CONSERVATION OF ENERGY The Law of Conservation of

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. 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: Equation: vf = 4. 0 m/s vf 2 = vi 2 + 2 a d d = 2. 8 m a = -9. 8 m/s 2 Substitute: (4. 0)2 = vi 2 + 2(-9. 8)(2. 8) Solve: vi = 8. 4 m/s

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 BEFORE Equation: AFTER KEf = ½mvf 2 Uf = mgh Substitute: KEi = ½mvi 2 Ui = 0 Wc = 0 Wn = 0 Ui + KEi + Wc = Uf + KEf + Wn KEi = Uf + KEf ½mvi 2 = mgh + ½mvf 2 ½mvi 2 = m(9. 8)(2. 8) + ½m(4. 0)2 ½mvi 2 = 27. 44 m + 8 m ½mvi 2 = 35. 44 m Solve: vi = 8. 4 m/s

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/s 2 Equations: viy = vi (sin θ) 40(sin 37) = 24 m/s vx = vi (cos θ) 40(cos 37) = 32 m/s vfy 2 = viy 2 + 2 ay dy 32. 0 41. 8 vfy 2 = (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 BEFORE Equation: AFTER KEi = ½mvi 2 Ui = mgh Ui + KEi + Wc = Uf + KEf + Wn KEi + Ui = KEf ½mvi 2 + mgh = ½mvf 2 Substitute: ½mvf 2 Wc = 0 Wn = 0 KEf = Uf = 0 ½m(40)2 + m(9. 8)(60) = ½mvf 2 vf = 52. 6 m/s

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. 513 m Fll Fg a ΣF = ma ΣF = Fll Given: vi= 10 m/s vf= 0 m/s a = -0. 513 m/s 2 Substitute: Equation: vf 2 = vi 2 + 2 a d (0)2 = (10)2 + 2(-0. 513)( d ) ma = Fll d = 97. 5 m ma = 0. 513 m a = 0. 513 m/s 2

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 BEFORE Equation: AFTER Ui + KEi + Wc = Uf + KEf + Wn KEf = 0 Uf = mgh KEi = Uf ½mvi 2 = mgh Substitute: KEi = ½mvi 2 Ui = 0 Wc = 0 Wn = 0 h d d (sin θ) = h d (sin 3) = 5. 1 d = 97. 5 m ½m(10)2 = m(9. 8)(h) h = 5. 1 m This is the height, determine d with trig

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 FT a ΣF = ma ma = FT - Fg ΣF = FT - Fg 1200 a = FT – 11, 760 Given: vi= 0 m/s 1200(1. 33) = FT – 11, 760 vf= 4. 0 m/s FT = 13, 360 N d = 6. 0 m Fg Equation: vf 2 = vi 2 + 2 a d Substitute: (4. 0)2 = (0)2 + 2(a)(6. 0 ) a = 1. 33 m/s 2

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: Work / NRG BEFORE AFTER Ui + KEi + Wc = Uf + KEf + Wn ½mvf 2 KEf = Uf = mgh WC = Uf + KEf FT d = mgh + ½mvf 2 Substitute: FT(6. 0) = (1200)(9. 8)(6. 0) + ½(1200)(4. 0)2 KEi = 0 Ui = 0 Wc = FT d Wn = 0 FT = 13, 360 N

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 a Given: vi = 0 m/s a = - 2. 10 m/s 2 d = - 3. 0 m Equation: vf 2 = vi 2 + 2 a d FT Fg 1 a FT Fg 2 ΣF = m 1 a ΣF = FT – Fg 1 ΣF = m 2 a ΣF = Fg 2 – FT Substitute: (vf)2 = (0)2 + 2(-2. 10)(-3. 0) v = - 3. 6 m/s m 1 a = FT – Fg 1 7 a = 68. 6 –(4. 5 a + 44. 1) 4. 5 a = FT – 44. 1 a = 2. 10 m/s 2 m 2 a = Fg 2 - FT 7 a = 68. 6 - FT FT = 4. 5 a + 44. 1

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 BEFORE AFTER Equation: Ui + KEi + Wc = Uf + KEf + Wn Ui = Uf + KEf KEi = 0 Ui = m 2 gh KEf = ½m 1 v 2 Uf = m 1 gh Wc = 0 Wn = 0 m 2 gh = m 1 gh + ½m 1 v 2 + ½m 2 v 2 KEf = ½m 2 v 2 Uf = 0 Substitute: (7)(9. 8)(3) = (4. 5)(9. 8)(3) + ½(4. 5)(v)2 + ½(7)(v)2 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 f Fll Fg a ΣF = ma ΣF = FT – Fll – f Fll = Fg(sin θ) = 588(sin 25) = 248 N Given: vi = 0 m/s a = 0. 78 m/s 2 d = 100 m Equation: vf 2 = vi 2 + 2 a d Substitute: vf 2 = (0)2 + 2(0. 78)(100) v = 12. 5 m/s ma = FT – Fll – f (60)a = 345 – 248 - 50 a = 0. 78 m/s 2

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. Equation: Work / NRG BEFORE AFTER KEf = ½mv 2 Uf = mgh Ui + KEi + Wc = Uf + KEf + Wn WC = Uf + KEf + Wn FTΔd = mgh + ½mv 2 + fΔd Substitute: (345)(100) = (60)(9. 8)(42) + ½(60)(v)2 + (50)(100) KEi = 0 Ui = 0 h d v = 12. 5 m/s Wc = FTΔd Wn = f Δd The d is 100 m → determine the height with trig d (sin θ) = h 100 (sin 25) = h h = 42 m