1. Use g = 10 m/s2 for these problems
A cart is loaded with a brick and pulled at constant speed along an inclined plane. If the mass of the loaded cart is 3.0 kg and the height of the incline is 0.45 meters, then what is the potential energy of the loaded cart at the height of the incline
UG = mgh → UG = (3)(10)(.45)
13.5 Joules

2. 2. If a force of 15.0 N is used to drag the loaded cart (from previous question) along the incline for a distance of 0.90 meters, then how much work is done on the loaded cart?
W= F•d → W = (15)(.9)
13.5 Joules
Note that the work done to lift the loaded cart up the inclined plane at constant speed is equal to the potential energy change of the cart. This is not coincidental! The reason for the relation between the potential energy change of the cart and the work done upon it is ___________________
Conservation of energy

3. 3. Determine the kinetic energy of a 1000-kg roller coaster car that is moving with a speed of 20.0 m/s.
KE = ½mv2 → KE = ½(1000)(20)2
200,000 Joules
4. If the roller coaster car in the above problem were moving with twice the speed, then what would be its new kinetic energy?
KE = ½mv2 → KE = ½(1000)(40)2
800,000 Joules
* Naturally you could have arrived at this answer by
THINKING !!!!
What did we do to the velocity? What effect would this have on KE?
doubled it
It would quadruple it [22]

4. 5. Missy Diwater, the former platform diver for the Ringling Brother's Circus had a kinetic energy of 15,000 J just prior to hitting the bucket of water. If Missy's mass is 50 kg, then what is her speed?
KE = ½mv2 → = ½(50)(v)2
24.5 m/s
6. A 750-kg compact car moving at 100 km/hr has approximately 290,000 Joules of kinetic energy. What is the kinetic energy of the same car if it is moving at km/hr? (HINT: use the kinetic energy equation as a “guide to thinking”)
What did we do to the velocity? What effect would this have on KE?
halved it
It would reduce to one-quarter it [(½)2]
72,500 Joules

5. 7. An escalator is used to move 20 passengers every minute from the first floor of a department store to the second. The second floor is located 5-meters above the first floor. The average passenger's mass is 60 kg. Determine the power requirement of the escalator in order to move this number of passengers in this amount of time.
W= F•d → W = (600)(5) = 3000 Joules per passenger
3000 Joules per passenger x 20 passengers = 60,000 Joules
P= W/t → P = (60,000)/(60)
1000 Watts

6. Us = ½kx2 Us = ½ (k)(0)2 = 0 J (cord not stretched) (¼ way down)
m = 70 kg
Us = ½kx2
13,335 = ½ (k)(11.8)2
k = N/m
top)
UG = mgh UG = (70)(9.8)(100) = 68,600 J
KE = ½mv2 KE = ½ (70)(0)2 = 0 J
Us = ½kx2 Us = ½ (k)(0)2 = 0 J (cord not stretched)
(¼ way down)
UG = mgh UG = (70)(9.8)(75) = 51,450 J
KE = the rest of the energy KE = 17,150 J
Us = ½kx2 Us = ½ (k)(0)2 = 0 J (cord not stretched)
(¾ way down)
UG = mgh UG = (70)(9.8)(25) = 17,150 J
KE = ½mv2 KE = ½ (70)(33)2 = 38,115 J
Us = the rest of the energy Us = 13,335 J

7. When velocity is doubled, KE is . . . . . . .
quadrupled (4 x 106 J)
When velocity is tripled, KE is
nonupled (9 x 106 J)
Total mechanical energy at the top of the platform is
15,000 J
Total mechanical energy everywhere in the problem is
15,000 J
KE’s are
3750 J; 7500 J; 11,250 J; 15,000 J

8. PE (U) is 30 J at the top of the wall; PE would be ______ J at the top of the ramp and _______ J at the top of the stairs. 30 30
If the ball fell off the back of the stairs, its KE just before striking the ground would be
30 J
Since UG = mgh, when the ball is on the second stair (⅔ as high), the PE would be
20 J

9. PE at the top for the pile driver is 104 J.
Total mechanical energy of this system is
104 J
Therefore the work done by the pile driver is
104 J
Total mechanical energy of this system is
50 J
Missing energies are
PEtop = 50 J
KEmiddle = 25 J
Total mechanical energy of this system is
10 J
Missing energies are
KEmoving down = 8 J
KEbottom= 10 J
PEtop= 10 J
KEtop= 0 J