1. Solving Problems with
Energy Conservation

2. KE1 + PE1 = KE2+ PE2 Conservation of Mechanical Energy  KE + PE = 0
(Conservative forces ONLY!! )
or E = KE + PE = Constant
For elastic (Spring) PE: PEelastic = (½)kx2
KE1 + PE1 = KE2+ PE2
 (½)m(v1)2 + (½)k(x1)2 = (½)m(v2)2 +(½)k(x2)2
x1 = Initial compressed (or stretched) length
x2 = Final compressed (or stretched) length
v1 = Initial velocity, v2 = Final velocity

3. (½)m(v1)2 + (½)k(x1)2 = (½)m(v2)2 + (½)k(x2)2
Example: Toy Dart Gun
Mechanical Energy Conservation
E =
A dart, mass m = 0.1 kg is pressed against the spring of a dart gun. The spring (constant k = 250 N/m) is compressed a distance x1 = 0.06 m & released. The dart detaches from the spring it when reaches its natural length (x = 0). Calculate the speed v2 it has at that point.
Figure Caption: Example 8-7. (a) A dart is pushed against a spring, compressing it 6.0 cm. The dart is then released, and in (b) it leaves the spring at velocity v2.
Solution: Use conservation of energy, with the initial energy as elastic potential energy; v = 3.0 m/s.
(½)m(v1)2 + (½)k(x1)2 = (½)m(v2)2 + (½)k(x2)2
( ½ )(250)(0.06)2 = (½)(0.1)(v2)2 + 0
Gives: v2 = 3 m/s

4. Example: Pole Vault
Estimate the kinetic energy & the speed required for a 70-kg pole vaulter to just pass over a bar 5.0 m high. Assume the vaulter’s center of mass is initially 0.90 m off the ground & reaches its maximum height at the level of the bar itself.
Figure Caption: Transformation of energy during a pole vault.
Solution: Assume the vaulter’s body is at rest at 9.0 m, and equate that total energy to the total energy when the running vaulter starts placing the pole. This gives a speed of about 9 m/s.

5. Example: Two Kinds of PE
v1 = 0
m = 2.6 kg, h = 0.55 m
Y = 0.15 m, k = ?
v2 = ?
A 2 step problem:
Step 1: (a)  (b)
(½)m(v1)2 + mgy1 = (½)m(v2)2 + mgy2
v1 = 0, y1 = h = 0.55 m, y2 = 0
Gives: v2 = 3.28 m/s
v3 = 0
Step 2: (b)  (c) (both gravity & spring PE)
 (½)m(v2)2 + (½)k(y2)2 + mgy2 =
(½)m(v3)2 + (½)k(y3)2 + mgy3
y3 = Y = 0.15m, y2 =  (½)m(v2)2 = (½)kY2 – mg
Solve & get k = 1590 N/m
ALTERNATE SOLUTION: (a)  (c) skipping (b)

6. Example: Bungee Jump Mechanical Energy Conservation
m = 75 kg, k = 50 N/m, y2 = 0
v1 = 0, v2 = 0, y1 = h = 15m + y
y =?
Mechanical Energy
Conservation
with both gravity & spring (elastic) PE
 (½)m(v1)2 + mgy1
= (½)m(v2)2 + mgy2 + (½)k(y)2
= 0 + mg(15+y)
= (½)k(y)2
Quadratic Equation for y:
Solve & get y = 40 m & -11 m
(throw away negative value) 
Δy =
+ 15m 
(a)  (c) directly!!

7. Other forms of energy; Energy Conservation
In any process, total energy is neither created nor destroyed.
Energy can be transformed from one form to another & from one body to another, but the total amount is constant.
 Law of Conservation of Energy
Again: Not exactly the same as the Principle of Conservation of Mechanical Energy, which holds for conservative forces only! This is a general Law!!
Forms of energy besides mechanical:
Heat (conversion of heat to mech. energy & visa-versa)
Chemical, electrical, nuclear, ..

