CONSERVATION OF ENERGY The Law of Conservation of Energy In an isolated system, the total amount of energy is conserved Equation: Ui + KEi + Wc = Uf + KEf + Wn Wc = work done by conservative forces (adds nrg to the system) Wn = work done by non-conservative forces (removes nrg from the system)

You throw a rock straight up at a frisbee resting in a tree You throw a rock straight up at a frisbee resting in a tree. If the rock’s speed as it reaches the frisbee is 4.0 m/s, what was its speed as it left your hand 2.8 m below? Kinematics / Dynamics Given: vf = 4.0 m/s d = 2.8 m a = -9.8 m/s2 Equation: vf2 = vi2 + 2a d Substitute: (4.0)2 = vi2 + 2(-9.8)(2.8) Solve: vi = 8.4 m/s

You throw a rock straight up at a frisbee resting in a tree You throw a rock straight up at a frisbee resting in a tree. If the rock’s speed as it reaches the frisbee is 4.0 m/s, what was its speed as it left your hand 2.8 m below? 40 kg Work / NRG Equation: Ui + KEi + Wc = Uf + KEf + Wn BEFORE AFTER KEi = Uf + KEf KEf = ½mvf2 Uf = mgh ½mvi2 = mgh + ½mvf2 Substitute: ½mvi2 = m(9.8)(2.8) + ½m(4.0)2 ½mvi2 = 27.44m + 8m ½mvi2 = 35.44m KEi = ½mvi2 Ui = 0 Wc = 0 Solve: vi = 8.4 m/s Wn = 0

A cannonball is shot off a 60 m tall cliff with a velocity of 40 m/s at 37 degrees. With what speed does it hit the ground? Kinematics / Dynamics Given: vi = 40 m/s @ 37° dy = -60 m ay = -9.8 m/s2 Equations: viy = vi (sin θ) vx = vi (cos θ) vfy2 = viy2 + 2ay dy 40(sin37) = 24 m/s 40(cos37) = 32 m/s 32.0 41.8 vfy2 = (24)2 + 2(-9.8)(-60) vfy = - 41.8 m/s vf = 52.6 m/s vfx = 32.0 m/s (since x-velocity is constant)

A cannonball is shot off a 60 m tall cliff with a velocity of 40 m/s at 37 degrees. With what speed does it hit the ground? Work / NRG Equation: Ui + KEi + Wc = Uf + KEf + Wn BEFORE AFTER KEi = ½mvi2 KEi + Ui = KEf Ui = mgh ½mvi2 + mgh = ½mvf2 Substitute: ½m(40)2 + m(9.8)(60) = ½mvf2 KEf = ½mvf2 Uf = 0 Wc = 0 vf = 52.6 m/s Wn = 0

A person on a bicycle traveling at 10 A person on a bicycle traveling at 10.0 m/s on a horizontal surface stops pedaling as she starts up a hill inclined at 3. How far up the incline will she travel before stopping? (neglect friction) Kinematics / Dynamics FN Fll = Fg(sin θ) = Fg(sin 3) = m(9.8)(sin 3) = 0.513m Given: vi= 10 m/s vf= 0 m/s a = -0.513 m/s2 Equation: vf2 = vi2 + 2a d Fll Fg Substitute: (0)2 = (10)2 + 2(-0.513)(d ) a ΣF = ma ΣF = Fll ma = Fll d = 97.5 m ma = 0.513m a = 0.513 m/s2

A person on a bicycle traveling at 10 A person on a bicycle traveling at 10.0 m/s on a horizontal surface stops pedaling as she starts up a hill inclined at 3. How far up the incline will she travel before stopping? (neglect friction) Work / NRG Equation: Ui + KEi + Wc = Uf + KEf + Wn BEFORE AFTER KEf = 0 KEi = Uf Uf = mgh ½mvi2 = mgh Substitute: ½m(10)2 = m(9.8)(h) KEi = ½mvi2 h Ui = 0 d d (sin θ) = h d (sin 3) = 5.1 h = 5.1 m Wc = 0 This is the height, determine d with trig Wn = 0 d = 97.5 m

