In our daily life, we tend to associate the amount of work we produce with physical effort. However, this is a very specific concept in physics, as we can only do work when we move an object. Regardless of our physical effort, we won't do any work if we push a wall as it won't move. Still, when we are pushing, our muscles use energy, so we get tired, we might begin to shake, and even sweat. It may sound a bit unfair that physicists don't consider all that effort, right? But there is a good reason for work's definition to be like that. Even if our body spends energy internally, no energy is transferred to the wall. This article will dig deeper into the curious concept of work in physics. We will also study the deep relationship between Kinetic Energy and Work.
where \(\vec{F}\) is the force causing the displacement of an object in newtons \(\left(\text{N}\right)\), \(\vec{r}\) is the displacement in meters \(\left(\mathrm m\right)\), and \(\theta\) is the angle between the force and the displacement in degrees \(\left(^\circ\right)\).
If the force is not constant we need to integrate:
\[\int_a^b\vec{F}\cdot\mathrm{d}\vec{r} = \int_a^bFdr\cos(\theta).\]
The integral is taken over the path from \(a\) to \(b\). It is important to note that the integral implies that the work can be found by determining the area under the curve of the parallel force as a function of the displacement.
Fig. 1 - The work done by a force is the area of a Force vs Displacement graph.
If the component of the force in the direction of the displacement, \(0^\circ\) or \(180^\circ\), it will contribute to the work, but the perpendicular component, \(90^\circ\), will not. Still, As we will see later a force that does work on the system can change the kinetic energy of the system. On the other hand, a force perpendicular to the displacement does no work but can change the direction of motion of the system.
Work and Kinetic Energy Relationship
As the definition of work states, there is a direct relationship between work and the mechanical energy of a system:
$$W=\Delta E,$$
We already know that the total energy of a closed system stays constant, such that its potential energy is converted into kinetic energy and vice versa. This means that there must a relationship between work and kinetic energy.
Its relationship with potential energy is dependent on the type of force that is exerted on the system. Conservative forces do work on a system, in a way that is independent of the object's path. It only depends on the displacement of the object. The work done by a conservative force will be equal to the negative change in the potential energy \(\left(U\right)\) of the system:
$$W_\text{cons} = -\Delta U.$$
In particular, if at the end of the object's path it returns to the initial position, the work done by the system and the change in potential energy will both be zero. On the other hand, the work done by a non-conservative force depends on the path taken by the object.
Work and Kinetic Energy Theorem
If we want to accelerate an object, a force must act upon the object. After work has been done, energy will have been transferred to the object and the object will have been displaced from its rest position. The energy that was transferred to the object in order to make it move it is called kinetic energy.
Kinetic energy is the type of energy that an object has by virtue of its motion.
The Work-Energy theorem states that the change in kinetic energy equals the net work.
The net work is the sum of the work made by all the forces acting on the object or system.
The kinetic energy depends on the object's mass and the square of its velocity. This means that the kinetic energy is a scalar quantity that will always be positive or zero,
$$K=\frac12mv^2.$$
A simple experiment that can be done in a lab to prove the Work-Energy Theorem is to set up a pulley and force sensor that would measure position and velocity. The sensor would be attached to a cart by a string which is pulled by masses hanging from the pulley. With this experiment, you can know the mass and velocity of the cart at different times and set up a graph of Kinetic Energy vs Position. To calculate the work done by the string on the cart, you must multiply the force acting on the cart by the change in position. In this case, the force acting on the cart is the weight of the pulley's hanging masses. This work should equal the kinetic energy in a graph for the corresponding change in position, proving the Work-Energy Theorem.
Fig. 3 - Set up for the experimental proof of the Work-Energy Theorem.
Work and Kinetic Energy Equation
The equation for the Work-Energy theorem is given by
$$\begin{align*}\Delta K&=\sum_iW_i,\\\Delta K&=\sum_iF_id_i.\end{align*}$$
It is important to remind that only the component of the force parallel/antiparallel to the displacement will contribute to the work done by the force. Thus, in the above equation \(F_i\) represent only forces parallel to the displacement \( d_i \). This is expected from the definition of work. If we consider a force that acts perpendicular to the system's displacement as it moves from location \(A\) to \(B\) we can see that:
$$\begin{align*}W&=\int_{A}^{B}\vec{F}\cdot\vec{\text{d}x},\\W&=\int_{A}^{B}|\vec{F}||\vec{\text{d}x}|\cancelto0{\cos{90^\circ}},\\W&=0.\end{align*}$$
Hence a force perpendicular to the displacement does no work and has no effect on the kinetic energy of the system.
