Work Energy Theorem

Work-Energy Theorem overview

In everyday life, we are used to the term work to mean anything that requires effort - muscular or mental. The definition in physics encapsulates this, but what you might not know is that the quantity of work in physics has units of energy, joules. Pushing a block, for example, causes a change in its displacement and also a change in its speed. Because the speed changes, the block has changed in kinetic energy. Let's recap what is meant by kinetic energy with the following definition.

The change in kinetic energy is equal to the work done on the block. This is very important in physics, as it makes many problems simpler, even those that we could solve already using Newton's Laws.

What is Work in physics?

In physics, work \(W\) is defined as energy that an object obtains from an external force causing the displacement of that object. Work will not only cause a change in displacement, but also a change in speed.

The equation for work along a straight line is

\[W = F s\tag{1}\]

where the object moves a displacement \(s\) by action of a force \(F\) in the same direction as the displacement. As can be seen by this equation, the work will increase whether it is the force or the displacement that increases. It has units of \(\text{force}\times\text{displacement} = 1\text{ N}\cdot\text{m} = 1\text{ J}\).

Fig. 1 - A box of mass \(m\) on a frictionless surface experiences a force \(F\) to the right.

Let's say we have a stationary box with mass \(m\) on a frictionless surface. When we look at the forces acting on it, there is weight \(w\) downwards, and the normal force \(n\) upwards. When we push it by exerting a force \(F\) on it to the right, the box will start sliding to the right. This is because the box will obey Newton's second law, and it will have an acceleration in the direction of the net force. Because acceleration is the rate at which velocity changes with time, the box will start speeding up. This also means that the work done on the object is positive because the direction of the displacement and the net force is the same.

Fig. 2 - In the image, a box moves to the right. As it moves, a net force is exerted on it in the opposite direction and the object slows down.

However, if you apply a force to the left while the box is moving to the right, the net force now is to the left, meaning that the acceleration is to the left as well. If velocity and acceleration are in opposite directions, this means the object will slow down! Also, if you realize that the direction of the net force and the displacement are opposite, you can conclude that the total work done on the object is negative.

What could we say about the total work done on the block if the force was applied at an angle to the displacement? In our case of the block, the displacement will still lie along a straight line. The work will be positive, negative or zero depending on the angle between the force \(\vec F\) and displacement \(\vec s\). Work is a scalar, and is given by the vector product of \(\vec F\) and \(\vec s\).

\[W = \vec F \cdot \vec s = Fs\cos\phi \tag{2}\]

Where \(\phi\) is the angle between the force \(\vec F\) and the displacement \(\vec s\).

Fig. 3 - A box of mass \(m\) moving at speed \(v\) experiences a vertical force.

If the box is moving to the right and a constant force is applied vertically downwards on the box, the net force is zero, and the work done by this force is zero. We can see this from the scalar product, as \(\vec F \cdot \vec s = Fs\cos 90^{\circ} = 0\). The acceleration will be zero as well, so there would be zero change in velocity. Therefore, in the absence of friction, the box keeps moving at the same speed in the same direction.

This may seem counterintuitive, but remember from our first image, the constant downward force in the image above will result in a normal force of the same magnitude but in the opposite direction. There will be no net downward force and, although there is a displacement \(s\), the product \(W = Fs = 0\). But if there were friction between the box and the surface, the frictional force would increase as it is proportional to the normal force (\(f = \mu N\)). There would be a quantity of work done by the frictional force in the opposite direction to the displacement and the block would slow down. This is because, by equation (2),

\[W_f = \mu N \cos 180^{\circ} = -\mu N = -f\]

You will see examples of the work-energy theorem with friction in a later section of this article.

See the article on work for more examples of work, and for cases where there are several forces acting on a body.

Work-Energy Theorem derivation

Fig. 4 - A block moving with initial speed \(v_1\), is acted upon by a force, \(\vec{F}_\text{net}\), over a displacement, \(s\), which increases its speed to \(v_2\).

