#### Transcript Work=Force x Distance Power = Work/Time

```Work
Essential Question:
What is work (in physics)?
Wake Up Warm Up
In a physical sense, what do you think is more
work?
b)Pushing really hard against a wall to the point of
exhaustion
floor or Pushing a 50kg box across the floor at a
constant velocity
d)Studying all night long for a physics test
Let’s create a definition
The only case that involves any work is the third case.
Why? What do you think is the definition of work?
Work is a force action over some distance to cause a
change in energy.
The amount of work done depends on two things:
a. The amount of force exerted and
b. The distance over which the force is applied.
There are two factors to keep in mind when deciding
when work is being done: something has to move
and the motion must be in the direction of the
applied force. Work can be calculated by using the
following formula:
Work=force x distance
Units of Work
Work = Force x Distance
 The unit of force is Newtons (N)
 The unit of distance is meters
 The unit of work is newton-meters
 One Newton-meter is equal to one joule
 So, the unit of work is a joule (J)
Work is done on
the books when
they are being
lifted, but no work
is done on them
when
they are
being
held
or
carried
horizontally.
In which of the following cases was
work done:
Someone applies a force to a wall and becomes exhausted.
2. A book falls off a table and free falls to the ground.
3. A waiter carries a tray full of meals above his head by one
arm straight across the room at constant speed.
4. A rocket accelerates through space.
1.
Let’s Practice
Work = Force x Distance
Calculate: If a man pushes a concrete block
10 meters with a force of 20 N, how much
work has he done?
Work = Force x Distance
Calculate: If a man pushes a concrete block 10
meters with a force of 20 N, how much work
has he done?
(W = 20N x 10m =200 joules)
Power and Work
 Power
is the rate at which work is
done.
 Force=Mass x Acceleration
 Work=Force x Distance
 Power = Work/Time
 The unit of power is the watt.
Practice Questions
1. Two students, Ben and Bonnie, are in the
weightlifting room. Bonnie lifts the 50 kg
barbell over her head (approximately 60
m) 10 times in one minute; Ben lifts the
50 kg barbell the same distance over his
head 10 times in 10 seconds.
• Which student does the most work?
• Which student delivers the most power?
Practice Questions
2. How much power will it take to move a 10 kg mass
at an acceleration of
2 m/s/s a distance of 10
meters in 5 seconds? This problem requires you to
use the formulas for force, work, and power all in
the correct order.
Force=Mass x Acceleration
Work=Force x Distance
Power = Work/Time
2. How much power will it take to move a 10 kg mass
at an acceleration of 2 m/s/s a distance of 10
meters in 5 seconds? This problem requires you to
use the formulas for force, work, and power all in
the correct order.
Force=Mass x Acceleration
Force=10 x 2
Force=20 N
Work=Force x Distance
Work = 20 x 10
Work = 200 Joules
Power = Work/Time
Power = 200/5
Power = 40 watts
What if the force is at an angle to the
displacement?
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
Force Scenarios
Scenario A: A force acts
rightward upon an object
as it is displaced
rightward.
Scenario B: A force acts
leftward upon an object
which is displaced
rightward
Scenario C: A force acts
upward on an object as it
is displaced rightward.
The least steep incline (30-degree incline angle) will require the
least amount of force while the most steep incline will require
the greatest amount of force. Yet, force is not the only variable
affecting the amount of work done by the car in ascending to a
certain elevation. Another variable is the displacement which is
caused by this force. A look at the animation above reveals that
the least steep incline would correspond to the largest
displacement and the most steep incline would correspond to
the smallest displacement.
Free Body Diagrams
W = (100 N) * (5 m)* cos(0 degrees) = 500 J
The force and the displacement are given in the problem statement. It is said (or shown
or implied) that the force and the displacement are both rightward. Since F and d are in
the same direction,the angle is 0 degrees.
W = (100 N) * (5 m) * cos(30 degrees) = 433 J
The force and the displacement are given in theproblem statement. It is said that the
displacement is rightward. It is shown that the force is 30 degrees above the horizontal.
Thus, the angle between F and d is 30 degrees.
W = (147 N) * (5 m) * cos(0 degrees) = 735 J
The displacement is given in the problem statement. The applied force must be 147 N
since the 15-kg mass (Fgrav=147 N) is lifted at constant speed. Since F and d are in the
same direction, the angle is 0 degrees.
Example: Force at an angle
A force of 50 N acts on the block at the angle
shown in the diagram. The block moves a
horizontal distance of 3.0 m. How much work
is done by the applied force?
W = F * d * cos(θ) W = (50 N) * (3 m) * cos (30o) = 129.9 J
Example force applied vertically
How much work is done by an applied force to lift a 15Newton block 3.0 meters vertically at a constant speed?
 To lift a 15-Newton block at constant speed, 15-N of force
must be applied to it (Newton's laws). Thus,
 W = (15 N) * (3 m) * cos (0 degrees) = 45 Joules
Example 2: Force at an angle
Calculate the work done by a 2.0-N force (directed at a 30°
angle to the vertical) to move a 500 gram box a horizontal
distance of 400 cm across a rough floor at a constant speed of
0.5 m/s. (HINT: Be cautious with the units.)
 Here is a good example of the
importance of understanding the
angle between F and d. In this
problem, the d is horizontal and the F
is at a 60-degree angle to the
horizontal. Thus, theta is 60 degrees.
W = (2.0 N) * (4.00 m) * cos (60
degrees) = 4.0 J
Bibliography
 Credit: MFA Resources dropbox
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