Date post: | 14-Jan-2016 |
Category: |
Documents |
Upload: | evelyn-harris |
View: | 228 times |
Download: | 0 times |
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:
• In physics, work is defined as a force acting upon an object to cause a displacement.
Definition and Mathematics of Work
Work is being done
Work is not being doneWork is not being done
Let’s practice – work or no work1. A student applies a force to a wall and
becomes exhausted. 2. A calculator falls off a table and free falls to
the ground.3. A waiter carries a tray full of beverages
above his head by one arm across the room
4. A rocket accelerates through space.
no work
work
no work
work
Calculating the Amount of Work Done by Forces
θF
d
• F - is the force in Newton, which causes the displacement of the object.
• d - is the displacement in meters• θ = angle between force and
displacement• W - is work in N m or Joule (J). 1 J = 1 ∙
N m = 1 kg m∙ ∙ 2/s2
• Work is a scalar quantity • Work is independent of time the force
acts on the object.θF
d
Fx
Fy
Only the horizontal component of the force (Fcosθ) causes a horizontal displacement.
cosFdW
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?.
cosFdW omNW 30cos)0.3)(0.50(
JW 130
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?
W = F dcos∙ θF = mg = (5.0 kg)(9.81 m/s2) cos0o = 49 NW = F d ∙ = (49 N) (1.2 m) = 59 J
Given: d = h = 1.2 meters; m = 5.0 kg; θ = 0Unknown: W = ?
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
30o
5.0 N
2.3 kgunknown:
W = ? J Solve:W = F d cos∙ ∙ θW = (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.
8.0 N
6.0 N
3.0 m
W = Fxdx
W = (8.0 N)(3.0 m) = 24 J
Given: Fx = 8.0 NFy = 6.0 N dx = 3.0 m
unknown: W = ? J F
Class work
• Page 170 practice #1-4
1. 1.50 x 107 J2. 7.0 x 102 J3. 1.6 x 103 J4. 1.1 m
The sign of work
When No work is done
0cos FdW
Force vs. displacement graph• The area under a force versus displacement
graph is the work done by the force.
Displacement (m)Fo
rce
(N)
work
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
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.
Fd
What is θ in each case?
Class work
• Page 171 #1-6