So you've made it to college and are moving to your new dorm! Congratulations! There is one downside to your excitement, however: boxes. Yep, the bane of every moving victim since the dawn of time itself. After staring at your boxes for a really long time, hoping they'll move themselves, you start with the box at rest and try sliding it on the floor. Next, you push the box; your friend helps by pulling the box, and yet the box barely moves. Even so, you and your friend still exert force on the box.
Let's assume that by some stroke of luck, the forces you and your friend exert on the box are constant. We also assume the forces you both exert are more significant than the net forces resisting the movement of the box. Then, in your darkest moment, when all hope seems to be lost, your physics teacher comes in and helps you move the box. As your hero leaves, she says, "Be careful not to let the kinetic energy of the box overwhelm you," in a very cryptic manner, as all physics teachers do.
But what is kinetic energy, and how does it relate to work? This is explained by the Work-Energy Principle/Theorem. This article will explore everything you need to know about the Work-Energy Principle for the AP Physics 1 Test. We will not only prove the Work-Energy Principle, but we will also do applicable examples, give an explanation of the Work-Energy Theorem, and define it with a neat formula.
What is Work? What is Energy?
Energy is the capacity to do work. Work is the transfer of energy into or out of a system. Now right about now, you're probably like, "Wait, a second...whoa, whoa whoa. Energy is work, and work is energy; how does that work?" This interrelation between work and energy allows the Work-Energy Principle to occur. But, more on that to come. Try to see work and energy as synonyms more than anything; doing that will help you later in the article.
Work and energy are scalar quantities; this means that work and energy have no direction associated with them: only a magnitude. However, they can have positive and negative values.
Work-Energy Principle Area of Application
Work \(W\) is done on an object when we apply a force \(F\) on it, as long as the force's component is in the direction of displacement \(d\). We can write this as:
Also, recall that work is the transfer of energy into or out of a system; this means that work equals a system's change in energy. Therefore, the equation above can be appropriately rewritten as
An outside force exerting on a system causes mechanical energy, the sum of a system's kinetic and potential energies, to be transferred into or out of the said system: so long as the force is parallel to the system's displacement. This energy-transferring process is equivalent to the work done.
Work is therefore equal to the area under a Force vs. Displacement Graph.
Fig. 2 - The work done is equal to the area under a Force vs. Displacement curve
There is no work done or \(W=0\) when:
- No displacement or force is acting on the object.
- The force is perpendicular to the motion (\(\cos{\theta} = 0\)).
Work done is negative when the force's component is in the opposite direction of the displacement. This is the case where \(\cos{\theta} = -1\) and \(\theta = 180^{\circ}\mathrm{.}\)Work has units of \(\mathrm{joules}\), \(\mathrm{J}\) or \(\mathrm{newton\,meter}\) \(\mathrm{N\,m}\). In SI base units, \(1\,\mathrm{J}=1\,\mathrm{N\,m}=1\,\mathrm{kg\,m^2/s^2}\mathrm{.}\)
Work done to a system is positive; work done by a system is negative. This article will only focus on work done by a constant force.
What is Kinetic Energy?
Recall that energy is the ability of an object to do work.
The kinetic energy of an object is its energy due to its motion.
You've likely seen kinetic energy written as
This is the equation for translational kinetic energy.
An object can have kinetic energy all on its own. Whereas potential energy requires a two-or-more-object system, kinetic energy only depends on an object's mass and velocity; therefore, a single-object system can have translational kinetic energy.
Explanation of the Work-Energy Principle
Any system with internal structure can have potential and kinetic energy.
A system's internal structure refers to its ability to interact within itself.
For example, an object-earth system can interact with itself because the earth would exert a gravitational force on the object. Understanding the internal structure of a system is essential because it allows us to explain the law of the conservation of energy.
Each system has internal energy, the energy within itself.
