1. 2. Positive and negative work
1. Work don by a constant force
Definition:
Units:
[W] = N*m = J
2. Positive and negative work
Work done by forces that oppose the direction of motion will be negative.
Centripetal forces do no work, as they are always perpendicular to the direction of motion.

2. Example: The object of mass m = 800 g is displaced by a constant the
force F = 20 N as shown below (the angle θ=60º). The force of friction is
f = 6 N. Find the work done by each of these forces and the total work.

3. N fk fk Δx Δx mg 90° 180° W = |fk| |Δx| cos(180°) = -|fk| |Δx| < 0
Example: A block slides down a rough inclined surface. The forces acting on the block are depicted below. The work done by the frictional force is:
A. Positive
B. Negative
C. Zero N fk Δx fk
180°
90° Δx mg
W = |fk| |Δx| cos(180°) = -|fk| |Δx| < 0
Work done by the normal force:
WN = |N| |Δx| cos(90°) = 0
0 < θ < 90°
Work done by weight:
Wmg = |mg| |Δx| cos(θ ) > 0

4. Work kinetic energy principle
W=K2 - K1
Definition:
Example: An 80-g arrow is fired from a bow whose string exerts an average force of 100 N on the arrow over a distance of 49 cm. What is the speed of the arrow as it leaves the bow?
m = 80 g
F = 100 N
d = 49 cm
v1= 0
v2 - ?

5. Example: Paul pushes a box along 10 m across the floor at a constant velocity by exerting a force of 200N. N Δx
FPaul fk mg
Work by Paul on the box: WPaul = (200 N)(10 m) = 2000 J
(energy coming from the biochemical reactions in his muscles)
Work by friction on the box: Wf = (-200 N)(10 m) = J
(released as thermal energy into the air and the floor)
WKE theorem: WPaul + Wf = Kf – Ki = 0
The weight of the box that Paul pushes along a horizontal surface does no work since the weight is perpendicular to the motion.
The normal by the floor does no work either.

6. We can use kinematics or… the WKE theorem
Example: A ball is dropped and hits the ground 50 m below. If the initial speed is 0 and we ignore air resistance, what is the speed of the ball as it hits the ground?
We can use kinematics or… the WKE theorem
Work done by gravity: mgh
Example: Two blocks (m1=2m2) are pushed by identical forces, each starting at rest at the blue vertical line (start). Which object has the greater kinetic energy when it reaches the green vertical line (finish)? 
A. Box1              
B. Box 2
C. They both have the same kinetic energy
Same force, same distance
Same work
Same change in kinetic energy

7. Example: What is the speed of the system after box 1 has fallen for 30cm?
Two approaches:
Use Newton’s laws to find acceleration; use kinematics to find speed. (long!!)
Use WKE theorem (smart!)
External forces doing work: m1g, fk
(The tensions are internal forces: their net work is zero)
m1g
m2g N fk T
m2 = 5.0 kg
μk = 0.2
m1 = 3.0 kg
V=1.2 m/s

8. Example: A person pulls a 10-kg block up a ramp of height h = 10 m and angle θ = 10 at constant speed. The rope makes an angle φ = 15 with the ramp and has a tension of 50 N. How many forces are doing work on the box? Find the work done by all the forces.
A B C D. 4 E.5
Wg < 0
Wf < 0
WT > 0
WN = 0 Δx T φ
WNet = 0
(Constant speed) h θ φ Δx T N fk mg
1. Work by the tension:

9. 90 N WN =0 2. Work by gravity: Of course!mg Δx
θ + π/2
Of course!
–h is the component of the displacement in the direction of weight!
3. Work by the normal: N Δx
90
WN =0

10. 1. Find fk with Newton’s laws and use it to find the work (long)
A person pulls a 10-kg block up a ramp of height h = 10 m and angle θ = 10 at constant speed. The rope makes an angle φ = 15 with the ramp and has a tension of 50 N. Find the work done by all the forces.
4. Work by the friction:
Two options:
1. Find fk with Newton’s laws and use it to find the work (long)
2. Use the WKE theorem (smart):