Transcript File
PE KE Work Review and Power
PE and KE Worksheet Solutions
Work
• Work is done on an object when a force causes a displacement of the object.
• Work is done only when components of a force are parallel to a displacement.
Indicate whether or not the following represent examples of work.
a. A teacher applies a force to a wall and becomes exhausted. NO b. A weightlifter lifts a barbell above her head.
YES
c. A waiter carries a tray full of meals across a dining room at a constant speed. NO d. A rolling marble hits a note card and moves it across a table. YES
A 100 N force is applied to move a 15 kg object a horizontal distance of 5 meters at constant speed. Calculate the work done.
W = (100N)(5m) = 500 J
Ben Pumpiniron applies an upward force to lift a 129-kg barbell to a height of 1.98 m at a constant speed. Determine the work done.
W = (129)(9.81)(1.98) = 2505.67 J
A toy car is moving along with 0.40 joules of kinetic energy. If its speed is doubled, then its new kinetic energy will be _______.
a.
b.
c.
d.
e.
0.10 J 0.20 J 0.80 J
1.60 J
still 0.40 J
Which would ALWAYS be true of an object possessing a kinetic energy of 0 joules?
a.
b.
c.
d.
e.
f.
g.
It is on the ground.
It is at rest.
It is moving on the ground It is moving. It is accelerating. It is above the ground. It is moving above ground level.
Which would ALWAYS be true of an object possessing a potential energy of 0 joules?
a.
b.
c.
d.
e.
f.
g.
It is on the ground.
It is at rest. It is moving.
It is accelerating. It is at rest above ground level It is above the ground.
It is moving above ground level.
The total mechanical energy of an object is the ______.
a.
KE minus the PE of the object b.
c.
PE minus the KE of the object
the initial KE plus the initial PE of the object
d.
final amount of KE and PE minus the initial amount of KE and PE
Calculate the kinetic energy of a 5.2 kg object moving at 2.4 m/s.
KE = ½(5.2kg)(2.4m/s ) 2 =14.976 J
Calculate the total mechanical energy of a 5.2 kg object moving at 2.4 m/s and positioned 5.8 m above the ground.
KE = ½(5.2)(2.4) 2 = 14.976 J PE = (5.2)(9.81)(5.8) = 295.8696 J ME = PE + KE
ME = 310.8456 J
honors
Calculate the speed of a 5.2 kg object that possesses 26.1 J of kinetic energy. 26.1 = ½ (5.2)v 2 26.1 = 2.6v
2 10.04 = v 2
V = 3.17 m/s
Power
The quantity
work
has to do with a force causing a displacement. Work has nothing to do with the amount of time that this force acts to cause the displacement. Sometimes, the work is done very quickly and other times the work is done rather slowly. For example, a rock climber takes an abnormally long time to elevate her body up a few meters along the side of a cliff. On the other hand, a trail hiker (who selects the easier path up the mountain) might elevate her body a few meters in a short amount of time. The two people might do the same amount of work, yet the hiker does the work in considerably less time than the rock climber. The quantity that has to do with the rate at which a certain amount of work is done is known as the power. The hiker has a greater power rating than the rock climber.
The standard metric unit of power is the Watt. As is implied by the equation for power, a unit of power is equivalent to a unit of work divided by a unit of time. Thus, a Watt is equivalent to a Joule/second. For historical reasons, the horsepower is occasionally used to describe the power delivered by a machine. One horsepower is equivalent to approximately 750 Watts.
Power
• The rate of energy transfer ▫ Energy used or work done per second Units = watts (W) Horsepower (hp) is a unit used 1.00 hp = 746 W
A hair dryer uses 72,000 joules of energy in 60 seconds. What is the power of this hair dryer?
Water flows over a section of Niagara Falls at the rate of 1.2 X 10 by the falling water?
6 kg/s and falls 50.0 m. How much power is generated
Two horses pull a cart. Each exerts a force of 250.0 N at a speed of 2.0 m/s for 10.0 min.
A. Calculate the power delivered by the horses.
B. How much work is done by the two horses?
Answers: 1.0 x 10 3 W and 6.0 x 10 5 J
An elevator lifts a 500-kg load up a distance of 10 meters in 8 seconds.
A.
Calculate the work done by the elevator.
B.
Calculate the elevator’s power.
W = (500)(9.81)(10) = 49,050 J P = 49,050 J/8 seconds = 6131.25 watts