Transcript File energy,conservation of energy,work,momentum

Energy and Energy
Conservation
Energy
Energy is the ability to do work.
Two types of Energy:
1. Kinetic Energy (KE) - energy of an
object due to its motion
2. Potential Energy (PE) - energy
associated with an object due to the
position of the object.
Kinetic Energy
Kinetic energy depends on the speed
and the mass of the object.
KE = ½ mv²
Sample Problem
What is the kinetic energy of a 0.15 kg
baseball moving at a speed of 38.8 m/s?
KE = ½ mv²
KE = (½)(0.15 kg)(38.8m/s)²
KE = 113 J
Work-Kinetic Energy Theorem
The net work done on an object is
equal to the change in kinetic energy
of an object.
Wnet = ΔKE
Wnet = ½mvf ² - ½mvi²
Sample Problem
A 50 kg sled is being pulled horizontally
across an icy surface. After being pulled
15 m starting from rest, it’s speed is 4.0
m/s. What is the net force acting on the
sled?
Vi = 0 m/s
Wnet = ½mvf ² - ½mvi²
∑Fd = ½mvf ²
Vf = 4.0 m/s
∑F ≈ 27 N
∑F = (½mvf ²)/d = [(½)(50kg)(4.0m/s)²]/ 15 m
Potential Energy
Potential energy (PE) is often referred to
as stored energy.
Gravitational potential energy (PEg)
depends on the height (h) of the object
relative to the ground.
PEg= mgh
Sample Problem
What is the gravitational potential energy
of a 0.25 kg water balloon at a height of
12.0 m?
PEg= mgh
PEg= (0.25 kg)(9.81 m/s²)(12.0 m)
PEg= 29.4 J
Potential Energy
Elastic potential energy (PEelastic) is the
potential energy in a stretched or
compressed elastic object.
PEelastic = ½ kx²
“X” is referred to as the distance of the
spring compressed or stretched.
“K” is the spring constant and is
expressed in N/m.
Sample Problem
Calculate the elastic potential energy of
a block spring, with a spring constant of
2.3 N/m, that has a compressed length of
0.15 m and a maximum stretch length of
0.55 m?
x = 0.55 m – 0.15 m = 0.40 m
PEelastic = ½ kx² = ½ (2.3 N/m)(0.40 m)²
PEelastic = 0.18 J
Conservation of Mechanical Energy
Mechanical energy (ME) is the sum of
kinetic and all forms of potential energy.
ME = KE +∑PE
Law of conservation of energy: Energy is
neither created or destroyed. It simply
changes form.
Total mechanical energy remains
constant in the absence of friction.
100 % PE
0 % KE
50 % PE
h
50 % KE
0 % PE
100 % KE
Sample Problem
Starting from rest, a child zooms down a
frictionless slide from an initial height of 3.00
m. What is the child’s speed at the bottom of
the slide? The child’s mass is 25.0 kg.
m = 25.0 kg
hi = 3.00 m
vi = 0 m/s
hf = 0 m
vf = ? m/s
m = 25.0 kg
hi = 3.00 m
vi = 0 m/s
hf = 0 m
vf = ? m/s
½ mvi² + mghi = ½ mvf² +mghf
(25.0 kg) (9.81 m/s²) (3.00 m) = (½)(25.0 kg) (Vf)²
736 J / (12.5 kg) = Vf ²
Vf ² = 58.9 m²/s²
Vf = 7.67 m/s
Mechanical Energy in the presence of friction
In the presence of friction, measured
energy values at start and end points will
differ.
KE
f
KE
Fapp
KE
Total energy, however, will remain
conserved.
KE
Work
Work
Any force that causes a displacement on
an object does work (W) on that object.
ΣF
d
W=∑Fd(cos θ)
Work
Work is done only when components of
a force are parallel to a displacement.
F
θ
ΣF
d
W=∑Fd(cos θ)
Work is expressed in Newton • meters
(N•m) = Joules (J)
Sample Problem
How much work is done on a box pulled
3.0 m by a force of 50.0 N at an angle of
30.0° above the horizontal?
50.0 N
30.0°
ΣF
d
W=∑Fd(cos θ)= (50.0 N x 3.0 m)(cos 30.0°)
W = 130 J
Efficiency
Efficiency is a measure of how much of
the work put into a machine is changed
into useful work by the machine.
Efficiency = (Wout/Win) x 100 %
Sample Problem
A man expends 200 J of work to
move a box up an inclined plane.
The amount of work produced is 40
J. What is the efficiency of the
inclined plane?
Efficiency = (Wout/Win) x 100 %
= (40 J/ 200 J) x 100 = 20 %
Momentum
and
Impulse
Momentum and Impulse
Momentum is a measure on how difficult
it is to stop a moving object.
Momentum is a vector quantity.
p = mν
Measured in kg • m/s
Objects with a high momentum can have
a greater mass, velocity, or both!
1
2
ν1 = ν 2
m1 > m2
Falling Object 1
Falling Object 2
ν1 > ν 2
m1 = m2
A change in momentum takes force and
time.
FΔt = Δp
This product of force and the time over
which it acts on an object is known as an
impulse (FΔt).
FΔt = mvf – mvi
Impulse-Momentum
Theorem
p1
p2
Wall exerts an
impulse on the
moving ball,
thereby
causing a
change in
momentum
Sample Problem
A 1400 kg car moving westward with a
velocity of 15 m/s collides with a pole
and is brought to rest in 0.30 s. What is
the magnitude of the force exerted on
the car during the collision? (Pg. 211)
FΔt = mνf – mνi
F (0.30 s) = (1400 kg)(0 m/s) – (1400 kg)(- 15 m/s)
F (0.30 s) = 21,000 kg • m/s
F = 7.0 x 104 N to the East
Δt = 0.30 s
m = 1400 kg
νf = 0 m/s
νi = -15 m/s