Transcript notes7

Chapter 7: Work and Energy
 Alternative method for the study of motion
 In many ways easier, gives additional
information
 Kinetic energy: consider an object of mass m
and speed v, we define the kinetic energy as
2
1
K
2
- a scalar, not a vector
- units kg m2/s2 = N m = Joule (J) in S.I.
(ft lb in B.E. and erg in CGS)
- like speed, gives a measure of an object’s
motion (a car and tractor-trailer may have
the same v, but different K)
E  mv  K
 Work: the work done on an object by an
applied constant net force F which results in the
object undergoing a displacement of s (or x or r)
 
W  F  s  Fs s  ( F cos  )s
 F ( s cos  )
- a scalar, units of N m = J
- if F and s are perpendicular, W=0
- work can be negative (>90°)

Fs

F

s
 Work-Energy Theorem: when a net external
force does work on an object, there is a change
in the object’s KE
W  K  K f  K o
W  mv  mv
1
2
2
f
1
2
2
0
Example
A crate on a incline is held in place by a rope. The
rope is released and the crate slides to the bottom.
Determine the total work done if the crate has a
mass of 100 kg, the incline has angle of 50.0°, the
coefficient of kinetic friction is 0.500, and
displacement of the crate is 10.0 m.
Solution:
Given: m = 100 kg,  = 50°, k = 0.500, s = 10.0 m
Approach: compute the work for each force
FN
fk
mg
y
FN
fk

s
x

mg
FBD

s
 Only force components along the direction of s
contribute (x-direction)
F
 FN  mg cos   0
FN  mg cos 

W f  F cos  s  f k cos 180 s   f k s
3
   k mg cos  s  -3.15x10 J

WN  FN cos 90 s  0
y
Wg  mg cos(90   ) s  mg sin  s
3
 7.51x10 J
Total work  W  Wg  W f
 mg sin  s   k mg cos  s
 mgs(sin    k cos  )
3
3
3
 7.51x10  3.15x10 J  4.36x10 J

 Or calculate the net force along s (x-direction)
F
 mg sin   f k  mg sin    k mg cos
 mg (sin    k cos )  Fs ( max )
W  Fs s  mgs(sin    k cos ) Same as above
x
 Now determine final velocity from work-energy
theorem, since v0 = 0, K0 = 0
W  K f  K 0  mv
2W
2(4360 J)
vf 

 9.34 ms
m
100 kg
2
f
1
2
 Check by kinematics
v  v  2a x x  v  2a s s
v f  2as s  2 g (sin   k cos ) s
v f  9.34 ms
2
f
2
0
2
0
Work Done by Gravity
 If one lifts an object of mass m from the floor
(y0=0) to a height yf=h, you have done work on
the object
W  F cos  s  mg ( y f  y0 )  mgh
 We have imparted energy to it, but it is at rest
(v=0). So, this energy is not kinetic energy. It is
called Potential Energy (PE or U), or in this
particular case, gravitational potential energy
 U is energy that is stored and which can be
converted to another kind of energy, K for example
U g  mgh
 PE is a scalar with units of J in S.I.
 h is the height above some reference point, e.g.
table, floor, …
Conservation of (Mechanical) Energy
 Total (mechanical) energy is constant within
some specified system
- total energy is conserved
- conservation principles are very important
in physics; we will see many others later