physics study formula sheet

Resonance in String Example A string in a musical instrument has a fixed length of 0.5m. It is tuned so the velocity of waves in 300m/s. Calculate the first 3 resonant frequencies (n = 1,2,3) /1 = (2L /1 = 1(0.5m) = 1.0m f1 = (300m/s)(1.0m) = 300hz /2 = (2L/2 = 2(0.5m) = 0.5m f2 = (300m/s)(0.05m) = 600hz /3 = (2L/3 = 3(0.65) = 0.33m f3 = (300m/s)(0.33m) = 900hz

/n = 2L/ N

Free body Diagram

1. Draw objects as simple points 2. identify all forces acting on objects and direction 3. Draw all forces as vector arrows away from point 4. Find components

Restoring Forces

Amplitude (A) the maximum distance an object moves away from equilibrium

finding the height with constant

Ei = 1/2 K ∆x^2 Ef = 1/2 k(0)^2 + mgyf = mgys Yf = Ei/Mg K=constant y= height E= energy ∆KE + ∆PE = 0 [if no non constant forces]

Buoyancy example: Calculate the buoyancy force for a cube of lead with sides d = 0.5m which is completely submerged in fresh water. Vdisplaced = d3 = (0.5m)3 = 0.125m^3 FB = ρgVdisplaced = (1000kg/m3)(9.8m/s^2)(0.125m3) = 1225N example: A 6.2kg fish hovers at a constant depth in fresh water. Calculate the volume of the fish. ΣFy = FB - mg = 0 FB = mg ρgVdisplaced = mg Vdisplaced = m /ρ = (6.2kg) / (1000kg/m^3) = 6.2x10-3m^3

FB = ρ gVdisplaced If FB > mg then: ΣFy = FB - mg = ma (positive - the object rises) If FB = mg then: ΣFy = FB - mg = 0 (neutral - the object floats at constant depth) If FB < mg then: ΣFy = FB - mg = ma (negative - the object sinks)

Tension pulling forces from rope example: A 15kg mass is tied to a string that passes over a pulley and attaches horizontally to the wall. Calculate the tension in the wire. Calculate the force of the wall acting on the string. Sum of the forces in the y direction for the hanging mass: ΣFy = FT - mg = 0 FT = mg = (15kg)(9.8m/s^2) = 150N Sum of the forces in the x direction for the wall and string contact: ΣFx = FT - FW = 0 FT = FW = 150N

FT = Pulling direction of rope

Circular Motion aka Centripetal forces example: A 2.5kg model airplane at the end of a cable flies in a circle with a radius of 5.0m at a constant speed of 10m/s as shown in figure 4.10(a). Calculate the acceleration of the plane and the tension in the cable. ac = v 2 / r = (10m/s)2 / (5.0m) = 20m/s^2 (toward the center) FT = Fc = mac = (2.5kg)(20m/s^2) = 50N The tension in the string is the centripetal force that pulls the plane toward the center. This tension is the force that creates the centripetal acceleration of the airplane

Fc = mac Fc = Centripetal Forces m = mass a = acceleration ac = v^2 / r r = radius v = velocity (speed) a = acceleration Fr = Fc = mac

Chapter 4 Weight Example: What is the weight of the average adult with a mass of 70kg? Fg = mg = (70kg)(9.8m/s^2) = 690N

Fg = mg g = 9.8 m/s^2 (gravity) m = mass F = force

Spring example A spring hanging vertically from the ceiling stretches down 0.15m when a 10kg mass is attached to the lower end. In this set up, the general displacement d will be the vertical ∆y. Calculate the spring constant k. ΣFy = FS - mg = 0 , FS = mg = - k ∆y k = - mg / ∆y = - (10kg)(9.8m/s^2) / (- 0.15m) = 650N/m

Fs = -kd k = constant proportionality d = displacement fs= static forces K = N/m

Kinetic Energy

KE = 1/2 mv^2 ∆KE = KEf - KEi [change in energy] Wnet = ∆KE

pressure example: Each foot pad of a 6000kg elephant has an average area of roughly 0.09m2. Calculate the average pressure created on the ground for a stationary elephant. Area A = (4 feet)(0.09m2/foot) = 0.36m^2 P = F / A = mg / A = (6000kg)(9.8m/s^2) / (0.36m^2) = 163,000N/m^2

