Transcript work and energy

Work and Energy
Work
The work done by a constant force is defined as the product of
the component of the force in the direction of the displacement
and the magnitude of the displacement.
 


W  F d
W  F d cos 
m2
W = Work
units of J or Nm or kg 2
s
The Joule is named after James Prescott Joule
 is the angle between the force and the displacement
Note that Fcos is the component of the force in the direction
of the displacement. If the angle is greater than ninety degrees
then the work will be negative (cos<0).


F
F




d
F cos 
Work is a scalar quantity. Energy is defined as the ability to do
work and therefore is a scalar quantity as well. Work can be
positive or negative but these signs are not direction. We will see
that they indicate a gain of kinetic energy or a loss of kinetic
energy respectively.
Negative work is done on an object when it is slowed by a force.
Positive work is done when an object is sped up by a force.
The area under a F-d graph is equal to the work done by an
applied force. Assume the force and displacement are co-linear.
The total work done on an object is the sum of all the work done
by individual forces.


W Fd


W  FR  d
a) Rousseau pushes with a force of 500 N on an immovable wall.
How much work is done on the wall?
b) David swings a rock around his head with a centripetal force of
250 N. The rock goes around his head 3 times in 0.56 s (the
radius of the circle is 0.8 m). What is the work done on the rock?
c) A 4 kg block is raised 5 m. How much work is done on the block
if one assumes it was lifted with a constant velocity?

Fa

Fg


F a work


W  Fg d cos 
F g work

W  Fg d cos 
W  mgd cos180
W  mgd cos 0
N
m
W  (4kg)(9.81 2 )(5m)( 1) W  (4kg)(9.81 )(5m)( 1)
kg
s
W  196.2 J
W  196.2 J
This means the work done on the ball is 0 J.
N.B.



W  Fnet  d
W  (0)(5)
W  0J
d) A sled (15 kg) is pulled with a 50 N [20o ath] force for 7.5 m.
The coefficient of friction is 0.21. Calculate the work done by each
force and total work done on the sled.

FN

Ff

Fa
Work done by Fa
 
20o

Fg
Work done by FN and Fg are zero
since they are perpendicular to
the displacement.
W  Fa d cos 
W  (50 N )(7.5m)(cos 20)
W  352.4 J
To calculate work done by friction
we must calculate FN.



0  F N  F ay  F g



F N   F ay  F g

FN
N
 50 sin 20  (15kg)(9.81 )
kg
Work done by Ff


W  F f d cos 
W  FN d cos 180
W  (0.21)(130.0)(7.5)( 1)
W  204.8 J

F N  130.0 N [up ]
Therefore the total work done on the sled is 147.6 J
WORK ENERGY THEOREM
For an object that is accelerated by a constant net force and
moves in the same direction . . .


W  Fnet d cos
W  mad cos
v2  v1
W  m(
)d cos
t
d
(1)
t
v v
W  m(v2  v1 )( 2 1 )
2
m
W  (v22  v12 )
2
W  m(v2  v1 )
mv22 mv12
W

2
2
2
mv
define kinetic energy as
2
W  Ek 2  Ek 1
(Ek)
W  Ek
The work done by the net force acting on a
body is equal to the change in the kinetic
energy of the body.
mv22 mv12
Fd cos  

2
2
A ball is dropped from rest at a height of 8.25 m. What will be
its speed when it hits the ground? (could solve this
kinematically but let’s do it using the work-energy theorem)
W  Ek

Fg
mv22 mv12
Fg d cos 

2
2
2
2
mv2 mv1
mgd cos 

2
2
2
v
m
(9.81 2 )(8.25m)(1)  2  0
s
2
m
v2  12.72
s
Therefore the speed of the
ball is 12.72 m/s when it hits
the ground.
GRAVITATIONAL POTENTIAL ENERGY
Conservative and Non-Conservative Forces
A ball is thrown upwards and returns to the thrower with the same
speed it departed with.
A block slides into a spring, compresses it and leaves the spring
with the same speed it first contacted it with.
A force is conservative if the kinetic energy of a particle returns to
its initial value after a round trip (during the trip the Ek may vary).
A force is non-conservative if the kinetic energy of the particle
changes after the round trip (Assume only one force does work on
the object). Gravitational, electrostatic and spring forces are
conservative forces.
Friction is an example of a non-conservative force. For a round
trip the frictional force generally opposes motion and only leads to
a decrease in kinetic energy.
We must introduce the concept of potential energy. This is energy
of configuration or position. As kinetic energy decreases the
energy of configuration increases and vice versa.
Ek  E p  0
Ep
Change in potential
energy
Ek  E g
Eg
Change in gravitational
potential energy
Ee
Change in elastic
potential energy
 E g  W
 E g  Fd cos
 E g  mg d ( 1)
or
 E g  mg d ( 1)
E g  mgh
define Eg as mgh
h is height relative to a
reference point
Gravity does work on an object as its height changes. As an
object increases its height gravity does negative work on the
object and the object’s kinetic energy decreases. This loss of
kinetic energy is a gain of potential energy.
mv22 mv12
(

