Introduction
In this lesson, we will pull all the knowledge about mass, motion, and forces together in a concept that, while very abstract, is at the essence of science and nature – energy, In this lesson, we will concentrate on a form of energy that characterizes mechanical properties, We will define mechanical energy and its two forms: kinetic and potential, We will wrap up our lesson with a study of the law of conservation of energy in a mechanical system.
Work
Work is a quantity that measures the effect of moving an object when a force is applied. When a constant force (no variation in time or space) is exerted but the object does not move, as in the case when you would push against a wall, we say that no work has been done.
To summarize in an expression the definition of work, we have:
W = (F · cos α) · d
The unit for work is Newtons times meters (N · m) and is known as a Joule (in honor of James Joule 1818–1889).Work is a scalar quantity although it is the result of two vectors—force and displacement. In other words: The most general expression of the work done is the dot product of force and displacement; and the result of a dot product, although starting with two vectors, is a scalar.
Work
Work is proportional to the product of the distance over which an object is moved, the displacement, and the constant force along that direction.
Example 1
Your car needs to be moved for a short distance, but it has no gas in the tank, so it will need to be pushed. You apply a force of 1,000 N and move it for about 180 em. What is the work done?
Solution 1
First, convert the units into SI (remember, 1 m = 100cm). Next, construct a diagram of the problem, an then proceed to find the answer.
d = 180 cm = 180 cm · 1 m/l00 cm = 1.8 m
F = 1,000 N
W = ?
As Figure 7.1 shows, the force and the displacement are in the same direction, so then the angle between F and d is zero. Since cos 0° = 1, the expression of the work becomes:
W = (F · cos α) · d = (F · cos 0°) · d = F · d
W = 1,000 N · 1.8 m = 1,800 J = 1.8 kJ
Example 2
Now imagine we are using the same force to push a big pile of boxes around, but the force is exerted so that the angle with the horizontal is 60°. Find out the work performed on the boxes by the force of 1,000 N in a distance of 180 cm.
Solution 2
First, we convert the units to SI (1 m = 100 cm). Next, construct a diagram of the problem, and then proceed to find the answer.
As Figure 7.2 shows, the force and the displacement are not in the same direction, but they make an angle of 60°. Because cos 60° = 0.5, the expression of the work becomes:
W = (F· cos α) · d = (F · cos 60°) · d = F · d/2
W = 1,000 N · 1.8 m/2 = 900 J = 0.9 kJ
Note
If you are to consider a vector force, the equation for work shows that only the component of the force along displacement is important in measuring the work done on an object.
Power
Any amount of work done can be performed in different intervals of time, and that will have an effect on how we perceive the total work done. How can we take this into account? Imagine you are taking the stairs up to the second level of a building and you take your time to go up. You are in no hurry, and if you run up the stairs, you will experience more tiredness. So, the shorter time you take to go up the stairs, the more your muscles work. So, how can you characterize this situation? We must define a new quantity: power.
There are other units that you might encounter for power, but the one referred to most often is horsepower (hp).
1 horsepower = 550 feet · pounds/seconds = 745.7 watts
Another way to determine power is through the expression of the work: For a constant force acting along the direction of the displacement, the previous definition becomes:
where the time rate of change of displacement defines the average speed:
and vo is the initial speed, where v is the final speed.
Power
Power is proportional to the work performed on an object and inversely proportional to the time taken to perform the work:
P stands for power and Δt for the interval of time. The unit for power is called a watt and is equal to joules per second (W = J/s), in honor of James Watt, 1736-1819. One watt is the power needed to perform one joule of work in a time of 1 second.
Example
Consider a car of 1,500 kg starting from the stop light and accelerating for a distance of 25 m with an acceleration of 3.5 m/s2. Calculate the average power generated by the force producing the acceleration.
Solution
First, check to see that the quantities are expressed in 51. Next, list the known data.
m = 1,500 kg
a = 3.5 m/s2
d = 25 m
p = ?
We can use the second expression of the power because we can determine the accelerating force and the average speed using Newton's second law for the first one, and we can use the equations of motion for the second quantity.
F = m · a = 1,500kg · 3.5m/s2 = 5.2 · 103 N
Now, to find the average speed, although we are given displacement, we do not have the time, so we have to use the equations of motion (Lesson 3) to make the connection between average speed, acceleration, and displacement.
By raising to the second power, we have:
And we can also write the displacement as:
And then the expression for the average speed becomes:
And the average speed is:
We can now calculate the power done by the force of acceleration to be:
Kinetic Energy
One result of doing work on objects, as we have seen in the examples before, might be motion, or it might be change in position. So, how are we characterizing the difference between these two situations— between motion and change in position? When an object is accelerated or decelerated because of work performed on it, we will measure the energy of motion known as the kinetic energy. Consider an accelerating force (net force) that changes the way an object moves. According to Newton's second law:
F = m · a
The force is acting over a distance d and along the displacement (α = 0°), then the work is:
W = F · d = m · a · d
From Lesson 3, we also know that:
v2 – vo2 = 2 · a · d
Then multiplying both sides with the mass m:
m · (v2 – vo2) = m · 2 · a · d
And then dividing by 2:
Or:
And again:
With the definition of kinetic energy, the expression of the work can also be written as:
KE – KEoz = W
ΔKE = W
This formula is called the work-energy theorem and expresses the fact that when a net external force does work on an object, the work is equal to the change in kinetic energy. This equation also defines the unit for kinetic energy, which is a joule, because the two quantities—work and the change in kinetic energy—are not only proportional but equal. Note that the work-energy theorem does not apply to individual forces but only to net external forces.