Work and Kinetic Energy Examples
An object of mass \(m=2.0\;\mathrm{kg}\) is pushed from the initial point \(\vec{r} = (x_i,z_i)=(0\;\mathrm m,\;0\;\mathrm m)\) upward to the final point \((x_f,z_f)=(3\;\mathrm m,\;3\;\mathrm m)\). We will consider two paths. Path 1 is a parabola \(z=x^2\), and Path 2 is a quarter circle whose equation is \({(x-1)}^2+{(z-3)}^2=4\). What is the amount of work done by gravity during this displacement for each path?
Solution
Gravity is a conservative force. Hence the amount of work won't depend on the path taken. Since gravity acts with a constant magnitude of \( mg, \) where \( g \) is the strength of Earth's gravitational field, \( g=10\,\mathrm{ \frac{N}{kg} }, \) we do not need to integrate.
$$W=\vec{F}\cdot\vec{r} = -mgz$$
Here, the minus sign indicates that gravity acts downwards and against the displacement. Also, note that we only care about the vertical component of displacement because only the parallel component of the force does work.
$$W=-mgz=-(2.0\;\mathrm{kg})\left(10\;{\textstyle\frac{\mathrm N}{\mathrm{kg}}}\right)(3\;\mathrm m)=-60\;\mathrm J.$$
Since the system gained potential energy
\[ \Delta U = U_f- U_i = mgz-mg\left(0\right) = mgz. \]
The work done by the gravitational force having negative sign matches our convention stating that it equals the negative of the change in potential energy \[ W_\text{cons}= -\Delta U = -mgz. \]
As the object is lifted from the ground, gravity does negative work and the object gains gravitational potential energy.
A block of mass, \(m,\) is attached to a spring and it is being held at the position \(x=A\). When released, it moves because of the spring's force. An additional friction force is also acting on the system in the direction of displacement, and it is given by the expression \({F}= b{x}\), where \(b\) is a positive constant. What is the block’s speed the first time it passes through the equilibrium position \(x=0\)?
(A) \(A\sqrt{\frac{(k+b)}m}\)
(B) \(A\sqrt{\frac{(k-b)}m}\)
(C) \(A\sqrt{\frac{(\frac12k+b)}m}\)
(D) \(A\sqrt{\frac{(\frac12k-b)}m}\)
(E) \(A\sqrt{\frac{\frac12(k-b)}m}\)
Solution
One of the forces doing work on the system is the friction force. Friction is a non-conservative force, so we need to take into consideration its path. However, the motion here occurs in a straight line, so we just integrate in the \( x\text{-}\)direction. As it slides from position \(A\) to \(0\):
$$\begin{align*}W_\text{friction}&=\int_A^0 \vec{F}\cdot\vec{\mathrm{d}{x}},\\W_\text{friction}&=\int_A^0|b{x}||{\text{d}x}|\cos{0^\circ},\\W_\text{friction}&= \int_A^0bx\text{d}x,\\W_\text{friction}&=\left.\frac12bx^2\right|_A^0,\\W_\text{friction}&=-\frac12bA^2.\end{align*}$$
The friction force does negative work, in contrast to the that it opposes the spring force which is doing positive work. We can be seen by looking at the form of the force \( \vec{F}= -k\vec{x},\) which is similar to that of friction but with an opposite sign. Spring force is converting the spring's potential energy to kinetic energy as the block moves from \(A\) to the equilibrium position.
Initial kinetic energy is zero as the object is initially at rest. And the initial potential energy of the spring is given by
$$U_{\text{spring}}=\frac12kx^2.$$
The final potential energy will be zero, as the final position is the equilibrium position. Now we can determine the block's speed when it passes through the equilibrium position using the definition of work:
$$\begin{align*}W_\text{friction}&=\Delta E,\\W_\text{friction}&=\;K_{\text{f}}-K_{\text{i}}+U_{\text{f}}-U_{\text{i}},\\\cancelto0{K_{\text{i}}}+U_{\text{i}}+W_\text{friction}&=\;K_{\text{f}}+\cancelto0{U_{\text{f}}},\\\frac12kA^2-\frac12bA^2&=\frac12mv^2,\\v&=A\sqrt{\frac{k-b}m}.\end{align*}$$
Which corresponds to option (B).
A \(11\,\text{kg}\) object is initially at rest on top of a ladder of \(10\,\text{m}\). Determine the work done on the object as it is dropped and reaches the ground if we know that the maximum potential energy of the object when on top of the ladder is \(98\,\text{J}\).
Initially, the potential energy is at its maximum of \(98\,\text{J}\) so that the initial kinetic energy of the object is zero. When the object reaches the ground its potential energy becomes zero and transforms into kinetic energy of \(98\,\text{J}\). The Work-Energy Theorem allows us to determine the net work done on the object, which is just the work done by gravity:
$$\begin{align*}W_{\text{net}}&=K_{\text{f}}-\cancelto0{K_{\text{i}}},\\W_{\text{net}}&=98\,\text{J}.\end{align*}$$
Gravity does positive work on the object so that its potential energy gets converted into kinetic energy as the object falls from the ladder.
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