8. total amount remains constant  Law of Conservation of Energy
The total energy is neither decreased
nor increased in any process.
Energy can be transformed from one form to another & from one body to another, but the
total amount remains constant
 Law of Conservation of Energy
Again: Not exactly the same as the Principle of Conservation of Mechanical Energy, which holds for conservative forces only! This is a general Law!!

9. Problems with Friction
We had, in general:
WNC = KE + PE
WNC = Work done by all non-conservative forces
KE = Change in KE
PE = Change in PE (conservative forces only)
Friction is a non-conservative force! So, if friction is present, we have (WNC  Wfr)
Wfr = Work done by friction
Moving through a distance d, friction force Ffr does work Wfr = - Ffrd

10. When friction is present, we have:
Wfr = -Ffrd = KE + PE = KE2 – KE1 + PE2 – PE1
Also now, KE + PE  Constant!
Instead, KE1 + PE1+ Wfr = KE2+ PE2
or: KE1 + PE1 - Ffrd = KE2+ PE2
For gravitational PE:
(½)m(v1)2 + mgy1 = (½)m(v2)2 + mgy2 + Ffrd
For elastic or spring PE:
(½)m(v1)2 + (½)k(x1)2 = (½)m(v2)2 + (½)k(x2)2 + Ffrd

11.  Ffr= 370 N Example: Roller Coaster with Friction
A roller-coaster car, mass
m = 1000 kg, reaches a vertical height of only
y = 25 m on the second
hill before coming to a momentary stop. It travels a total distance d = 400 m.
Figure Caption: Example 8–10. Because of friction, a roller-coaster car does not reach the original height on the second hill. (Not to scale)
Solution: The difference in the initial and final energies is the thermal energy produced, J. This is equal to the average frictional force multiplied by the distance traveled, so the average force is 370 N.
Calculate the work done by friction (the thermal energy produced) & calculate the average friction force on the car.
m = 1000 kg, d = 400 m, y1 = 40 m, y2= 25 m, v1= y2 = 0, Ffr = ?
(½)m(v1)2 + mgy1 = (½)m(v2)2 + mgy2 + Ffrd
 Ffr= 370 N

12. Power
Power  Rate at which work is done or rate at which energy is transformed:
Average Power:
P = (Work)/(Time) = (Energy)/(time)
Instantaneous power:
SI units: Joule/Second = Watt (W) 1 W = 1J/s
British units: Horsepower (hp). 1hp = 746 W
A side note:
“Kilowatt-Hours” (from your power bill). Energy!
1 KWh = (103 Watt)  (3600 s) = 3.6  106 W s = 3.6  106 J

13. Example: Stair Climbing Power
A 60-kg jogger runs up a long flight of stairs in 4.0 s. The vertical height of the stairs is 4.5 m.
a. Estimate the jogger’s power ss output in watts and horsepower.
b. How much energy did this ss require?
Figure 8-21.
Solution: a. The work is the change in potential energy, and the power is the work divided by time: P = 660 W = 0.88 hp.
b. The energy is 2600 J.

14.  F (d/t) = F v = average power v  Average speed of the object
Its often convenient to write power in terms of force & speed. For example, for a force F & displacement d in the same direction, we know that the work done is:
W = F d So
 F (d/t) = F v = average power
v  Average speed of the object

15. Example: Power needs of a car
Calculate the power required for a 1400-kg car to do the following: a. Climb a 10° hill (steep!) at a steady 80 km/h b. Accelerate on a level road from 90 to 110 km/h in 6.0 s Assume that the average retarding force on the car is FR = 700 N.
a. ∑Fx = 0
F – FR – mgsinθ = 0
F = FR + mgsinθ
P = Fv l
b. Now, θ = 0
∑Fx = ma
F – FR= 0
v = v0 + at
Figure Caption: Example 8–15: Calculation of power needed for a car to climb a hill.
Solution: a. If the speed is constant, the forces on the car must add to zero. This gives a force of 3100 N. Then the power is 6.8 x 104 W = 91 hp.
b. The car must exert enough force to reach the given acceleration, as well as overcome the retarding force. This gives F = 2000 N. The maximum power is required at the maximum speed, 6.12 x 104 W or 82 hp.