1. Aim: How do we explain work done by a constant force?

2. Definition of Work (non-calculus)
The work W done by an agent exerting a constant force on a system is the product of the component FcosΘ along the direction of the displacement of the point of application of the force and magnitude of Δx. W=F Δx cos Θ

3. Characteristics of Work
Is work a vector or scalar? Scalar
What are the units of work? Joules
3) A person slowly lifts a heavy box of mass m a vertical height h, and then walks horizontally at constant velocity a distance d while holding the box.
Determine the work done by (a) by the person and (b) by the gravitational force on the box
a) mgh b) -mgh

4. When will the following occur:
Negative work done on an object
Positive work done on an object
0 work done on an object
The angle between force and displacement must be between 90 and 270 degrees
The angle between the force and displacement must either be between 0 and 90 degrees or 270 and 360 degrees
90 degrees and 270 degrees

5. Mountain
Climber A and Climber B have equal weights. They both climb to the top of the mountain. Climber A takes a path which has a gentle slope. Climber B takes a path which has a steep slope. Ignoring friction, who does more work in climbing the mountain? SAME WORK

6. Mr. Clean problem
A man cleaning his apartment pulls a vacuum cleaner with a force of magnitude F = 50.0 N. The force makes an angle of 30 degrees with the horizontal. The vacuum cleaner is displaced 3.00 m to the right. Calculate the work done by the 50 N force on the vacuum cleaner.

8. Work Problems
2) Find the work done by the man on the vacuum cleaner if he pulls 3.00 m with horizontal force of 32.0N. 3) If a person lifts a 20 kg bucket slowly from a well and does 6000J work work on the bucket, how deep is the well? 4) A 65 kg woman climbs a flight of 20 stairs, each 23 cm high. How much work is done by the gravitational force in the process?
2) 96 J 3) 30.6 m 4) kJ

10. Work-Energy Theorem
When work is done on a system and the only change in the system is in its speed, the work done by the net force equals the change in kinetic energy of the system. W = ΔKE where KE = ½ mv2

11. Work-Energy Thm Example
A 6 kg block initially at rest is pulled to the right along a horizontal frictionless surface by a constant, horizontal force of 12.0 N. Find the speed of the block after it has moved 3 m.
A 0.6 kg particle has a speed of 2.00 m/s at point A and kinetic energy of 7.50 J at point B.
What is its kinetic energy at A?
It speed at B?
The total work done on the particle as it moves from A to B?
3.46 m/s

13. Dot Product of Two Vectors
W = F ˚Δr Work is the dot product of a force vector and displacement vector. W=|F||r|cosᶿ or W=Fxrx +Fyry What is true about the dot product of two vectors and the product of the magnitudes of the two vectors?

14. Dot product Examples
1. A force F = 2i + 3j causes an object to have a displacement,
Δr = -i + 2j.
Calculate the work done on the object.
Find the angle Θ between the force and displacement vectors.
60.5 degrees

16. Dot Product Examples
2) As a particle moves from the origin to (3i – 4j) m, it is acted upon by a force given by (4i-5j) N. Calculate the work done by this force. W=Fxrx +Fyry = (3)(4) + (-4)(-5) = 32 J
32 J

17. Conclusion A 3 kg object has an initial velocity vi =(6i-2j) m/s
What is its kinetic energy at this time?
Find the total work done on the object if its velocity changes to (8i+4j)m/s
v= √40 so KE=1/2mv2=1/2(3)(40)= 60 J
V=√80 so KE=1/2mv2=1/2(3)(80) =120 J
So 60 J of work must have been done because this was the change in kinetic energy
a) 60 J b) 60 J