Download Forms of Energy

Chapter 2: Conservation of Energy
Work – is said to be done when an object moves in the
direction of the force applied.
Work is a scalar quantity.
W = Fd W = Work (N.m or J)
F = Force (N)
d = Displacement (m)
Ex: A force of 20 N is applied to move a 5.0 kg object a
distance of 50 cm. Find the amount of work done.
W = Fd
= 20 N x 0.50 m
= 10 J
Ex: A 200 g object is lifted through a height of
2.0 m. Find the work done.
W = Fd
= (0.200 kg x 9.81 m/s2) x 2.0 m
= 3.9 J
Work done at an angle
W = Fcosθ.d
Ex: An object is pulled by a force of 100 N at an angle of 60˚
a distance of 2.0 m. Find the work done.
W = Fcosθ.d
= 100 N cos 60˚ 2.0 m
= 1.0 x 102 J
Work done by a changing force
Work done is the area under the graph.
Power - is the rate at which work is done.
P = W/t W = Work (J)
t = Time (s)
P = Power (J/s or Watt-W)
Ex: A 200 kg mass is lifted by a machine through a distance
of 2.0 m in 50 s. Find the power.
P = W/t
= (200 kg x 9.81 m/s2 x 2.0 m) / 50 s
= 78 W
Solve page 213-214, Q#(1-5) and (9-13)
Energy – the capacity to do work.
Energy Consumed = Work Done
Energy Stored = Work Done
Change in Object’s Energy = Work Done
ΔEm = Work = Fd
ΔEk = Work = Fd
ΔEp = Work = Fd
Forms of Energy
Mechanical Energy (Em) is the sum of the potential and
kinetic energy an object has.
Em = Ep + Ek = mgh + ½mv2 = Work done = Fd
a)Potential energy is the energy possessed by a
body by virtue of its position or condition. Conserved
energies can be recovered. Friction in non
recoverable force.
i. Gravitational Potential Energy (Ep) –energy due
to gravity (position above ground, only the vertical
distance).
-energy is taken with respect to a reference point.
Ep = mgh
Ep = Potential Energy (J)
m = mass (kg)
g = 9.81 m/s2
h = height (m)
ii. Elastic Potential Energy – Energy stored in an
ideal spring
Hooke’s Law - the extension of a spring is proportional
to the force applied.
F/x = k,
Ep = ½kx2
Ep = Potential Energy (J)
k = Spring Constant (N/m)
x = Extension/Compression(m)
b)Kinetic Energy – the energy possessed by a body
due to its movement
Ek = ½mv2
Ek = Kinetic Energy (J)
m = Mass (kg)
v = Velocity (m/s)
Ex: A spring (k = 200 N/m) is extended 5.0 cm. What is its
elastic potential energy?
Ep = ½kx2
= 1/2(200 N/m)(5.0 x 10-2 m)2
= 0.25 J
Example
A 60 kg person goes up an escalator 20 m long. The
escalator makes an angle 20o with the horizontal. Find the
work done.
d=20xsine 20(vertical height)
W=Fd=60x9.81x6.84=4.0x103 J
Law of Conservation of Energy
Energy cannot be created or destroyed but can be converted
from one form into another. Mass is another form of energy.
Energy Conversions
Mechanical energy is the sum of potential energy and kinetic
energy.
Mechanical Energy = Ek + Ep
Mechanical Energy = 1/2mv2 + mgh
Conversion between gravitational potential energy and
kinetic energy
100m
60m
0m
Find the potential, kinetic and total mechanical energy at
100m,60m, and 0m if an object of
5.0 kg is dropped from a height of 100 m.
An object of mass 2.0 kg is dropped from a height of 20
m. Find the speed of the object just as it reaches the
ground.
Example
A ball of mass 0.50 kg is thrown upward with
a speed of 30 m/s. How high will it rise?
Example
A 5.00 kg object is dropped from a height above the ground.
When the object is 4.00 m from the ground, it has a speed of
9.00m/s. The potential energy of the object is chosen to be zero
at ground level and the effects of air resistance
are ignored.
(a) What is the total mechanical energy of the falling object?
(b) What was the initial height of the object?
(c)What is the speed of the object just as it
reaches the ground?
Roller-coaster problems
1. A 1000 kg roller coaster, with its passengers, starts from rest at
point A on a frictionless track whose profile is shown in the
diagram.
(a) What is its maximum speed?
(b) With what speed does the roller coaster arrive at point E?
(c)What constant braking force would have to be applied to the
roller coaster at point E, to bring it to rest in a horizontal distance
of 5.0m?
2. A 250 kg roller-coaster car starts at position 1 and continues
through the entire track. The brakes are applied after the car passes
position 4.Assume that the effects of friction are negligible on the
roller-coaster.
(a)What is the roller-coaster’s speed at position 4?
(b)After the car passes position 4, the brakes stop the car in 3.03 s.
What is the magnitude of the average frictional force applied by
the brakes to stop the car?
3. A 30 kg girl goes down a slide at an amusement park, reaching
the bottom with a velocity of 2.5 m/s. The slide is 10.0 m long and
the top end is 4.0 m above the bottom end, measured vertically.
(a)What is her gravitational potential energy at the top of the slide,
relative to the bottom?
(b)What is her kinetic energy when she reaches the bottom?
(c)How much energy is lost due to friction?
(d)Calculate the average frictional force acting on her as she goes
down the slide?
Archery Problems
Work=Energy
Ex: An archer puts a 0.25 kg arrow to the bowstring. An
average force of 200 N is used to draw the string back 1.2
m.
a)Assuming no frictional loss, at what speed does the
arrow leave the bow?
