Download SPH 3U – Unit ~ Energy, Work, and Power

SPH 3U – Unit ~ Energy, Work, and Power
Lesson 4 & 5: Conservation of Energy
The word ‘conservation’ in physics implies that something is to remain constant. So, that
means that ‘Conservation of Energy’ implies that Energy is to remain constant. The type of
energy can change (ie. Gravitational potential energy becomes kinetic energy) but, the total
amount of energy in the system will not change. This also means that the energy can be
transferred from one object to the next within a system.
Conservative Forces
A conservative force is one that does work on an object, but the work done is
independent of the path taken. For example, the force of gravity is a conservative force because
no matter what direction you lift an object, the amount of work done is still the same.
Non-Conservative Forces
A non-conservative force is one that does work on an object, but the amount of work
done changes depending on the path taken. For example, the amount of work done on a textbook
if you push it straight across a table is less than the amount of work done if you were to push it
back and forth as you move it down the table. This means that the force of friction is a nonconservative force.
On page 240 of your textbook there is a table of the conservative and non-conservative
forces.
What does this mean?
Work done by conservative forces result in energy changes that do not rely on the path
taken, therefore they are reversible. However, work done by non-conservative forces are path
dependent and therefore, are not reversible.
Mechanical Energy
Mechanical Energy can be described as the sum of Gravitational Potential Energy and
Kinetic Energy. From this, we can get the Law of Conservation of Energy, which states, the
total mechanical energy of a system always remains constant if work is done by conservative
forces.
Equation
ET = Eg + Ek
Where ET is the total mechanical energy of the system, Eg is the gravitational potential
energy of the system, and Ek is the kinetic energy of the system. Because this is Energy, it is
measured in Joules.
Example 1
A textbook of mass 2.00 kg is lifted at a constant velocity to a height of 3.00 m and then
is dropped back on the ground. Find the work done on the textbook, the gravitational potential
energy of the car at its highest point, and the velocity of the textbook just before it strikes the
ground. (Neglect friction of air)
m = 2.00 kg
∆h = 3.00 m
g = 9.81 m/s2
If the textbook is lifted at a constant velocity then the applied force and the force of
gravity must be equal.
W = Fd
W = mgd
W = (2.00 kg)(9.81 m/s2)(3.00 m)
W = 58.9 J
Eg = mg∆h
Eg = (2.00 kg)( 9.81 m/s2)(3.00 m)
Eg = 58.9 J
Since we know that gravity is a conservative force, ET = Eg + Ek. But, at its highest point,
the textbook is not moving and has Gravitational Potential Energy, but not Kinetic Energy. So,
at the top, Ek = 0 J. Therefore,
ET = Eg(top) + Ek(top)
ET = 58.9 J + 0 J
ET = 58.9 J
We also know that since gravity is a conservative force, the total energy in the system is
constant and ET = Eg(bottom) + Ek(bottom). Also, as the object is falling there is no gravitational
potential energy.
ET = Eg(bottom) + Ek(bottom)
58.9 J = 0 J + ½ mv2
58.9 J = ½ (2.00 kg)v2
v = 7.67 m/s
Example 2
You throw a ball directly upward, giving it an initial velocity of 10.0 m/s. Neglecting
friction, what would be the maximum height of the ball?
v1 = 10.0 m/s
v2 = 0 m/s
You know that when you throw the ball there is Kinetic Energy, but no Gravitational
Potential Energy. You also know that energy is conserved because gravity is a conservative
force.
ET = Eg(bottom) + Ek(bottom)
ET = 0 J + ½ mv12
ET = ½ mv12
Since energy is conserved,
ET = Eg(top) + Ek(top)
But at the top of its arc, the ball stops moving and therefore has to Kinetic Energy.
ET = mg∆h + 0 J
If you equate the two equations, you can solve for the height.
½ mv12 = mg∆h
½ (10.0 m/s)2 = (9.81 m/s2)∆h
∆h = 5.10 m
Therefore, the ball reaches a maximum height of 5.10 m.
Example 3
A crane is lifting a 120 kg beam at a height of 96.0 m above the ground. The cord
holding the beam snaps and it falls to the ground. Neglecting friction of air, at what speed would
the beam hit the ground?
m = 120 kg
∆h = 96.0 m
We know that gravity is a conservative force, so, energy will be conserved. We also
know that while the beam is suspended there is no Kinetic Energy, and as the beam is falling
there is no Gravitational Potential Energy.
ET = Eg(top) + Ek(top)
ET = mg∆h + 0 J
ET = (120 kg)(9.81 m/s2)(96.0 m)
ET = 1.13 x 105 J
ET = Eg(bottom) + Ek(bottom)
ET = 0 J + ½ mv2
ET = ½ (120 kg)v2
1.13 x 105 J = ½ (120 kg)v2
v = 43.4 m/s
Therefore the beam hits the ground at a speed of 43.4 m/s.
Review
If a force is conservative, then any work done by that force in a system is conserved. This
means that energy remains constant in the system, it is not lost or gained. The energy can,
however change forms (eg. Gravitational Potential Energy and transform to Kinetic Energy). If
work is done by a non-conservative force, then energy is not conserved (eg. Force of Friction).
The Law of Conservation of Energy
Energy is neither created nor destroyed. It simply changes form or is transferred from
one body to another. The total energy of an isolated system, including all forms of energy
always remains constant.
Example 1
A boulder is sitting on the edge of a cliff. It goes over the edge and falls to the ground.
The speed of the boulder is measured to be 45 m/s just before the boulder hits the ground.
Neglecting air friction, how high above the ground is the boulder before it falls off the cliff?
v = 45 m/s
g = 9.81 m/s2
Energy is conserved so,
Eg(top) + Ek(top) = Eg(bottom) + Ek(bottom)
mgh + 0 = 0 + ½ mv2
2gh = v2
2(9.81 m/s2)h = (45 m/s)2
h = 103 m
Therefore, the boulder is at an initial height of 103 m.
Example 2
An amusement park has a slide for which participants are given a cloth sack to sit on.
The top of the slide is 6.0 m high. Determine the speed attained at the bottom of the slide by a
30 kg child. Assume that the child starts from rest and friction is ignored.
h = 6.0 m
m = 30 kg
g = 9.81 m/s2
mgh + 0 = 0 + ½ mv2
2(9.81 m/s2)(6.0 m) = v2
v = 10.8 m/s
Therefore, the child reaches a final speed of 10.8 m/s at the bottom of the slide.