Chapter 7
Work and Kinetic Energy
Reading and Review
Vertical circular motion
A
B
C vertical (down)
vertical (up)
horizontal
Centripetal acceleration must be
Condition for falling: N=0
at C:
(now apparent weight is in the opposite direction to true weight!)
So, as long as:
at the top, then N>0 and pointing down.
Barrel of Fun
A rider in a “barrel of
fun” finds herself stuck with her back to the wall. Which diagram correctly shows the forces acting on her?
a b c d e
The normal force of the wall on the
rider provides the centripetal force
needed to keep her going around in a
circle. The downward force of gravity
is balanced by the upward frictional
force on her, so she does not slip
vertically.
Barrel of Fun
A rider in a “barrel of
fun” finds herself stuck with her back to the wall. Which diagram correctly shows the forces acting on her?
Follow-up: What happens if the rotation of the ride slows down?
a b c d e
a) centrifugal force is pushing you into the door
b) the door is exerting a leftward force on you
c) both of the above
d) neither of the above
Around the CurveAround the CurveYou are a passenger in a car, not wearing a seat belt. The car makes a sharp left turn, and you find yourself hitting the passenger door. What is the correct description of what is actually happening?
a) centrifugal force is pushing you into the door
b) the door is exerting a leftward force on you
c) both of the above
d) neither of the above
The passenger has the tendency
to continue moving in a straight
line. There is a net centripetal
force, provided by the door, that
forces the passenger into a
circular path.
Around the CurveAround the CurveYou are a passenger in a car, not wearing a seat belt. The car makes a sharp left turn, and you find yourself hitting the passenger door. What is the correct description of what is actually happening?
Atlas holds up the world. Sisyphus pushes his rock
up a hill.
Who does more work?(a)
(b)
Working Hard... or Hardly Working
Atlas holds up the world. Sisyphus pushes his rock
up a hill.
(a)
(b)
Working Hard... or Hardly Working
With no displacement, Atlas does no work
Work Done by a Constant ForceThe definition of work, when the force is parallel to the displacement:
SI unit: newton-meter (N·m) = joule, J
Friction and Work IFriction and Work I
a) friction does no work at all
b) friction does negative work
c) friction does positive work
A box is being pulled
across a rough floor
at a constant speed.
What can you say
about the work done
by friction?
f
N
mg
Displacement
Pull
Friction acts in the opposite direction to
the displacement, so the work is
negative. Or using the definition of
work (W = F (Δr)cos ), because =
180º, then W < 0.
Friction and Work IFriction and Work I
a) friction does no work at all
b) friction does negative work
c) friction does positive work
A box is being pulled
across a rough floor
at a constant speed.
What can you say
about the work done
by friction?
Can friction ever
do positive work? a) yes
b) no
Friction and Work IIFriction and Work II
Can friction ever
do positive work? a) yes
b) no
Consider the case of a box on the back of a pickup
truck. If the box moves along with the truck, then it is
actually the force of friction that is making the box
move.
Friction and Work IIFriction and Work II
Forces not along displacement
If the force is at an angle to the displacement:
Convenient notation: the dot productThe work can also be written as the dot product of the force and the displacement:
Force and displacement
The work done may be positive, zero, or negative, depending on the angle between the force and the displacement:
Sum of work by forces = work by sum of forces
If there is more than one force acting on an object, we can find the work done by each force, and also the work done by the net force:
Units of Work
Lifting 0.5 L H2O up 20 cm = 1 J
1 kcal = 1 Cal = 4.186 kJ
In a baseball game, the
catcher stops a 90-mph
pitch. What can you say
about the work done by
the catcher on the ball?
a) catcher has done positive work
b) catcher has done negative work
c) catcher has done zero work
Play Ball!Play Ball!
In a baseball game, the
catcher stops a 90-mph
pitch. What can you say
about the work done by
the catcher on the ball?
a) catcher has done positive work
b) catcher has done negative work
c) catcher has done zero work
The force exerted by the catcher is opposite in direction to the displacement of the ball, so the work is negative. Or using the definition of work (W = F (Δr)cos ), because = 180º, then W < 0. Note that the work done on the ball is negative, and its speed decreases.
Play Ball!Play Ball!
Follow-up: What about the work done by the ball on the catcher?
Tension and WorkTension and Work
a) tension does no work at all
b) tension does negative work
c) tension does positive work
A ball tied to a string is
being whirled around in
a circle. What can you
say about the work
done by tension?
Tension and WorkTension and Work
a) tension does no work at all
b) tension does negative work
c) tension does positive work
A ball tied to a string is
being whirled around in
a circle. What can you
say about the work
done by tension?
v
T
No work is done because the force
acts in a perpendicular direction to
the displacement. Or using the
definition of work (W = F (Δr)cos ),
because = 90º, then W = 0.
Follow-up: Is there a force in the direction of the velocity?
Work by gravity
Fg a
h
A ball of mass m drops a distance h. What is the total work done on the ball by gravity?
N
Fg
A ball of mass m rolls down a ramp of height h at an angle of 45o. What is the total work done on the ball by gravity?
h
a
θ
Fgx = Fg sinθ
h = L sinθ
W = Fd = Fgx x L = (Fg sinθ) (h / sinθ)
W = Fg h = mgh
W = Fd = Fg x h
W = mgh
Path doesn’t matter when asking “how much work did gravity do?”
Only the change in height!
Path independence
• If a force depends on POSITION only then the work done by it on an object moving from
to will NOT depend upon the path.
• Such a force is called a Conservative Force
1r
2r
Motion and energy
When positive work is done on an object, its speed increases; when negative work is done, its speed decreases.