P = F / A P = pressure N/m^2 F = force A = area

Fluid energy example: Fresh water flows through a horizontal pipe that changes its area from Ā = 0.06m^2 to Á = 0.02m^2 as shown. The input pressure is P1 = 200,000N/m^2 and the input velocity is v1 = 1.2m/s. Use Bernoulli's equation to solve for P2. (ρ =1000kg/m^3) 2 = v1 ( Ā / Á ) = (1.2m/s)(0.06m^2) / (0.02m^2= v1 ( Ā / Á ) = (1.2m/s)(0.06m^2) / (0.02m^2) = 3.6m/s^2 = 3.6m/s P1 + ρgy1 + 𝝆v𝟏^𝟐/𝟐 = P2 + ρgy2 + 𝝆v𝟐^𝟐/2 P1 + 𝜌v1^2/2= P2 +𝜌v2^2/2 p2 = p1 + 𝜌v1^2/2 - 𝜌v2^2/2 P2 = 110,000N/m^2 + (1000kg/m^3)(1.2m/s)^2 / 2 - (1000kg/m^3)(3.6m/s)^2 / 2 = 104,000N/m^2

P1 + ρgy1 + 𝝆v𝟏^𝟐/𝟐 = P2 + ρgy2 + 𝝆v𝟐^𝟐/2 P = F / A P = Fd / Ad. fd = work done Ad = volume J/m^3

Springs example: A spring with a constant k = 450N/m is stretched 0.45m from equilibrium. Calculate the potential energy stored in the spring. µs =𝒌x^2/𝟐 = (450N/m)(0.45m)^2 / 2 = 46Nm = 46J

PE = 1/2 K (∆X^2) k = constant [N/M] ∆X = change in height [∆x = how far we are from equilibrium]

Gravitational PE

Peg = mgy Y= height

Elastics - energy is conserved in elastics - do not conserve energy Conservation of Momentum. The two people at rest on the ice in figure 8.1 push with a force of 100N for 1.5s. The large person B has a mass of 50kg and the small person A has a mass of 30kg. Calculate the change in momentum and change in velocity for each person. ∆pTotal = Σpf - Σpi = 0 ∆pA = IA = FA ∆t = (- 100N)(1.5s) = - 150kgm/s ∆pB = IB = FB ∆t = (+ 100N)(1.5s) = +150kgm/s ∆pA = ∆(mAvA) = (mAvA)f - (mAvA)i = (mAvA)f - 0 (solve for vfA ) vfA = ∆pA / mA = (- 150kgm/s) / (30kg) = - 5m/s ∆pB = ∆(mBvB) = (mBvB)f - (mBvB)i = (mBvB)f - 0 (solve for vfB ) vfB = ∆pB / mB = (+150kgm/s) / (50kg) = +3m/s Σpi = 0 Σpf = +150kgm/s - 150kgm/s = 0

Pif + P2i = Pif + P2f collisions

Power example: How long will it take to perform 1200J of work on an object if you can only deliver 50 Watts of power? P = W / t, t = W / P = (1200J) / (50W) = 24s

Power = work/ ∆T = ∆KE/∆T [used when velocity isn't constant] P= Fv(cos ⁡θ) [if forces and velocity is constant

Continuity of flow example: The average adult aorta has a radius of about 0.01m and blood flows through it at about 1.0m/s. Calculate the total cross-sectional area of capillaries if the average speed of blood is only 0.0013m/s through them. Á = Ā (v1 / v2) Á = πr1^2 (v1 / v2) = π(0.01m)^2 (1.0m/s) / (0.0013m/s) = 0.24m^2 Ā = π(0.01m)^2 = 0.0003m^2

Rin = Rout R = mass flow rate kg/s R = ρ Av ( ρ Av )in = ( ρ Av )out Āv1 = Áv2

Waves in Strings example: A wire has a tenth L=8.2 & a mass m=0.01kg. calculate the velocity v of wavelength in this wire when a tension Ft=118N is applied // = m/L = (0.01kg 8.2m) = 0.0012Kg/m v = √Ft/µ v = √ 118N / 0.012kg/m = 314m/s

Vstrings = √Ft/µ Ft = tension

chapter 7: Conservation of energy Finding Height A ball is thrown at an initial speed of 20m/s from an initial height of 1.7m. Use the conservation of mechanical energy to calculate the speed of the ball when it hits the ground. (Kf - Ki) + (Uf - Ui ) = 0 yf = 0, so Uf = 0: Kf - Ki - Ui = 0 (𝑚v^2/2)𝑓 - (𝑚v^2/2)𝑖 - mgyi = 0 vf^2 - vi^2 - 2gyi = 0 vf^2 = vi^2 + 2gyi = (20m/s)^2 + 2(9.8m/s^2)(1.7m) = 433m2/s^2 vf = 21m/s

combined KE & PE 1/2 mv^2f + mgyf since no mass is given 1/2 mv^2f + mgyf = 1/2 mvi^2 1/2vf^2 = 1/2 vi - gyf vf = √vi^2- 2gyf