In the image, a block with mass \(m\) has initial speed \(v_1\) and position \(x_1\). A constant net force \(\vec F\) acts to increase its speed to \(v_2\). As its speed increases from \(v_1\) to \(v_2\) it undergoes a displacement \(\vec s\). Because the net force is constant, the acceleration \(a\) is constant and is given by Newton's second law: \(F = ma_x\). We can use the equation of motion with constant acceleration, that relates final speed, an initial speed, and displacement.

\[{v_2}^2={v_1}^2+2 a_x s\]

Rearranging for the acceleration:

\[a_x = \frac{{v_2}^2-{v_1}^2}{2s}\]

Inputting these into Newton's Second Law

\[F = ma_x = m \frac{{v_2}^2-{v_1}^2}{2s}\]

The work done by the force over a displacement \(s\) is then

\[W = F s = \frac{1}{2}m {v_2}^2 - \frac{1}{2}m {v_1}^2, \]

which is just the final kinetic energy minus the initial kinetic energy of the block, or the change in kinetic energy of the box after it is accelerated.

This leads us to the following definition:

Work-Energy Theorem equation

In our definition of work in the first section, we have said that the object speeds up if the work done is positive and slows down if it is negative. When an object has speed it also has kinetic energy. According to the work-energy theorem, the work done on an object is equal to the change in kinetic energy. Let's investigate by using our equation (3) which we derived in the previous section.

\[W_{\text{tot}} = K_2 - K_1 = \Delta K\]

For work to be positive, \(K_2\) should be larger than \(K_1\) which means the final kinetic energy is larger than the initial kinetic energy. Kinetic energy is proportional to speed, so the final speed is bigger than the initial speed. That means our object speeds up.

Work-Energy Theorem constant force examples

Here will look at some examples of the application of the work-energy theorem for the specific case that the force under consideration has a constant value.

Work-energy theorem without friction

Work-energy theorem with friction

Work-energy Theorem for a varying force

Previously we discussed work done by constant forces and applied the work-energy theorem.

Here we discuss the work-energy theorem as applying only to point particles, or point masses. As the later general proof will demonstrate, the work-energy theorem is applicable to forces that vary in magnitude, or direction, or both!

An example of the opposite would be the human body, where different parts of the body move in different ways. We call that a composite system. The total kinetic energy of a composite system can change without work done to the system, but the total kinetic energy of a point particle will only change by an external force doing work on it.

To show that the theorem also applies for a varying force, let's consider a force that varies with position \(x\), \(F_x\). You have met the concept of work as the area under the force-displacement curve in the article Work.

We divide the area under the curve into narrow columns of width \(\Delta x_i\) and height \(F_{i,x}\), as shown. The area of these is given by \(F_{i,x}\Delta x_i\). As we take the width \(\Delta x_i\) to be smaller and smaller, we obtain the following integral for a varying force along a straight line displacement from \(x_1\) to \(x_2\),\[W = \int^{x_2}_{x_1} F_x\; dx\tag{4}\]

We can apply this to a spring, which requires more force to compress or stretch as the displacement from its natural position increases. The magnitude of force to stretch/compress a spring is

\[F_x = kx\]

Where \(k\) is the force constant in \(\text{N/m}\). To stretch or compress a spring therefore involves

\[\begin{align}W &= \int^{x_2}_{x_1} k\;x\; dx \\ &= \left[\textstyle\frac{1}{2}kx^2\right]_{x_1}^{x_2} \\ & = \textstyle\frac{1}{2}k{x_2}^2- \textstyle\frac{1}{2}k{x_1}^2.\end{align}\]

The work done by the force on the spring is equal to the area of the triangle with base \(x_2-x_1\) and height \(kx_2\).

Work Done by a Varying Force Along a Straight Line

Consider you are having to move a point-like mass in the \(x\)-direction, but the resistance to movement changes along the way, so the force you apply is varying with position. We might have a force that varies as a function of \(x\), ie. force = \(F(x)\)

Work-energy theorem with varying force - work done on a spring

Work done by a varying force along a curved line

The work-energy theorem can be generalized to a curved path and a variable force. If we follow the path shown in the figure, the direction of \(\vec F\) in relation to the displacement vector \(\vec s\) at a point will be continually changing. We can divide the path into smaller and smaller displacements \(\delta \vec s\), where \(\delta \vec s = \delta x\;{\hat{\textbf{i}}} + \delta y\;{\hat{\textbf{j}}}\).

Fig. 8 - Curved path split into small elements of displacement due to the presence of varying force.

The line integral of \(\vec F\) along the path above is approximated by a sum of the contributions from each of the small displacements \(s_i\).

Recall our definition of work in terms of the scalar product - equation (2): \(W = \vec F \cdot \vec s = Fs\cos\phi\) - and our integral definition of work in equation (4).