Kinetic Energy
The kinetic energy of an object of mass m and speed v is defined by:
Gravitational Potential Energy
Consider a body being launched up an inclined plane, and suppose the friction is so small that we can neglect it altogether. If one calculates the work done to raise the object on the inclined plane, one has to consider only the work done by the force along the direction of the plane. That is — W · sin α where α is the angle the inclined plane makes with the horizontal, and W is the weight of the object (the minus shows that although they are on the same axis, the displacement and the force are in opposite directions). To avoid confusion, we will use the expression of the weight with respect to mass: W = m · g. Therefore, the force performing work is (see Figure 7.4):
– m · g · sin α
And the displacement on the plane is h/sin α. Hence, the work done by the weight on the object is:
W = F · d = – m · g · sin α · (h/sin α) = – m · g · h
We can also write this expression as:
W = 0 – m · g · h
Or:
W = –(m · g · h – 0) = –(m · g · h – m · g · 0)
This helps us define a new form of energy that is not related to motion, or to speed, but to the position from the ground—height (h).
We can now rewrite the expression of the work as the negative of the change in potential energy:
W = – (m · g · h – m · g · ho)
W = –(PE – PEo)
W = –ΔPE
In Figure 7.4, you can see that the same result can be obtained if you find the work done, by weight, on raising the object directly on the vertical:
W = m · g · h
In this case, the work performed is independent of the chosen path. This is called a conservative force. A conservative force is also defined as the force that determines zero work when it acts on a closed path. An example of a nonconservative force is friction.
Gravitational Potential Energy
The gravitational potential energy of an object of mass m is defined by:
PE = m · g · h
where h is the distance from the surface of the earth that is considered zero potential energy.
Example
The first swing of a pendulum raises the ball at the end of a string 30.0 cm long at an angle of 60.0° with the vertical. Find the potential energy at the top of the swing if the mass of the ball is 0.25 kg.
Solution
The data given in the problem is in SI units. List the known data and draw a diagram of the problem (see Figure 7.5).
l = 30.0 cm
α = 60.0°
m = 0.25 kg
PE = ?
Because the problem does not specify, the choice of the ground level is ours. So, we will consider 0 level as the position where the ball starts moving. At that level, the potential energy is also 0: PE0 = 0 J. The final potential energy will be given by the distance with respect to this ground level.
PE = m · g · h
From the diagram in Figure 7.5, we can see that the height of the ball with respect to the ground is l – x. So, we need to find x first and then final potential energy (PE).
x = l · cos α
h = l – l · cos α = l · (1 – cos · α)
PE = m · g · l · (1 – cos α) = 37J
Conservation of Mechanical Energy
The forms of energy discussed previously give the mechanical energy of an object and encompass both the energy of motion and the potential energy:
E = KE + PE
And considering the work-energy theorem discussed previously, we have:
W = ΔKE
W = –ΔPE
Then:
ΔKE = –ΔPE
KE – KEo = –(PE – PEo)
And rearranging the terms so that each side contains only initial or final states, we write:
KEo + PEo = KE + PE
In other words, the initial and final mechanical energies are equal as long as there are no nonconservative forces in the process or if the work they perform is zero. This establishes the principle of conservation of mechanical energy.
Example
Consider an object falling freely from a height of 22 m and stopping on a cliff at a distance of 10 m from the starting point. Find out the speed of the object when it lands on the cliff.
Solution
All data is given in the SI units. Therefore, list the data and draw a diagram for this example.
ho = 22 m
h = 22 m – 10 m 12 m
Because the object moved 100 m away from where it was released, it must be closer to the earth.
v = ?
And because this is a conservative process, the energy is conserved, and the mechanical energy at the top and at the time the object hits the cliff are equal. And finally, because the object is falling freely, it means that it started with zero speed:
v0 = 0 m/s
m · g · h0 + m · v02/2 = m · g ·: h + m · v2/2
m · g · h0 = m · g · h + m · v2
The mass can be canceled out through the entire expression:
g · h0 = g · h + v2/2
And we need to solve for v:
v2/2 = g · h0 – g · h
v2 = 2 · g · (h0 – h)
v2 = 2 · g · (22 – 12)
Conservation of Mechanical Energy
The total mechanical energy of an object remains constant provided that the net work done by the external nonconservative forces is zero.
Einitial = Efinal
Practice problems of this concept can be found at: Mechanical Energy Practice Questions
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