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```Work
1. Work don by a constant force
Definition:
Units:

W  F|| d  Fd cos   Fd
[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.
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.
y
Fy F

Fx
f
a
x
5m
WF  Fx x  ( F cos ) x  (20 N )(cos 600 )(5 m)  50 J
W fr  f (cos1800 ) x   fx  (6 N )(5 m)   30 J


Wtot  F cos  f x  20 N cos 600  6 N (5m)  20 J
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
fk
C. Zero
N
fk
90°
Δx
Δx
180°
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
Work kinetic energy principle
v22  v12  2a( x2  x1 )
mv22 mv12

 ma( x2  x1 )  Fd
2
2
Definition:
K
mv 2
mv22 mv12
W

2
2
W=K2 - K1
2
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 - ?
W  Fd
K1  0
mv22
K2 
2
mv22
Fd 
2
2 Fd
v2 
m
2 100 N  49 102 m
v2 
 35m / s
3
80 10 kg
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
FPaul  f k  ma  0
FPaul  f k
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) = -2000 J
(released as thermal energy into the air and the floor)
•WKE theorem: WPaul + Wf = 0
Kf – K i = 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.
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
W  K

mgh  12 mv 2  0
v  2 gh  2(9.8 m/s2 )(50 m)  31 m/s
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
Example: What is the speed of the system after box 1 has fallen for 30cm?
=x
Two approaches:
1. Use Newton’s laws to find acceleration; use kinematics to find speed. (long!!)
2. Use WKE theorem (smart!)
External forces doing work: m1g, fk
(The tensions are internal forces: their net work is zero)
1
1
2
2
KE  mv

m
v
0
1
2
2
2
Wnet  m1 gx  fk x
 m1 gx  k m2 gx
1
m1  k m2  gx  m1  m2 v 2
2
N
m2 = 5.0 kg
T
fk
m1  k m2
v 
2 gx
m1  m2
T
m1 = 3.0 kg
V=1.2 m/s
m1g
m2g
μk = 0.2
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. 1
B. 2
Wg < 0
C. 3
Wf < 0
D. 4
E.5
WT > 0
WN = 0
WNet = 0
Δx
(Constant speed)
φ
T
Δx
1. Work by the tension:
WT T x cos 
 (50 N)(56.7 m) cos15
 2740 N
N
φ
T
h
fk
θ
x  h / sin 
mg
2. Work by gravity:
Δx


Wg  mg x cos   
2

 mg x sin
 mgh  980 J
θ + π/2
mg
Of course!
–h is the component of
the displacement in the
direction of weight!
3. Work by the normal:
WN =0
90
Δx
N
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): Wnet  KE  0
Wnet  Wg WN WT Wf  0

Wf   Wg WN WT

   980  0  2740  1760 J
```
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