Normal Forces

example: Calculate the Normal force for the crate in while someone pushes horizontally on it with an external force of 220N. Sum of the forces in the x direction: ΣFx = F = max Rearranging: ax = F / m = (220N) / (120kg) = 1.8m/s2 Sum of the forces in the y direction: ΣFy = FN - Fg = 0 where: Fg = mg Rearrange: FN = mg = (120kg)(9.8m/s^2) = 1180N

Chapter 3 Newtons 1st law

if net force = 0 then ∆V = 0 and A=0 ∑ = "sum of"

Friction (oppose direction of motion) example: Calculate the frictional force f on the steel 50kg crate when 20N are applied horizontally, but it remains stationary. Sum of the forces in the x direction: ΣFx = F - f = 0, therefore: F = f = 20N

kinetic friction Fk = Uk FN fk = Friction forces Uk = constant Proportionaly FN = normal Force Us > Uk Static Friction (not moving) Fs = Us FN fs = static Friction Us= Constant proportionally FN = normal Forces Fs<Us FN

Chapter 10 Harmonic Motion period & Frequency example: When a large bell is struck, it walls more in & out with a frequency of 100 cycles.calculate the period of oscillation T= 1/f = 1/80Hz = 0.0056s when calculating frequency f = 1/T = 1/4s = 0.25Hz

period = T Frequency = f hertz = Hz T = 2π√m/k k = constant T = oscillation T = 2π√L/g L = length of the pendulum g = acceleration of gravity

Chapter 11 : waves example: A set of waves near shore have a speed v=1.2m/s. A surfer counts 16s between swells (the height of each wave). Calculate the wavelength v = // T // = vT (1.2m/s)(16s) = 19m

v = distance / time = λ/T T = period λ = wave v = λ(1/T)λƒ ƒ = frequency v = velocity v = // ƒ // = wavelength

Dynamics of Harmonic Motions

v = vas cos (2π × f × t) Vmas= max (2π × f) amax = yams (2π × f)2

Chapter 5: work of energy Definition of work example: A constant net force of 100N is applied to a large crate as it moves across a frictionless surface a distance of 10m. Calculate the work done on the crate. W = Fd = (100N)(10m) = 1000J A 1.2kg ball falls straight down a height h = 3.2m. Calculate the work done on the ball by the constant force of gravity. W = Fd = Fgd = (-mg)(-h) = -(1.2kg)(9.8m/s2)(-3.2m) = 38J A constant net force F of 140N pulls on a wagon at an angle θ = 32o above the horizontal. The wagon is pulled for a horizontal distance d = 12m. Calculate the work done on the wagon. W = (F cosθ ) d = (140N)(cos32o)(12m) = 1400J

w = Fd [used when no ⁡θ is given] f = frictionless force d = distance w = work work = J (joules) N times M = J W = F(cos ⁡θ)d [used only when given ⁡θ or motion of friction] W= fd = fdg = (-mg)(-h)

Rotation and Simple Harmonic Motions Example A spring with k=50N/m and a 10kg mass hanging on its end stretches down to its equilibrium position & is stationary there. If it is pulled down 0.02m from this position & released, the mass oscillates around this equilibrium position in simple harmonic. Calculate the frequency & use to describe the motion. Calculating T T= 2π(m/k)^1/2 = 2(10kg / 50n/m)^1/2 = 2.8s f = 1/T = 1/2.8s = 0.36Hz y = ymax sin(2π×f×t) = (0.02m)sin(2π(0.36hz)t y = (0.02m) sin(2.2)t

y = ymax sin(2π× f × t)

Chapter 12: Static Fluids density Example The density of gold is 19.3g/cm^3. the dimensions of a standard gold bar are 4.45cm × 9,21cm× 17,8cm. Calculate the mass of standard gold bar v = (4.45cm)(9.21cm)(17.8cm) = 729.5cm^3 m = ρV = (19.3g/cm^3)(729.5cm3) = 14,080g = 14.1kg

ρ = m / V ρ = density

Chapter 9 : Torques τ > 0 : counter clockwise (positive) τ < 0 : clock wise (negative) Suppose you attach a 0.4m long wrench to a bolt and grab the wrench 0.3m from the center of the bolt and apply 85N of force. If the angle between the force and the wrench is 65o, calculate the proportion of force that would cause rotation. r = 0.3m F┴ = Fy = F sin θ = (85N) sin (65o) = 77N adding torques A 2.0m long steel bar is used as a lever. One end is placed under a 1200kg mass and a pivot point is created just 0.01m from this end of the bar. If you ignore the mass of the bar, how much force is required at the other end of the bar to just begin to lift the mass? Assume the bar is horizontal. Στ = 0 Στ = τ mass + τ force = 0 Στ = (F┴ r) mass + (F┴ r) force = 0 Στ = (1200kg)(9.8m/s2)(0.01m) - F(1.99m) = 118Nm - F(1.99m) = 0 F = 118Nm / 1.99m = 59N

