The elevator in the world’s tallest building in Taiwan will
carry a passenger up to the outdoor observations deck on
the 89th floor with an acceleration of 0.629 m/s2 . As the
elevator goes downwards, the acceleration of the elevator is
0.232m/s2. Philip, who weighs 150. lbs, stands on a
bathroom scale in this elevator.
1. What would the scale reading be when the elevator is
at rest?
2. What would the scale reading be when he accelerates
up to the observation deck on the 89th floor?
3. What would the scale reading be when the elevator is
accelerating from the observation deck back to the
ground?
4. What would the scale reading be when the elevator is
travelling at a constant speed?
Warm-Up – (8 points)
During the investigation of a traffic accident,
police find skid marks along the snowy road that
are 125 m long. They determine the coefficient
of friction between the car’s tires and the
roadway to be 0.28 for the prevailing conditions.
Estimate the speed of the car when the brakes
were applied. If the speed limit was 55mph,
should the driver be ticketed for speeding,
ticketed for careless driving (>20 mph over), or
not ticketed at all because the speed was safe for
the snowy road
conditions?
Bellwork: In one situation, the ball falls off the top of
the platform to the floor. In the other situation, the
ball rolls from the top of the platform along the
staircase-like pathway to the floor. Indicate what
type of forces are doing work upon the ball and
whether the energy of the ball is conserved and
explain why. Finally, fill in the blanks for the 2-kg ball.
Bellwork Solution
• The only force doing work is gravity. Since it is an
internal or conservative force, the total mechanical
energy is conserved. Thus, the 100 J of original
mechanical energy is present at each position. So the
KE for A is 50 J.
• The PE at the same stairstep is 50 J (C) and thus the KE
is also 50 J (D).
• The PE at zero height is 0 J (F and I). And so the kinetic
energy at the bottom of the hill is 100 J (G and J).
• Using the equation KE = 0.5*m*v2, the velocity can be
determined to be 7.07 m/s for B and E and 10 m/s for
H and K.
Work Done by a Variable Force
Potential Energy, Spring Force and Hooke’s Law
Objectives:
1. Define and
understand
potential energy
2. Calculate the work
done by a spring
force.
3. Study the WorkEnergy Theorem
and apply it in
solving problems.
Potential Energy: Energy that is stored
in an object as a result of its position
or condition.
• Chemical – energy stored in bonds between
atoms
• Gravitational – energy stored as a result of an
object’s vertical position, or height
• Elastic – energy stored in material as a result
of stretching or compression
Gravitational Potential Energy:
depends on the mass and
the height of the object.
• Potential energy = mass x gravity x height
PEgrav  m  g  h
• zero height position must first be assigned.
• Typically, the ground is considered to be a
position of zero height.
Example 1: A cart is loaded with a brick and pulled at
constant speed along an inclined plane to the height of
the top of a chair. If the mass of the loaded cart is 3.0
kg and the height of the seat top is 0.45 meters, then
what is the potential energy of the loaded cart at the
height of the seat-top?
Solution to Example 1
• Given:
– m = 3 kg
– g = 9.8 m/s2
– h = 0.45 m
• PE = m*g*h
• PE = (3 kg ) * (9.8 m/s/s) * (0.45 m)
• PE = 13.2 J
Example 2: If a force of 14.7 N is used to drag
the loaded cart (from previous question) along
the incline for a distance of 0.90 meters, then
how much work is done on the loaded cart?
Solution to Example 2
Given: F= 14.7 N
d= 0.9 m
q = 0o
• W = F * d * cos Theta
• W = 14.7 N * 0.9 m * cos (0 degrees)
• W = 13.2 J
Elastic Potential Energy: Spring Forces
• Springs are a device which can store elastic
potential energy due to either compression or
stretching.
• A force is required to compress a spring; the
more compression there is, the more force which
is required to compress it further.
• For certain springs, the amount of force is
directly proportional to the amount of stretch or
compression (x); the constant of proportionality
is known as the spring constant (k).
Calculating the Spring Force
Hooke's Law
Fspring = k * x
where k is the spring constant
and x is the amount of stretch or compression.
• If a spring is not stretched or compressed, then
there is no elastic potential energy stored in it.
• The spring is said to be at its equilibrium position.
The equilibrium position is the position that the
spring naturally assumes when there is no force
applied to it.
Calculating Elastic Potential Energy
The amount of elastic potential energy is
related to the amount of stretch (or
compression) and the spring constant. The
equation is
The work done by an external force in stretching
or compressing a spring (to overcome the spring
force) is calculated by:
1 2
W  kx
2
Where k is the spring constant and x is the stretch
or compression distance from the equilibrium
position
Example 3: A spring of spring constant
20 N/m is to be compressed by 0.10 m.
• A) what is the maximum force required?
• B) What is the work required?
Solution to Example 3
• Given:
– k = 20 N/m
– x = -0.10 m
• For a compression, F = -kx
• W = ½ kx2
A) F = -(20 N/m)(-0.10 m) = 2.0 N
B) W = ½ (20 N/m)(-0.10 m)2 = 0.10 J
Round the Clock Spring Problems
• Draw a circle, or “clock” on your paper and
divide it into 4ths.
• Set 4 appointments with your classmates; one
at 12 o’clock, one at 3 o’clock, one at 6 o’clock,
and one at 9 o’clock.
• Solve the following problems at your
“appointment”.
Round the Clock Spring Problems
3 o’clock: When a 13.2-kg mass is placed on top of a
vertical spring, the spring compresses 5.93 cm. Find the
force constant of the spring.
6 o’clock: If a spring has a spring constant of 400 N/m, how
much work is required to compress the spring 25.0 cm
from its undisturbed position?
9 o’clock: A compressed spring that obeys Hooke's law has
a potential energy of 18 J. If the spring constant of the
spring is 400 N/m, find the distance by which the spring
is compressed.
12 o’clock: To compress spring 1 by 0.20 m takes 150 J of
work. Stretching spring 2 by 0.30 m requires 210 J of
work. Which spring is stiffer?
Kinetic Energy: energy of motion
• An object which has motion - whether it be
vertical or horizontal motion - has kinetic energy.
– Vibrational : the energy due to vibrational motion
– Rotational: the energy due to rotational motion
– Translational: the energy due to motion from one
location to another
• The amount of translational kinetic energy an
object has depends upon two variables: the mass
(m) of the object and the speed (v) of the object.
• The following equation is used to represent the
kinetic energy (KE) of an object.
3 o’clock: Determine the kinetic energy of a 625-kg
roller coaster car that is moving with a speed
of 18.3 m/s.
6 o’clock: If the roller coaster car in the above
problem were moving with twice the speed,
then what would be its new kinetic energy?
9 o’clock: 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?
12 o’clock: 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.
The Work-Energy Theorem
• The net work done on an object is equal to its
change in kinetic energy.
W = K – K0 = DK
• Work is a measure of energy transfer.
• Energy is the capacity to do work.
• The net force acting on an object causes the
object to accelerate, changing its velocity:
Combining Kinematic Equations and
Newton’s Second Law
W  F  d  ma  d
v
W  m
d