Kinetic Energy
After algebraic manipulations of the equations of motion, we find:
Therefore, we define the kinetic energy:
Work-Energy Theorem
Work-Energy Theorem: The total work done on an object is equal to its change in kinetic energy.
(True for rigid bodies that remain intact)
Lifting a BookLifting a Book
You lift a book with your hand
in such a way that it moves up
at constant speed. While it is
moving, what is the total work
done on the book?
a) mg ∆ r
b) FHAND ∆ r
c) [FHAND + mg] ∆ r
d) zero
e) none of the above
mg
∆r FHAND
v = consta = 0
Lifting a BookLifting a Book
You lift a book with your hand
in such a way that it moves up
at constant speed. While it is
moving, what is the total work
done on the book?
The total work is zero because the net force
acting on the book is zero. The work done
by the hand is positive, and the work done
by gravity is negative. The sum of the two
is zero. Note that the kinetic energy of the
book does not change either!
a) mg ∆ r
b) FHAND ∆ r
c) (FHAND + mg) ∆ r
d) zero
e) none of the above
mg
∆r FHAND
v = consta = 0
Follow-up: What would happen if FHAND were greater than mg?
By what factor does the
kinetic energy of a car
change when its speed
is tripled?
a) no change at all
b) factor of 3
c) factor of 6
d) factor of 9
e) factor of 12
Kinetic Energy IKinetic Energy I
By what factor does the
kinetic energy of a car
change when its speed
is tripled?
a) no change at all
b) factor of 3
c) factor of 6
d) factor of 9
e) factor of 12
Because the kinetic energy is mv2, if the speed
increases by a factor of 3, then the KE will increase by a
factor of 9.
Kinetic Energy IKinetic Energy I
Slowing DownSlowing Down
a) 20 m
b) 30 m
c) 40 m
d) 60 m
e) 80 m
If a car traveling 60 km/hr can
brake to a stop within 20 m,
what is its stopping distance if
it is traveling 120 km/hr?
Assume that the braking force
is the same in both cases.
F d = Wnet = ∆KE = 0 –
mv2,
and thus, |F| d = mv2.
Therefore, if the speed doubles,
the stopping distance gets four
times larger.
Slowing DownSlowing Down
a) 20 m
b) 30 m
c) 40 m
d) 60 m
e) 80 m
If a car traveling 60 km/hr can
brake to a stop within 20 m,
what is its stopping distance if
it is traveling 120 km/hr?
Assume that the braking force
is the same in both cases.
Application: ball on a trackhow high must I place the ball so
that it can complete a loop?
Condition: Fcp > mg at top of loop
Fcp = mv2/r = mg v2 = gr
KE = mv2 / 2 = mgr/2Gravity must provide this energy Wg = mg∆h = KE
∆h = r/2 above the top of the loop!
Work Done by a Variable Force
If the force is constant, we can interpret the work done graphically:
Work Done by a Variable Force
If the force takes on several successive constant values:
Work Done by a Variable Force
We can then approximate a continuously varying force by a succession of constant values.
Work Done by a Variable Force
The force needed to stretch a spring an amount x is F = kx.
Therefore, the work done in stretching the spring is
Application: work by a spring
Hooke’s Law: F = - kx k = (3kg)(9.8 m/s2) / (3.9 cm)k = 760 N/m
Loaded spring: W = kx2/2 = (760 N/m) (0.04m)2/ 2
W = 0.61 J
How fast?: v = d/t = (0.020 m) (0.020 s) = 1 m/s
KE = mv2/2 = (1kg)(1m/s)2 / 2KE = 0.55 J
Kinetic Energy:
Power
Power is a measure of the rate at which work is done:
SI unit: J/s = watt, W
1 horsepower = 1 hp = 746 W
ave
WP
t
Power
Power
If an object is moving at a constant speed in the face of friction, gravity, air resistance, and so forth, the power exerted by the driving force can be written:
Question: what is the total work per unit time done on the object?
F x xP F Fv
t t
a) energy
b) power
c) current
d) voltage
e) none of the above
Electric BillElectric Bill
When you pay the electric company
by the kilowatt-hour, what are you
actually paying for?
We have defined: Power = energy / time
So we see that: Energy = power × time
This means that the unit of power × time
(watt-hour) is a unit of energy !!
Electric BillElectric Bill
When you pay the electric company
by the kilowatt-hour, what are you
actually paying for?
a) energy
b) power
c) current
d) voltage
e) none of the above
A block rests on a horizontal frictionless surface. A string is attached to the block, and is pulled with a force of 45.0 N at an angle above the horizontal, as shown in the figure. After the block is pulled through a distance of 1.50 m, its speed is 2.60 m/s, and 50.0 J of work has been done on it. (a) What is the angle (b) What is the mass of the block?
The pulley system shown is used to lift a 52 kg crate. Note that one chain connects the upper pulley to the ceiling and a second chain connects the lower pulley to the crate. Assuming the masses of the chains, pulleys, and ropes are negligible, determine (a) the force F required to lift the crate with constant speed, and(b) the tension in two chains
(a) the force F required to lift the crate with constant speed, and(b) the tension in two chains
(a) constant velocity, a=0, so net force =0.
2T - (52kg)(9.8m/s2) = 0
T = 250 NF = -250 Ny
(b) upper pulley doesn’t move:Tch - 2Trope = 0 Tch = 500 N
lower pulley has constant accelerationTch -2Trope =0 Tch = 500 N
Mechanical Advantage!
(a) how much power is applied to the box by the chain?(b) how much power is applied on the rope by the applied force?
What about work?
Trope = 250 NTchain = 500 NF = -250 Ny
(a) P = Fv = 500 N * vbox
(b)P = Fv = 250 N * vhand
hand moves twice as fast
hand moves twice as far