This energy comprises the potential and kinetic energies within the system. The law of energy conservation requires that a closed system's energy (a system that does not interact with its environment) remains constant. Therefore, a change in the system's potential energy can change its kinetic energy to keep the overall internal energy constant.
This now brings us to the Work-Energy Principle.
The Work-Energy Principle is the physical phenomenon that the work done on a system is equal to the change in the system's kinetic energy.
Work-Energy Principle Formula
Written mathematically, it looks something like this:
where \(W\) is the work done, and \(\Delta K\) is the change in the kinetic energy.
The Work-Energy Principle works because of energy conservation; it's just an outstanding citizen who wants to keep the balance of energy in the universe.
Proving the Work-Energy Principle
As we saw above, the Work-Energy Principle relates work to kinetic energy, but there must be some basis in math for why work and kinetic energy can be related. So, let's see if we can't prove the Work-Energy Theorem. First, we'll start with the definition of kinetic energy:
We will also define force \(F\) as a product of mass \(m\) and acceleration \(a\). You may recognize this as one way to write Newton's Second Law:
Recall our definition of work,
where we define \(F\) as the force and \(x\) as the object's displacement due to its motion.
Let's assume that our force is completely parallel to the displacement of our object, \(\theta = 0\). Using our definition of force, we get:
This is three variables multiplied; it is not the word "max."
From the kinematics for constant acceleration, we can use the equation
Where the subscript \(\mathrm{f}\) refers to the final velocity and subscript \(\mathrm{i}\) refers to the initial velocity. Let's rearrange this to define acceleration as
and then plug this value into our definition of work and get
Then, we can cancel out the \(x\)'s and distribute the \(\frac{1}{2}\\m\) to each of our \(v\)'s on the right side of the equation to get a new work equation of
\begin{align}W &= \frac{1}{2}m(v_\mathrm{f} ^2 - v_\mathrm{i} ^2)\mathrm{.} \\W &= \frac{1}{2}mv_\mathrm{f} ^2 - \frac{1}{2}mv_\mathrm{i} ^2 \\W &= K_\mathrm{f} - K_\mathrm{i} \mathrm{.} \\W &= \Delta K\end{align}
From our proof above, we see that work equals the change in the kinetic energy, proving the Work-Energy Theorem. We can find applications of the Work-Energy Principle in the examples below.
Work-Energy Principle Examples
You can't have a physics article without loads of examples, so let's dive right in.
A \(1\,500.0\,\mathrm{kg}\) vehicle traveling \(30.0\,\mathrm{m/s}\) applies its breaks and skids to rest. Let the frictional force between the sliding tires and pavement be \(5\,000.0\,\mathrm{N}\). How far does the vehicle skid before coming to rest?
First, let's write our work equation
Then we'll plug in our variables
Let
therefore,
\begin{align}x &= \frac{W}{-F} \\x &= \frac{-67.5\,\mathrm{kJ}}{-5\,000.0\,\mathrm{N}} \\x &= 135\,\mathrm{m}\end{align}
Thus, our vehicle skids \(135\,\mathrm{m}\) before coming to rest.
Suppose a \(10.0\,\mathrm{kg}\) box slides across the floor at \(5.00\,\mathrm{m/s}\).
- How much work does friction do to stop the box?
- What is the force required to bring the box to a stop if the box moves a distance of \(2.00\,\mathrm{m}\)?
a) We only need to calculate the change in kinetic energy for the box:
\begin{align}W &= \Delta K \\W & =K_\mathrm{f} -K_\mathrm{i} \\W &= \frac{1}{2}mv_\mathrm{f} ^2 - \frac{1}{2}mv_\mathrm{i} ^2 \\W &= \frac{1}{2}m\left( v_\mathrm{f}^2-v_\mathrm{i}^2\right) \\W &= \frac{1}{2}(10.0\,\mathrm{kg})[(0\,\mathrm{m/s})^2-(5.00\,\mathrm{m/s})^2] \\W &= 0\,\mathrm{J}-125\,\mathrm{J} \\W &= -125\,\mathrm{J.}\end{align}
As the box slows down, it has lost \(125\,\mathrm{J}\) from friction.