τ=rF⊥ r = distance from center of rotation F = force ⊥ = perpendicular units: N × M τ=rF⊥sin⁡θ [angle between r to F]

Energy in Harmonic Motion Example On a frictionless surface a horizontal spring with k=600N/M is fixed at one end a 10kg mass is attached to the free end. It is stretch 0.2m and released a) Calculate the max velocity at the equilibrium points b) Calculate how long it takes to reached the equilibrium point from full extension a) xi = 1.2m xƒ = 0 (equilibrium), vi = 0 ∆K + µs= 0 (10kg)vf^2 / 2 - 0 + 0-(500n/m)(0.2m)^2/2 = 0 vƒ^2 = (500n/m)(0.2m)^2 / (100kg) = 2.0m^2s^2 Vƒ = 1.4m/s b) T = 2pi√m/k = 2pi×10kg/600n/m = 0.84 time it takes t1 full cycle. time it takes to make max extinction T = T/4 = 0.8/4 = 0.20s

∆K + ∆µs = 0

CHAPTER 13: TEMPERATURE and HEAT example: Calculate the energy needed to heat a 23kg block of ice at -10oC to a liquid at 35oC. (cice = 2090J/kgoC), (Lf = 3.35x105J/kg) QTotal = Qice + Qf + Qwater = micecice∆T + mLf + mwatercwater∆T Energy needed to warm the ice: Qice = (23kg)(2090J/kgoC)(0oC - (-10oC)) = 4.8x105J Energy needed to melt the ice: Qf = (23kg)(3.35x105J/kg) = 7.7x106J Energy needed to warm the water: Qwater = (23kg)(4186J/kgoC)(35oC - 0oC) = 3.37x106J QTotal = 4.8x105J + 7.7x106J + 3.37x106J = 1.16x107J

∆L = α Li ∆T units = 1/C or 1/K ∆V = α Li ∆T Li= initial length α = Expansion coefficient different for different material ∆T = change of temp Q = mc∆T ΣQ = Q A + Q B + Q C + ... = 0 Qf = mLf (heat infusion) Qv = mLv (Latent heat of vaporization)

Chapter 6: Potential Energy Change in Potential Energy example: A 10kg rock is lifted from a 0.8m high desk to a 1.5m high shelf. a. Calculate the change in potential energy of the rock. ∆U = (mgyf - mgyi ) = (10kg)(9.8m/s^2)(1.5m) - (10kg)(9.8m/s^2)(0.8m) = 69J b. Calculate the change in potential energy if the rock falls from the shelf to the floor. ∆U = (mgyf - mgyi ) = (10kg)(9.8m/s2)(0m) - (10kg)(9.8m/s^2)(1.5m) = -150J c. Calculate the work done by gravity as the rock falls from the shelf to the floor. Wg = - ∆U = - (mgyf - mgyi ) = - [(10kg)(9.8m/s^2)(0m) - (10kg)(9.8m/s2 )(1.5m)] = +150J

∆PE = mg∆T m = mass g = gravity [negative numbers only when height is negative] [+m & +g = + numbers]

Static Equilibrium

∑Fx = 0 ∑Fy = 0 ∑Fτ = 0

Newtons 2nd law kg= mass m/s^2 = gravity N = neutons ( force) Kg & m/s^2 makes N example: If the box in figure 3.4 had a mass of 50kg and the person pushed with a force of Fx = -25N, calculate the acceleration. Rearrange ax = ΣFx / m = (-25N) / (50kg) = -0.5m/s^2

∑f = ma F = force m = mass a = acceleration Fnet x = ∑fx = F1x + F2x + F3x + .... F1x + F2x + F3x = when theres a lot of components

Newtons 3rd Law (different directions)

∑fx = max in the x direction ∑fy = may in the y direction

Chapter 8: Momentum P is conserved NO NET FORCES A 0.15kg baseball is thrown at 40m/s. The batter hits the ball directly back to the pitcher at 50m/s. The ball is in contact with the bat for 0.002s. Calculate the impulse given to the ball and the force it experiences.First, assume the batter hits the ball in the +x direction. Therefore vi = - 40m/s Expanding equation 8.5: I = ∆p = ∆(mv) = (mv)f - (mv)i I = (0.15kg)(+50m/s) - (0.15kg)(- 40m/s) = (+7.5kgm/s) - (-6.0kgm/s) = +13.5kgm/s F = I / t F = I / t = (+13.5kgm/s) / (0.002s) = +6750N

∑pi = ∑pf 0 = m1v1 + m2v2 [when 2 masses given and trying to find 1 velocity] vector p = m(vector V) direction matters + = right - = left + = up - = down M1v1 = M2V2×Vf