A 1200 kg elevator must be lifted by a cable that causes the elevator’s speed to increase from zero to 4.0 m/s as it rises 6.0 m. What is the tension in the cable? Kinematics / Dynamics ma = FT - Fg ΣF = ma ΣF = FT - Fg FT 1200a = FT – 11,760 a 1200(1.33) = FT – 11,760 Given: vi= 0 m/s vf= 4.0 m/s d = 6.0 m FT = 13,360 N Equation: vf2 = vi2 + 2a d Fg Substitute: (4.0)2 = (0)2 + 2(a)(6.0 ) a = 1.33 m/s2

A 1200 kg elevator must be lifted by a cable that causes the elevator’s speed to increase from zero to 4.0 m/s as it rises 6.0 m. What is the tension in the cable? Equation: Ui + KEi + Wc = Uf + KEf + Wn Work / NRG BEFORE AFTER WC = Uf + KEf KEf = ½mvf2 FTd = mgh + ½mvf2 Uf = mgh Substitute: FT(6.0) = (1200)(9.8)(6.0) + ½(1200)(4.0)2 KEi = 0 Wc = FTd FT = 13,360 N Ui = 0 Wn = 0

A 7. 00 kg mass is connected to a 4 A 7.00 kg mass is connected to a 4.50 kg mass by a rope passed over a frictionless pulley. The system is released from rest and the 7.00 kg mass falls 3.0 meters. What is its speed at this point? Kinematics / Dynamics Given: vi = 0 m/s a = - 2.10 m/s2 d = - 3.0 m Substitute: (vf)2 = (0)2 + 2(-2.10)(-3.0) Equation: vf2 = vi2 + 2a d v = - 3.6 m/s a FT ΣF = m1a ΣF = FT – Fg1 m1a = FT – Fg1 7a = 68.6 –(4.5a + 44.1) Fg1 4.5 a = FT – 44.1 a = 2.10 m/s2 a ΣF = m2a ΣF = Fg2 – FT FT m2a = Fg2 - FT FT = 4.5a + 44.1 7a = 68.6 - FT Fg2

A 7. 00 kg mass is connected to a 4 A 7.00 kg mass is connected to a 4.50 kg mass by a rope passed over a frictionless pulley. The system is released from rest and the 7.00 kg mass falls 3.0 meters. What is its speed at this point? Work / NRG Equation: Ui + KEi + Wc = Uf + KEf + Wn BEFORE AFTER Ui = Uf + KEf + KEf KEf = ½m1v2 m2gh = m1gh + ½m1v2 + ½m2v2 KEi = 0 Uf = m1gh KEi = 0 Ui = m2gh KEf = ½m2v2 Ui = 0 Uf = 0 Substitute: (7)(9.8)(3) = (4.5)(9.8)(3) + ½(4.5)(v)2 + ½(7)(v)2 Wc = 0 Wn = 0 v = 3.6 m/s

A rope exerts a 345 N force pulling a 60 kg skier upward along a hill inclined at 25 degrees. If the skier started from rest, calculate her speed after moving 100 m up the slope. The snow exerts a frictional force of 50 N on the skier. Kinematics / Dynamics FT FN Fll = Fg(sin θ) = 588(sin 25) = 248 N f Given: vi = 0 m/s a = 0.78 m/s2 d = 100 m Fll Fg Equation: vf2 = vi2 + 2a d Substitute: vf2 = (0)2 + 2(0.78)(100) a v = 12.5 m/s ΣF = ma ΣF = FT – Fll – f ma = FT – Fll – f (60)a = 345 – 248 - 50 a = 0.78 m/s2

A rope exerts a 345 N force pulling a 60 kg skier upward along a hill inclined at 25 degrees. If the skier started from rest, calculate her speed after moving 100 m up the slope. The snow exerts a frictional force of 50 N on the skier. Work / NRG Equation: Ui + KEi + Wc = Uf + KEf + Wn BEFORE AFTER KEf = ½mv2 WC = Uf + KEf + Wn Uf = mgh FTΔd = mgh + ½mv2 + fΔd Substitute: (345)(100) = (60)(9.8)(42) + ½(60)(v)2 + (50)(100) KEi = 0 h v = 12.5 m/s Ui = 0 d Wc = FTΔd d (sin θ) = h 100 (sin 25) = h h = 42 m Wn = f Δd The d is 100 m → determine the height with trig