1. 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)

2. 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

3. 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

4. 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 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 = m/s
vf = 52.6 m/s
vfx = 32.0 m/s (since x-velocity is constant)

5. 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

6. 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 = 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 = m/s2

7. 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

8. 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

9. 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

10. 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 = m/s2
d = m
Substitute:
(vf)2 = (0)2 + 2(-2.10)(-3.0)
Equation:
vf2 = vi2 + 2a d
v = m/s a FT
ΣF = m1a
ΣF = FT – Fg1
m1a = FT – Fg1
7a = 68.6 –(4.5a )
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
7a = FT
Fg2

11. 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

12. 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 –
a = 0.78 m/s2

13. 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