W = Ek
Fd = 1/2mv2
200 N (1.2 m) = 1/2(0.25 kg)v2
= 44 m/s
b)If the arrow is shot straight up, how high
does it rise?
W = Ep
Fd = mgh
200 N (1.2 m) = 0.25 kg (9.81 m/s2)h
= 98 m
Conversion of Elastic Potential energy to other forms of
energy and vice versa
EP(e) = 1/2kx2
k = F/x
Ex: A linear spring can be compressed 10.0 cm by an
applied force of 5.0 N. A 4.5 kg crate of apples, moving at
2.0 m/s, collides with the spring. What will be the maximum
compression of the spring?
Ek  EP(e)
N/m
k = F/x = 5.0 N \(10.0 x 10-2 m) = 50
1/2mv2 = 1/2kx2
1/2(4.5 kg)(2.0 m/s)2 = 1/2(50 N/m)x2
x = 0.60 m
Ex: A ball bearing of mass 50 g is sitting on a vertical
spring(k=120 N/m). By how much must the spring be
compressed so that, when released, the ball rises to a
maximum height of 3.1 m above its release position?
Ep(e)  Ep(g)
1/2kx2 = mgh
1/2(120 N/m)x2 = (50 x 10-3 kg)(9.81 m/s2)(3.1 m)
x = 0.16 m
Ex: A 3.0 kg ball is dropped from a height of
0.80 m onto a vertical spring of force constant
1200 N/m. What is the maximum compression of the
spring?
Ep(g)  Ep(e)
mgh = 1/2kx2
3.0 kg(9.81 m/s2)(0.80 m) = 1/2(1200 N/m)x2
= 0.20 m
Pendulum Type Problems
Ex: A pendulum is dropped from a height of 0.25 m above
the equilibrium position. What is the speed of the pendulum
as it passes through the equilibrium position?
Ep + Ek= Ep + Ek
mgh + 0.0 = 0.0 + 1/2mv2
9.81 m/s2(0.25 m) = 1/2v2
= 2.2 m/s
Ex: A pendulum is 1.20 m long. What is the speed of the
pendulum bob as it passes through the equilibrium position if
it is pulled aside until it makes an angle of 25 .0° to the
vertical position?
Cos 25° = y/1.20
= 1.0875 m
Height = 1.20 – 1.0875
= 0.1125 m
Ep + Ek= Ep + Ek
mgh + 0.0 = 0.0 + 1/2mv2
9.81 m/s2 (0.1125 m) = 1/2v2
= 1.49 m/s
Elastic and Inelastic Collisions
An elastic collision is one in which both momentum and
kinetic energy are conserved.
Elastic collisions occur at the atomic level. In ordinary
collisions, kinetic energy is converted into sound, heat, and
light.
Inelastic collisions are collisions in which only momentum is
conserved.
Ex: A 5.0 kg metal ball at rest, is hit by a metal ball (1.0 kg)
moving at 4.0 m/s. The 5.0 kg ball moves forward at 1.0 m/s
and the 1.0 kg ball bounces back at 2.0 m/s. Is this an
elastic collision?
In an elastic collision, kinetic energy is conserved.
Ek1 + Ek2 = Ek1’ + Ek2’ if collision is elastic
1/2m1v12 + 1/2m2v22 =
1/2m1v1,2 + 1/2m2v2,2=
1/2(5.0 kg)(0 m/s)2 + 1/2(1.0 kg)(4.0 m/s)2=8.0 J
1/2(5.0 kg)(1.0 m/s)2 + 1/2(1.0 kg)(2.0 m/s)2=4.5 J
Kinetic energy before collision(8.0 J) is not equal to
to kinetic energy after collision(4.5J).
Collision is inelastic due to loss of energy in the form of heat,
light, and sound.
Example
A 6.0 kg ball of putty moving at 10.0 m/s runs
head on into another 6.0 kg ball of putty at rest They stick
together and move ahead at 5.0 m/s. Is this an elastic
collision? Show calculations
Ballistic pendulum problems
The ballistic pendulum is a device used to
measure the speed of a projectile, such as
a bullet. The projectile is fired into a large
block(of wood or other material) which
is suspended like a pendulum. As a result
of the collision, the pendulum and projectile together swing up
to a maximum height.
By using the conservation laws, the
speed of the projectile can be found.
A bullet’s speed may be determined by firing
it into a sandbag pendulum and measuring
the vertical height to which it rises. The bullet stays
in the sandbag.
(a)What is the change in the gravitational potential
energy of the sandbag and bullet during the
swing?
(b)What is the velocity of the sandbag-bullet
combination at the start of the swing?
(c) What was the original velocity of the bullet?
(d) Is the collision between the bullet and the
sandbag elastic or inelastic? Justify your answer.
Assignment
1.A chronograph is a sophisticated device commonly
used by technicians to find the speed of a bullet fired from a gun.
Before this technology was developed, the speed of a bullet was
determined
using different methods. A simplification of one method of
determining the speed of a bullet requires
shooting the bullet into the end of a stationary wooden block
suspended by two long strings. This method is illustrated below.
In this case, a bullet(m=10.89 g) is shot from a handgun and
becomes embedded in a 2.20 kg
block of wood. The block of wood swings upward
a vertical distance of 69.1 cm. Using the information provided, the
Law of Conservation of Energy, and the Law of Conservation of
Momentum, calculate the speed of the bullet just before the impact
with the block.
2. A 56.7-g steel ball is shot from a spring
gun into a 203-g pendulum of a ballistic
pendulum. The pendulum swings aside
and is kept at its maximum height by
mechanical means. The change in height
is measured to be 13.1 cm. What is the
velocity of the steel ball when fired from
the spring gun?
Review
Momentum(p)=mv
Change in momentum
Impulse
Law of conservation of momentum