)  (mgh2  mgh1 )  0
2
2
mv12
mv22
 mgh1 
 mgh2
2
2
Em1  Em2
define Em as the
mechanical energy
Mechanical Energy is
conserved when an object
is acted upon by
conservative forces.
LAW OF CONSERVATION OF ENERGY
Energy may be transformed from one kind to another, but it cannot be created
or destroyed: the total energy is constant. There are many forms of energy such
as electromagnetic, electrical, chemical, nuclear, and thermal.
Ea  Eb  Ec  Ed ....  0
a) A ball is launched from a height of 2 m with an initial
velocity of 25 m/s [35o ath]. What is the speed of the ball
when its height is 7.5 m?
mv12
mv22
 mgh1 
 mgh2
2
2
v12
v22
 gh1   gh2
2
2
v22  v12  2 gh1  2 gh2
m 2
m
m
 (25 )  2(9.81 2 )( 2m)  2(9.81 2 )( 7.5m)
s
s
s
m
v2  22 .73
s
The speed of the ball is 22.73 m/s
v22
The energy approach doesn’t calculate the velocity but it is
quicker. The kinematics approach is longer but more precise.
ELASTIC POTENTIAL ENERGY
Hooke’s Law (Robert Hooke 1678)
The magnitude of the force exerted by a spring is directly
proportional to the distance the spring has moved from its
equilibrium position. An ideal spring obeys Hooke’s Law because
it experiences no internal or external friction.
breaking point
F (N)
slope = k
non-elastic region
elastic limit
x (m)
The linear region is sometimes called Hooke’s Law region. It
applies to many elastic devices.
Hooke’s Law
F = force exerted on the spring (N)
k = force constant of spring (N/m)
F  kx
x = position of spring relative to the equilibrium (deformation) (m)
The direction of compression on the spring is negative while the
direction of elongation is positive (for F and x). The spring exerts
an equal and opposite force on the object.
Derivation of Elastic Potential Energy
A spring exerts a conservative force on a object. An object will
have the same kinetic energy after a round trip with a spring. The
spring will begin at its equilibrium position with zero potential
energy.
Ek  E p  0
 Ee  Ek
 Ee  W


 ( Ee 2  Ee1 )  F  d cos 
kx
 Ee 2  x(1)
2
2
kx
Ee 2 
2
zero since at equilibrium
the force on the object and its
d have opposite directions
For an object interacting with an ideal spring.
F  kx
1 2
Ee  kx
2
The potential energy of this object must be considered in the
mechanical energy.
2
2
1
2
2
2
mv1
mv2
kx
kx
 mgh1 

 mgh2 
2
2
2
2
Remember a reference height is needed for height. The
direction of x is not important unless solving for x. If it is known
that the answer is compression then –x is correct. If the answer
is elongation then +x is correct.
If one form of energy is not present then it need not be included
in the equation.
a) A 2 kg ball is dropped from a height of 10 m onto a spring that
is 0.75 m in length and has a spring constant of 1000 N/m. How
far will the ball compress the spring? What force is exerted on the
ball at its lowest point?
10 m
initial
v1=0
h1=10 m
x1=0
final
v2=0
h2=0.75+x2
0.75 m
x2
0.75+x2
2
2
1
2
2
2
mv1
mv2
kx
kx
 mgh1 

 mgh2 
2
2
2
2
kx22
mgh1  mgh2 
2
kx22
mgh1  mg (0.75  x2 ) 
Use the quadratic formula to solve.
2
x2   0.5832 m or x2   0.6224 m
 the spring compresses 0.6224 m
The force exerted by the spring on the ball at this is calculated by. . .
F  kx
N
F  (1000 )(0.6224 m)
m
F  622.4 N [up]