b) Therefore, we can use that value to calculate the force to stop the box:
\begin{align}W &=Fd\cos{\theta} \\F &= \frac{W}{d\cos{\theta}} \\F &= \frac{-125\,\mathrm{J}}{(2.00\,\mathrm{m})(\cos{0^{\circ}})} \\F &= -62.5\,\mathrm{N.}\end{align}
Fig. 4 - A \(25\,\mathrm{kg}\) box on a \(10\,\mathrm{m}\) incline at \(30^{\circ}\)
A box of mass \(25\,\mathrm{kg}\) slides down a \(30^{\circ}\) incline. The box accelerates at \(2.0\,\mathrm{m/s}\) and the incline is \(10.0\,\mathrm{m}\) long.
a) What is the kinetic energy of the crate when it reaches the bottom of the incline?
b) How much work is done in overcoming friction?
The work done by gravity is \(P=W_g = mgh\) where \(m\) is the mass, \(g\) is the acceleration due to gravity, and \(h\) is the height/vertical distance from where the box descends.
c) What is the magnitude of the frictional force acting on the box as it slides?
Three-part questions are rough, so let's break them up.
a) The crate has mass \(m\) and acceleration \(a\). It slides a distance \(d\) from rest. So:
\begin{align}K &= \frac{1}{2}mv^2 \\K &= \frac{1}{2}m(2ad) \\K &= mad \\K &= (25\,\mathrm{kg})(2.0\,\mathrm{m/s})(10.0\,\mathrm{m}) \\K &= 500\,\mathrm{J.}\end{align}
b) Recall that \(W_\mathrm{g} =mgh\). The incline is \(30^{\circ}\), so \(h=d\sin{\theta}\).
Combining work and the vertical distance gives
\begin{align}W_\mathrm{g} &= mgd\sin{\theta} \\W_\mathrm{g} &= (25\,\mathrm{kg})(9.8\,\mathrm{m/s^2})(10.0\,\mathrm{m})(\sin{30^{\circ}}) \\W_\mathrm{g} &= 1225\,\mathrm{J.}\end{align}
The other force that does work on the box is friction. Let \(W_\mathrm{fr} \) be the work done by friction.
Therefore,
and with some simple algebra, we get
for a final answer of
Therefore, the work spent overcoming friction is \(725\,\mathrm{J}\).
c) The magnitude of the frictional force can be calculated from work: \(W=Fd\), and so,
Work Energy Principle - Key takeaways
- Energy is the capacity to do work. Work is the transfer of energy into or out of a system.
- Work and energy are scalar quantities; this means that work and energy have no direction associated with them: only a magnitude. However, they can have positive and negative values.
- Work \(W\) is done on an object when we apply a force \(F\) over displacement \(d\). Note that the force's component is in the direction of displacement \(F_{||} \). We can write this as
- An outside force exerting on a system causes mechanical energy, the sum of a system's kinetic and potential energies, to be transferred into or out of the said system: so long as the force is parallel to the system's displacement. This energy-transferring process is equivalent to the work done:
- Work is equal to the area under a Force vs. Displacement Graph curve.
- Each system has internal energy, the energy within itself. This energy comprises the potential and kinetic energies within the system. The law of energy conservation requires that a closed system's energy (a system that does not interact with its environment) remains constant. Therefore, a change in the system's potential energy can change its kinetic energy to keep the overall internal energy constant.
- The Work-Energy Principle is the physical phenomenon that the work done on a system is equal to the change in the system's kinetic energy: \(W=\Delta K\). It can be derived from Newton's second law \(F=ma\) and the definition of kinetic energy \(K=\frac{1}{2}\\mv^2\).
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