2
x


1 2 1 2
W  mv  mv0
2
2
2
2
 v0 

Example 4: The opposing kinetic
friction force between a 60.0 kg object
and a horizontal surface is 50.0 N. If
the initial speed of the object is 25.0
m/s, what distance will it slide before
coming to a stop?
Fnorm
Ffric
Fgrav
Solution to Example 4
• Given:
– m = 60.0 kg
– v0 = 25.0 m/s
Ffric = 50.0 N
v = 0 m/s
W = K-K0
W = (Fcosq)d = (Ffriccos180)d
W = K-K0= ½ mv2 – ½ mv02
(50.0 N)(cos180)(d ) = ½(60.0kg)(0m/s)2-1/2(60.0 kg)(25.0 m/s)2
d = 3.75 x 102 m
Law of Conservation of Energy:
First Law of Thermodynamics
• If the working forces (the forces doing nonzero
work) in a system are conservative, the total
mechanical energy of the system is conserved.
• Conservative Forces: independent of the path but
dependent on only the initial and final locations.
• Nonconservative Forces: the work done by or
against it depends on the path (friction, for
example)
E = KE + PE
KE0+PE0 = KE + PE
½ mv2 + mgh = ½ mv02 + mgh0
Independent Practice
• Conservation of Energy – Work with your Best
Partner to solve the problems related to total
Mechanical Energy.
E = KE + PE
• Homework: “Work Done by a Variable Force”