
Angular velocity (ω) * “Rate of spin”
ω=2πff= frequency (Hz)ω=v/rv=tangential velocity (m/s)*r=radius (m)
Circumference of a circle
C=2πr
Momentum
p=mv*Always conserved in an isolated system
Impulse * “average force”
Impulse=Δpor =mΔv
Elastic collisions *(bounce off)**Momentum & KE are BOTH conserved
1/2m1v1² + 1/2m2v2² = 1/2m1v1² + 1/2m2v2²(KE of object one before + KE of object two before= KE of object one after + KE of object two after)
Inelastic collisions**Momentum is conserved but KE is NOT*Energy goes to creating breakage, deformity, etc.
m1v1+m2v2 = m1v1 + m2v2YOU MUST USE SIGNS. Any velocity vector to the left or down must be given a negative signPerfectly inelastic: m1v1+m2v2=(m1+m2)v3
Stress
Force/Area
Strain
Δdimension/original dimension
Moduli of Elasticity (ME)
ME=stress/strain(force/area) / (Δdimension/original dimension)
Thermal Expansion
When solids are heated, they expand. When they are cooled, they shrinkΔL = αL0ΔTα: the Coefficient of linear expansionL0: Initial length of the objectΔL: Length change of the objectΔT: Temperature change of the object
Kinetic Energy
1/2mv²
Gravitational Potential Energy
PE= -Gm1m2/r ormgh (near earth)
Elastic Potential energy
PE = 1/2k∆x²
Electrical Potential energy
kqq/r orqEdorqV
Potential Energy stored by a capacitor
1/2 CV²*can be used with C=Q/V(Capacitance)
Mechanical Energy
KE+PE*In the absence of friction, drag etc., ME is always conserved
Work* Joules=Nmor(kgm²)/s²
∆Energyor W = Fdcosθ or W=(Force)(Displacement)
Thermodynamics: work & heat are the only 2 ways Energy can be transferred into or out of a system
∆E=W + Q
Work-Energy Theorem*If a net force does work on a rigid object, the work done on that object is equal to the change in the KE of the object
W= KE final - KE initial
Force necessary to lift any mass w/o a machine (at constant velocity)
F necessary=mg
Ramps
Fm=mg(h/d)h=height of rampd=distance of hypotenuse*Machines reduce the amount of FORCE necessary, but NEVER reduce the amount of work needed to be done!
Levers
Fm=mg (L1/L2)L1=lever arm for massL2=lever arm for the applied force
Pulleys
Fm=mg/ (# of vertical ropes directly lifting the mass)
Hydraulic lifts
Fm=mg (h1/h2)h1=distance traveled by large plungerh2=distance traveled by small plungerorF=mg(A1/A2)A1=cross-sectional area of small plungerA2= of large plunger
Power
P=∆E/tP=W/t (work/time)P=Fdcosθ/tPi=Fvcosθ (Instantaneous power)Units: watts (J/s)Power of air resistance: P=Fv = -mgv