energy and energy conservation. energy two types of energy: 1. kinetic energy (ke) - energy of an...
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Energy and Energy Conservation
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EnergyEnergy
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.
Energy is the ability to do work.
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Kinetic EnergyKinetic Energy
Kinetic energy depends on the speed and the mass of the object.
KE = ½ mv²
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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
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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
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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?
Wnet = ½mvf ² - ½mvi²
Vi = 0 m/s Vf = 4.0 m/s
∑Fd = ½mvf ²
∑F = (½mvf ²)/d = [(½)(50kg)(4.0m/s)²]/ 15 m
∑F ≈ 27 N
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Potential EnergyPotential 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
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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
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Potential EnergyPotential 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.
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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?
PEelastic = ½ kx²
x = 0.55 m – 0.15 m = 0.40 m
= ½ (2.3 N/m)(0.40 m)²
PEelastic = 0.18 J
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Conservation of Mechanical Energy
Law of conservation of energy: Energy is neither created or destroyed. It simply changes form.
Mechanical energy (ME) is the sum of kinetic and all forms of potential energy.
ME = KE +∑PE
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h
100 % PE
0 % KE
50 % PE
50 % KE
0 % PE
100 % KE
Total mechanical energy remains constant in the absence of friction.
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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.
hi = 3.00 mm = 25.0 kg
hf = 0 m
vi = 0 m/s
vf = ? m/s
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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
hi = 3.00 mm = 25.0 kg
hf = 0 m
vi = 0 m/s
vf = ? m/s
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Mechanical Energy in the presence of friction
In the presence of friction, measured energy values at start and end points will differ.
f Fapp
KE
KE
KE
KETotal energy, however, will remain conserved.
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Work
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WorkWork
W=∑Fd(cos θ)
Any force that causes a displacement on an object does work (W) on that object.
ΣF
d
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Work is done only when components of a force are parallel to a displacement.
W=∑Fd(cos θ)
WorkWork
F
θ
d
ΣF
Work is expressed in Newton • meters (N•m) = Joules (J)
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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?
W=∑Fd(cos θ)
50.0 N
30.0°
d
ΣF
= (50.0 N x 3.0 m)(cos 30.0°)
W = 130 J
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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 %
EfficiencyEfficiency
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Sample ProblemSample 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 %
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Momentum and
Impulse
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Momentum and ImpulseMomentum 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
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Objects with a high momentum can have a greater mass, velocity, or both!
1
2
ν1 = ν2
m1 > m2
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Falling Object 1 Falling Object 2
ν1 > ν2
m1 = m2
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A change in momentum takes force and time.
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
FΔt = Δp
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Wall exerts an impulse on the moving ball, thereby causing a change in momentum
p1
p2
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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)
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m = 1400 kg
νi = -15 m/s
FΔt = mνf – mνi
Δt = 0.30 s
νf = 0 m/s
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