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physics skill check on work
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Gravity
Terms in this set (28)
work is a ( ) acting over some amount of ( ) to cause a change in ( )
force
distance
energy
A teacher applies a force to a wall and becomes exhausted. is there work being done?
No
While there is a force, the force does not cause
a displacement. of any object.
A weightlifter lifts a barbell above her head. is there work being done?
Yes
There is a force applied to the barbell and the
force causes a displacement of the barbell.
A waiter carries a tray full of meals across a dining room at a
constant speed. is there work being done?
No
There is a upward force applied to the tray
but the tray moves horizontally. Upward forces do not cause
horizontal displacement.
A shotputter launches the shot. is there work being done?
Yes
The shotputter applies a force to displace the
shot from a compressed arm position to an outstretched arm
position.
Work is a ______________; a + or  sign on a work value indicates information about _______.
scalar; whether the work adds or removes energy from the object
Which sets of units represent legitimate units for the quantity
work? Circle all correct answers.
Joule
Foot x pound
N x m
kg x m2/sec2
info on work formula
The amount of work (W) done on an object by a given force can be calculated using
the formula
W = F d cos Q
where F is the force and d is the distance over which the force acts and Q is the
angle between F and d. It is important to recognize that the angle included in the
equation is not just any old angle; it has a distinct definition that must be
remembered when solving such work problems.
For each situation below, calculate the amount of work done by the applied force.
A 100 N force is applied to
move a 15 kg object a
horizontal distance of 5 meters
at constant speed.
W = (100 N)•(5 m)•cos(0°)
W = 500 J
For each situation below, calculate the amount of work done by the applied force.
A 100 N force is applied at an
angle of 30o to the horizontal to
move a 15 kg object at a
constant speed for a horizontal
distance of 5 m.
W = (100 N)•(5 m)•cos(30°)
W = 433 J.
For each situation below, calculate the amount of work done by the applied force.
An upward force is applied to
lift a 15 kg object to a height of
5 meters at constant speed.
The F value is equal to m•g
since the speed is constant.
W = (147 N)•(5 m)•cos(0°)
W = 735 J
Indicate whether there is positive (+) or negative () work being done on the object. An eastwardmoving car skids to a stop across dry pavement.

Indicate whether there is positive (+) or negative () work being done on the object.
A freshman stands on his toes and lifts a World Civilization book to the top
+
Indicate whether there is positive (+) or negative () work being done on the object.
At Great America, a roller coaster car is lifted to the peak of the first hill on the Shock Wave.
+
Indicate whether there is positive (+) or negative () work being done on the object.
A catcher puts out his mitt and catches the baseball.

Indicate whether there is positive (+) or negative () work being done on the object.
A falling parachutist opens the chute and slows down.

Before beginning its initial descent, a roller coaster car is always pulled up the first hill to a high
initial height. Work is done on the car (usually by a chain) to achieve this initial height. A coaster
designer is considering three different angles at which to drag the 2000kg car train to the top of the
60meter high hill. Her big question is: which angle would require the most work?
__All angles
result in the same work.__ Show your answers and explain. This conclusion is supported by the
calculations below. In each case, the angle Q in the work equation is 0°; this is the angle between the
F vector and the displacement (not the incline angle); since these two directions are parallel to each
other, the angle is 0°. The work value is just force•distance•cosine(0°). There is little to no difference
between the three resulting values. The difference falls outside the level of precision to which the
given F and d values have been expressed.
Angle Force Distance Work
35° 1.15
104 N 105 m 1.21
* 106 J
45° 1.41
104 N 84.9 m 1.20
* 106 J
55° 1.64
104 N 73.2 m 1.20
* 106 J
how do you know work was done?
when there's a displacement
what are the three main things needed for work
force, displacement, and cause
the force must cause the
In order for a force to qualify as having done work on an object, there must be a displacement and the force must cause the displacement.
in every example of work being done, there is...
In each case described here there is a force exerted upon an object to cause that object to be displaced.
A waiter carries a tray full of meals above his head by one arm straight across the room at constant speed. (Careful! This is a very difficult question that will be discussed in more detail later.)
No.
This is not an example of work. There is a force (the waiter pushes up on the tray) and there is a displacement (the tray is moved horizontally across the room). Yet the force does not cause the displacement. To cause a displacement, there must be a component of force in the direction of the displacement.
W = F • d • cos Θ
where F is the force, d is the displacement, and the angle (theta) is defined as the angle between the force and the displacement vector. Perhaps the most difficult aspect of the above equation is the angle "theta." The angle is not just any 'ole angle, but rather a very specific angle. The angle measure is defined as the angle between the force and the displacement.
when the force and displacement are:
in the same direction, the angle (O) is zero, making cos 0=1. w=F dcos(1)
exactly opposite directions, the angle (0) is 180, making cos 0= 1. W=Fdcos(1)
perpendicular, the angle (0) is 90, making cos 0= 0. W=Fd(0). No work. no work is done when the force is PERPENDICULAR to displacement. EX: A force acts upward on an object as it is displaced rightward. In such an instance, the force vector and the displacement vector are at right angles to each other. Thus, the angle between F and d is 90 degrees. A vertical force can never cause a horizontal displacement; thus, a vertical force does not do work on a horizontally displaced object!!
negative work
The Meaning of Negative Work
On occasion, a force acts upon a moving object to hinder a displacement. Examples might include a car skidding to a stop on a roadway surface or a baseball runner sliding to a stop on the infield dirt. In such instances, the force acts in the direction opposite the objects motion in order to slow it down. The force doesn't cause the displacement but rather hinders it.
Joule (abbreviated J)
One Joule is equivalent to one Newton of force causing a displacement of one meter
The Joule is the unit of work.
1 Joule = 1 Newton * 1 meter
1 J = 1 N * m
In fact, any unit of force times any unit of displacement is equivalent to a unit of work.
Nonstandard Units of Work
foot•pound
kg•(m/s2)•m
kg•(m2/s2
In summary, work is done when a force acts upon an object to cause a displacement. Three quantities must be known in order to calculate the amount of work. Those three quantities are force, displacement and the angle between the force and the displacement.
...
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