Transcript Work and energy

Work and energy
Objectives
1. Recognize the difference between the scientific
and the ordinary definitions of work.
2. Define work, relating it to force and displacement.
3. Identify where work is being performed in a
variety of situations.
4. Calculate the net work done when many forces are
applied to an object.
Homework:
Definition and Mathematics of Work
• In physics, work is defined as a force acting
upon an object to cause a displacement.
Work is being done
Work
Workisisnot
notbeing
beingdone
done
Let’s practice – work or no work
1. A student applies a force to a wall and
becomes exhausted.
no work
2. A calculator falls off a table and free falls to
the ground.
work
3. A waiter carries a tray full of beverages
above his head by one arm across the room
no work
4. A rocket accelerates through space.
work
Calculating the Amount of Work Done by Forces
W  Fd cos
F
θ
• W - is work in N∙m or Joule (J). 1 J = 1
N∙m = 1 kg∙m2/s2
F
θ
d
• d - is the displacement in meters
• θ = angle between force and
displacement
d
Fy
• F - is the force in Newton, which causes
the displacement of the object.
Fx
• Work is a scalar quantity
• Work is independent of time the force
acts on the object.
Only the horizontal component of the
force (Fcosθ) causes a horizontal
displacement.
Example 1
• How much work is done on a vacuum cleaner
pulled 3.0 m by a force of 50.0 N at an angle
of 30o above the horizontal?.
W  Fd cos
W  (50.0 N )(3.0m) cos 30
W  130J
o
Example 2
• How much work is done in lifting a 5.0 kg box
from the floor to a height of 1.2 m above the
floor?
Given: d = h = 1.2 meters; m = 5.0 kg; θ = 0
Unknown:
W=?
W = F∙dcosθ
F = mg = (5.0 kg)(9.81 m/s2) cos0o = 49 N
W = F∙d = (49 N) (1.2 m) = 59 J
Example 3
• A 2.3 kg block rests on a horizontal surface. A constant force of
5.0 N is applied to the block at an angle of 30.o to the
horizontal; determine the work done on the block a distance
of 2.0 meters along the surface.
Given:
F = 5.0 N;
m = 2.3 kg
d = 2.0 m
θ= 30o
unknown:
W
=?J
5.0 N
30o
2.3 kg
Solve:
W =
F∙d∙cosθ
W = (5.0 N)(2.0 m)(cos30o) = 8.7 J
Example 4
• Matt pulls block along a horizontal surface at constant
velocity. The diagram show the components of the force
exerted on the block by Matt. Determine how much work is
done against friction.
Given:
Fx = 8.0 N
Fy = 6.0 N
dx = 3.0 m
unknown:
W=?J
6.0 N
F
W = Fxdx
8.0 N
W = (8.0 N)(3.0 m) = 24 J
3.0 m
Class work
• Page 170 practice #1-4
1.
2.
3.
4.
1.50 x 107 J
7.0 x 102 J
1.6 x 103 J
1.1 m
The sign of work
When No work is done
W  Fd cos  0
Force vs. displacement graph
Example: a block is pulled along a
table with 10. N over a distance of 1.0
m.
W = Fd = (10. N)(1.0 m) = 10. J
height
base
area
Force (N)
• The area under a force versus displacement
graph is the work done by the force.
work
Displacement (m)
The angle in work equation
• The angle in the equation is the angle between the
force and the displacement vectors.
F & d are in the same direction, θ is 0o.
d
F
What is θ in each case?
Class work
• Page 171 #1-6