1. Chapter 7 Work and Energy
Chapter Opener. Caption: This baseball pitcher is about to accelerate the baseball to a high velocity by exerting a force on it. He will be doing work on the ball as he exerts the force over a displacement of several meters, from behind his head until he releases the ball with arm outstretched in front of him. The total work done on the ball will be equal to the kinetic energy (1/2 mv2) acquired by the ball, a result known as the work-energy principle.

2. Problem 29
29.(II) What will a spring scale read for the weight of a 53-kg woman in an elevator that moves (a) upward with constant speed 5.0 m/s, (b) downward with constant speed 5.0 m/s, (c) upward with acceleration 0.33 g, (d) downward with acceleration 0.33 g, and (e) in free fall?

3. 6-4 Satellites and “Weightlessness”
More properly, this effect is called apparent weightlessness, because the gravitational force still exists. It can be experienced on Earth as well, but only briefly:
Figure Caption: Experiencing “weightlessness” on Earth.

4. 7-1 Work Done by a Constant Force
The work done by a constant force is defined as the distance moved multiplied by the component of the force in the direction of displacement:
W= Fd cos θ
Figure 7-1. Caption: A person pulling a crate along the floor. The work done by the force F is W = Fd cos θ, where d is the displacement.
If the force and the displacement are in the
same direction θ =0 and W=Fd

5. 7-1 Work Done by a Constant Force
In the SI system, the units of work are joules:
Work is a scalar
As long as this person does not lift or lower the bag of groceries, he is doing no work on it. The force he exerts has no component in the direction of motion.
Figure 7-2. Caption: The person does no work on the bag of groceries since FP is perpendicular to the displacement d.

6. Total Work or Net Work
When more than one force acts on an object, the total work is the sum of the work done by each force.
Airplane Example N
Another way to look at it is….
Fdrag mg
Airplane Example

7. A boy pulls a sled across the snow 10 m
A boy pulls a sled across the snow 10 m. The work done on the sled by the normal force is
Positive
Negative
Zero
Question
Normal

8. A boy pulls a sled across a level field of snow
A boy pulls a sled across a level field of snow. The work done on the sled by gravity is
Positive
Negative
Zero
Question
Gravity

9. A boy pulls a sled across the snow 10 m
A boy pulls a sled across the snow 10 m. The work done on the sled by the Boy is
Positive
Negative
Zero
Force Boy
Question

10. Question
A pendulum bob is swinging back and forth in an arc in the plane of the page as seen below. The Tension force is 20 N, and the arc length is 5 m.
The Work done by the Tension force in one quarter swing (up or down) is
A) Positive
B) Negative
C) Zero
D) Depends on the direction of the swing T

11. Alice lifts a 1 kg box 2 meters off the ground at a constant speed where gravity = 10m/s2
Work done by gravity is:
A) +20 J
B) -20 J
C) 0
D) -5 J
E) +5 J
Question
2 m

12. 7-1 Work Done by a Constant Force
Example 7-1:Work done on a crate.
A person pulls a 50-kg crate 40 m along a horizontal floor by a constant force FP = 100 N, which acts at a 37° angle as shown. The floor exerts a friction force Ffr=50N. Determine (a) the work done by each force acting on the crate, and (b) the net work done on the crate.
Figure 7-3. Caption: Example 7-1. A 50-kg crate is pulled along a smooth floor.
Solution: The only work that is done is done by the horizontal component of FP. W = 3200 J.

13. Problem 9
9.(II) A box of mass 6.0 kg is accelerated from rest by a force across a floor at a rate of 2m/s2 for 7.0 s. Find the net work done on the box.

14. Problem 10
10. (II) (a) What magnitude force is required to give a helicopter of mass M an acceleration of 0.10 g upward? (b) What work is done by this force as the helicopter moves a distance h upward?

15. 7-1 Work Done by a Constant Force
Example 7-2: Work on a backpack.
(a) Determine the work a hiker must do on a 15.0-kg backpack to carry it up a hill of height h = 10.0 m, as shown. Determine also (b) the work done by gravity on the backpack, and (c) the net work done on the backpack. For simplicity, assume the motion is smooth and at constant velocity (i.e., acceleration is zero).
Figure 7-4.
Solution: Only the vertical component of the hiker’s force and of gravity do any work. For the hiker, W = 1470 J; for gravity, W = J. The net work is zero.

16. 7-1 Work Done by a Constant Force
Solving work problems:
Draw a free-body diagram.
Choose a coordinate system.
Apply Newton’s laws to determine any unknown forces.
Find the work done by a specific force.
To find the net work, either
find the net force and then find the work it does, or
find the work done by each force and add.

17. 7-2 Scalar Product of Two Vectors
Definition of the scalar, or dot, product:
Therefore, we can write:
Figure 7-6. Caption: The scalar product, or dot product, of two vectors A and B is A·B = AB cos θ. The scalar product can be interpreted as the magnitude of one vector (B in this case) times the projection of the other vector, A cos θ, onto B.

18. Dot Product or Scalar Product
The scalar product is commutative
The scalar product obeys the distributive law of multiplication
The scalar (dot) product is a scalar not a vector

19. Dot Products of Unit Vectors
Using component form with and