As we shrink these displacements to infinitesimal displacements \(d\vec s\) until they are approximately straight-line segments, tangent to the path at a point, we obtain the following integral

\[W = \int_{\text{path}} \vec F\; d \vec s = \int^{P_2}_{P_1} F \cos \phi \; ds\tag{5}\]

The force is practically constant over an infinitesimal segment \(d\vec s\), but may vary in space. The change in kinetic energy over the whole path is equal to the work; that is, it is equal to the integral in (5). As for our earlier examples, it is only the force acting along the displacement that does the work and changes the kinetic energy.

The below example involves calculating a vector line integral.

Work-Energy Theorem Proof

The work-energy theorem is applicable when the force varies with position and in direction. It is also applicable when the path takes any shape. In this section is a proof of the work-energy theorem in three dimensions. Consider a particle moving along a curved path in space from \((x_1,y_1,z_1)\) to \((x_2,y_2,z_2)\). It is acted on by a net force \[\vec F = F_x\;{\hat{\textbf{i}}} + F_y\;{\hat{\textbf{j}}} + F_z\;{\hat{\textbf{k}}}\]

where \(F_x = F_x(x)\), \(F_y = F_y(y)\) and \(F_z=F_z(z)\).

The particle has initial velocity

\[\vec v = v_x\;{\hat{\textbf{i}}} + v_y\;{\hat{\textbf{j}}} + v_z\;{\hat{\textbf{k}}}\]

where \(v_x = v_x(x)\), and the path is divided into many infinitesimal segments \[d\vec s = dx\;{\hat{\textbf{i}}} + dy\;{\hat{\textbf{j}}} + dz\;{\hat{\textbf{k}}} \]

For the \(x\)-direction, the \(x\)-component of work \(W_x = F_x dx\), and is equal to the change in kinetic energy in the \(x\)-direction, and the same for the \(y\)- and \(z\)-directions. The total work is the sum of the contributions of each path segment.

The force varies with position, and as \(\text{Force} = \text{mass$\; \times\; $acceleration}\), it also varies with velocity.

Making a change of variable and using the chain rule for derivatives, for the \(x\)-direction, we have:

\[a_x = \frac{dv_x}{dt}=\frac{dv_x}{dx}\frac{dx}{dt}=v_x\frac{dv_x}{dx}\]

Likewise for the other directions, \(a_y = v_y\frac{dv_y}{dy}\) and \(a_z = v_z\frac{dv_z}{dz}\).

For the \(x\)-direction, and taking \(v_{x_1} = v_x(x_1)\) for example:

\[\begin{align}W_x &= \int_{x_1}^{x_2} m\;a_x\;dx \\ &=m\int_{x_1}^{x_2}v_x\frac{dv_x}{dx}\;dx\\&=m\int_{x_1}^{x_2} v_x\;dv_x\\&=\textstyle\frac12 m \left[{v_x}^2\right]_{x_1}^{x_2}\\&=\frac12 m {v_{x_2}}^2-\frac12 m {v_{x_1}}^2\end{align}\]

We obtain equivalent for the \(y\)- and \(z\)-directions.

Therefore

\[\begin{align}W_\text{tot} = \displaystyle\int_{x_1, y_1, z_1}^{x_2, y_2, z_2} &\vec F \cdot d\vec l \\ \\ = \int_{x_1, y_1, z_1}^{x_2, y_2, z_2}&F_x dx +F_y dy + F_z dz \\ &= \int_{x_1}^{x_2} F_x dx + \int_{y_1}^{y_2} F_y dy + \int_{z_1}^{z_2} F_z dz \\ \\ &=\;\;\frac12 m {v_{x_2}}^2-\frac12 m {v_{x_1}}^2 \\ &\;\;\;+ \;\;\frac12 m {v_{y_2}}^2-\frac12 m {v_{y_1}}^2 \\&\;\;\;+ \;\; \frac12 m {v_{z_2}}^2-\frac12 m {v_{z_1}}^2\\ \\&=K_2-K_1. \end{align}\]

Since we use Newton's second law to derive the work-energy theorem here, note that this particular derivation only applies in inertial frames of reference. But the work-energy theorem itself is valid in any reference frame, including non-inertial reference frames, wherein the values of \(W_\text{tot}\) and \(K_2 - K_1\) may vary from one inertial frame to another (due to the displacement and speed of a body being different in different frames). To account for this, in non-inertial frames of reference, pseudo-forces are included in the equation to account for the extra acceleration that each object seems to have attained.