Work and Kinetic Energy

When we apply a force on the block, we have the feeling that we add something to the system; intuitively, we say that “we are adding energy,” although this is not necessarily true. If the force is not enough to break the static friction, then it is not possible to say whether the block has changed in any way. Certainly, we are adding something to the system, an “effort,” so to speak, but it is also true that the static friction returns that “effort” in an opposite and complementary direction. However, when the static friction is broken, then it is possible to distinguish a change in the system: it is now in a different state of motion. To produce that new state of motion, we must add “something new” to the system: that “something” is what we call kinetic energy.

Now we have two physical quantities that describe the state of a physical object or system: the linear momentum, which we know well and represents the state of motion, and the kinetic energy that we will begin to study and which, for now, will represent what we have had to add to take the body or system from rest to that state of motion.

The relationship between work and kinetic energy

Let’s go back to the block and the force that sets it in motion. If the force does not produce movement, then we say that it has not added anything to the system, while if it does, we say that it adds kinetic energy. When movement occurs, the body necessarily follows a certain path, and while it does, the force will add energy. This action of adding or removing kinetic energy is called “performing mechanical work,” thus the mechanical work element dW is defined through the equation

\begin{array}{lr} dW =\vec{F} \cdot d\vec{r} & (1) \end{array}

where \vec{F} is the applied force and d\vec{r} is the displacement element on which the force has acted. Since this force is applied to a body of mass m, using the second law of Newton we can write

\begin{array}{lr} \displaystyle\vec{F} =\frac{d\vec{p}}{dt} = m\frac{d\vec{v}}{dt} & (2) \end{array}

Thus, from equations (1) and (2) we will have

\begin{array}{llr} dW & \displaystyle =m\frac{d\vec{v}}{dt} \cdot d\vec{r} = m\frac{d\vec{r}}{dt} \cdot d\vec{v} = m\vec{v} \cdot d\vec{v} & (3) \end{array}

Integrating this last expression to obtain the total work done, we will have:

\begin{array}{llr} W & = \displaystyle {\int_{i}^{f}} m\vec{v} \cdot d\vec{v} = \left.\frac{1}{2}m \|\vec{v}\|^2 \right|_i^f & \\ \\ & \displaystyle = \frac{1}{2}m \|\vec{v}_f\|^2 - \frac{1}{2}m\|\vec{v}_i\|^2 & (4) \end{array}

From this reasoning, we can see that mechanical work is equivalent to a difference of the same magnitude in two different states, one corresponding to a final state and another to the initial one. Such magnitude corresponds to what we have had to add (or remove) to change the state of motion, and it is what we call Kinetic Energy, and consequently, it is defined through the equation:

\begin{array}{llr} E_{cin} & \displaystyle = \frac{1}{2}m\|\vec{v}\|^2 & (5) \end{array}

And therefore, we have that

\begin{array}{llr} W & = \Delta E_{cin} & (6) \end{array}

This is what is known as the Living Forces Theorem.

Kinetic Energy and Braking Distance

A very common mistake made by motorists is to intuit that the braking distance is directly proportional to speed: if we double the speed, the braking distance also doubles. In this section, we will review the error behind that intuition and demonstrate that, in reality, the braking distance is proportional to the square of the speed.

Let’s suppose we have a block of mass m moving on a horizontal plane with initial speed v_i\hat{x} and where there is a kinetic friction coefficient \mu_c. From this, we see that there is a friction force opposite to the motion \vec{F}_{roce}=-\mu_c mg\hat{x} that will act until the body stops after traveling the braking distance x_{fre}. The work done by this force is given by:

\displaystyle W_{roce}= \int_{0}^{x_{fre}} \vec{F}_{roce} \cdot d\vec{l} = \int_{0}^{x_{fre}} -\mu_c mg dx = -\mu_cmgx_{fre}

On the other hand, the variation of the kinetic energy of a body that starts with initial speed v_i and reaches rest with final velocity v_f=0 is:

\displaystyle \Delta E_{cin}= \frac{1}{2}m (\underbrace{\color{red}{v_f^2}}_{= 0} - v_i^2) = - \frac{1}{2}mv_i^2

So that if all the kinetic energy is dissipated by friction until bringing the body to rest, we will have that:

\displaystyle \begin{array}{rrl} & W_{roce} & \displaystyle = \Delta E_{cin} \\ \\ \equiv & -\mu_cmg x_{fre} & \displaystyle = - \frac{1}{2}mv_i^2 \\ \\ \equiv & x_{fre} & = \frac{1}{2} \frac{v_i^2}{\mu_c g} \end{array}

From this, we have what we wanted to demonstrate, that the braking distance is proportional to the square of the speed.

Work and Potential Energy

Let’s imagine we have a body of mass m that falls from a height h_i to a final height h_f (with h_f \leq h_i). Then we will have that the work done by the gravitational force is in the form:

\displaystyle W_g = \int_{h_i}^{h_f} \vec{F}_g \cdot d\vec{l} = \int_{h_i}^{h_f} -mgdz = -mg(h_f - h_i)

From this, we have that if the initial height is h_i = h and the final height is at ground level h_f = 0, then we will have:

\displaystyle W_g = -mg(0 - h) = mgh

With this, we have that when a body falls from a height h, energy associated with its relative position in space is released. This energy is what we call Potential Energy.

\displaystyle \begin{array}{rr}{E_{pot} = mgh} & (7)\end{array}

Potential Energy and Free Fall

Let’s recall the problem of free fall. Using now the potential and kinetic energy we have studied, we can now find the falling speed with a much simpler procedure. Energy is another of those physical magnitudes that have the characteristic of being conservative; that is, it is not created or destroyed, only transformed. When we have a body set at a height h from the ground, when it falls by the action of gravity until it reaches the ground, its potential energy does not disappear but transforms into another form of energy: kinetic energy, so we will have:

\begin{array}{rl} E_{pot,inicial} & = E_{cin, final} \\ \\ mgh &\displaystyle = \frac{1}{2}mv^2 \\ \\ v^2 & = 2gh \\ \\ v& =\sqrt{2gh}\end{array}