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Kinetic and Potential Energy
Physics
Ms. Li
Kinetic Energy
The energy of motion
The net work done on an object is equal to
the change in kinetic energy of an object.
Work = ΔKE
Fd = mad= m(1/2(vf^2-vi^2)=1/2mvf^21/2mvi^2= KEf - KEi
Kinetic Energy
KE = ½mv2
This equation reveals that the kinetic
energy of an object is directly
proportional to the square of its
speed. That means that for a twofold
increase in speed, the kinetic energy
will increase by a factor of four. For a
threefold increase in speed, the
kinetic energy will increase by a factor
of nine.
Energy is scalar
Kinetic energy is a scalar quantity; it does
not have a direction. Unlike velocity,
acceleration, force, and momentum, the
kinetic energy of an object is completely
described by magnitude alone.
Check Your Understanding
1. Determine the kinetic energy of a 625kg roller coaster car that is moving with a
speed of 18.3 m/s.
2. If the roller coaster car in the above
problem were moving with twice the
speed, then what would be its new kinetic
energy?
Check Your Understanding
1. KE = 0.5*m*v2 KE = (0.5) * (625 kg) *
(18.3 m/s)2
KE = 1.05 x105 Joules
2. If the speed is doubled, then the KE is
quadrupled. Thus, KE = 4 * (1.04653 x 105
J) = 4.19 x 105 Joules. or
KE = 0.5*m*v2
KE = 0.5*625 kg*(36.6 m/s)2
KE = 4.19 x 105 Joules
Examples
3.Missy Diwater, the former platform diver
for the Ringling Brother's Circus, had a
kinetic energy of 12 000 J just prior to
hitting the bucket of water. If Missy's mass
is 40 kg, then what is her speed?
4. A 900-kg compact car moving at 60
mi/hr has approximately 320 000 Joules of
kinetic energy. Estimate its new kinetic
energy if it is moving at 30 mi/hr.
Answers
3.KE = 0.5*m*v2 12 000 J = (0.5) * (40 kg) * v2
300 J = (0.5) * v2
600 J = v2
v = 24.5 m/s
4. KE = 80 000 J
The KE is directly related to the square of the
speed. If the speed is reduced by a factor of 2
(as in from 60 mi/hr to 30 mi/hr) then the KE will
be reduced by a factor of 4. Thus, the new KE is
(320 000 J)/4 or 80 000 J.
Example
How much work is required to accelerate a
1000 kg car from 20 m/s to 30 m/s?
Example
5.A car traveling 60 km/hr can brake to a stop
within a distance of 20 m. If the car is going
twice as fast, 120 km/h, what is its stopping
distance?
What is Potential Energy?
Stored energy.
Can be used later to do work
Depends on:
– Position of object (ex. Height)
– Configuration of object(ex. Stretched spring)
Examples of PE.
Wound up clock spring.
A book about to fall off a table
Gasoline
Fuel
Gravitational PE
Depends on height
Depends on mass
Depends on
acceleration due to
gravity
PEgrav= mgh
Work and PE
When work is done
on an object often the
PE of an object
changes.
If an object is lifted
upward the PEgrav
increases
Hmm… that guy
has a lot of PE!
Knowing that the potential energy at the top of
the tall platform is 50 J, what is the potential
energy at the other positions shown on the stair
steps and the incline?
Elastic PE
Often when a spring is compressed it
gains potential energy.
The formula for the force of a spring is
F = -kx
F = force(N)
k = spring constant(N/m)
x = displacement of spring (m)
Elastic PE
When a spring is stretched or compressed
there must be work done on the spring.
This work done gives the spring PE
Equation:
PEelastic= ½kx2
Example
A 1000 kg roller
coaster moves from
point A to point B and
then to point c. What
is the change in PE
as the car moves
form B to C?
B
10m
A
15 m
C
Mechanical Energy
The total amount of potential and kinetic
energy an object has.
If all the forces acting on an object are
conservative then the total mechanical
energy of the system can be given as:
E = KE + PE + Q
E = ½mv2 + PE +Q
Conservation of Energy
The total mechanical energy of a system
neither increases nor decreases in any
process. It stays constant. This is true
only if only conservative forces are acting
in the system.
In other words : If an object falls all of its
potential energy turns into kinetic energy
by the time it reaches zero height.
Problem solving
Remember that an objects PE changes
into KE as it loses height.
For example, a ball rolls down a
frictionless incline plane of height H. It
then rolls up another frictionless incline
plane of height 2H.
The height that the ball reaches must be
only halfway up the second ramp. (H)
Examples
Estimate the kinetic energy and the
velocity required for a 70 kg pole vaulter
to pass over a bar 5.0 m high. Assume
the pole vaulter’s center of mass is
initially 0.90 m off the ground and
reaches its maximum height at the level
of the bar itself.
On board