Work Done

After long hours of doing your physics homework, you might feel pretty tired, as you have done a lot of work. However, because you did your homework, you now know that 'work' is a physical quantity! Have you actually been doing work in the physical sense?

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Jetzt kostenlos anmeldenAfter long hours of doing your physics homework, you might feel pretty tired, as you have done a lot of work. However, because you did your homework, you now know that 'work' is a physical quantity! Have you actually been doing work in the physical sense?

**Work** is **t**he amount of energy transferred to an object by an external Force when it is moved over a certain distance by that Force.

The **work done** on an object is the amount of energy transferred to an object through work.

When you are exerting a force on an object that** **causes its position to change in the same direction as that of the force,** y**ou are **doing**** work** on this object. Work done on an object is made up of two main components**: **force on and displacement of the object. The displacement of an object **must** happen along the line of action of the force in order for the force to do work on the object.

Work has units of energy because it is defined as an amount of (transferred) energy, so work usually has units of \(\mathrm{J}\) (joules).

The equation that describes the work \(W\) done on an object that moves a distance \(s\) while a force \(F\) is acting on it in the same direction as the object's movement is given by

\[W=Fs.\]

Work is measured in joules, the force is measured in newtons, and the displacement is measured in metres. From this equation, we can conclude that

\[1\,\mathrm{Nm}=1\,\mathrm{J}.\]

This is an important conversion to be able to do!

This conversion is easy to remember once you remember the equation describing the work done in terms of the product of a force and a distance.

As you know, a force is a vector, which means that it has three components. We can choose these components such that one is exactly along the direction of the movement of the object it is working on, and such that the other two components are perpendicular to that movement. To illustrate this, we will discuss vectors in two dimensions, so one component will be along the direction of movement and the other one will be perpendicular to it.

Let's take the movement of our object to be in the \(x\)-direction. Looking at the figure below, we see that the **horizontal component **\(F_x\) of the force \(F\) is calculated using the formula:

\[F_x=F\cos\left(\theta\right),\]

where \(\theta\) is the angle that the force makes with the direction of motion of the object. The work being done on the object is done only by this component of the force that is parallel to the direction of travel of the object, so the work \(W\) done on an object moving a distance \(s\), acted upon by a force \(F\) that makes an angle \(\theta\) with the direction of motion of the object is

\[W=Fs\cos\left(\theta\right).\]

We see that a force that is perpendicular to the direction of motion of the object indeed does no work on the object because \(\cos\left(90^\circ\right)=0\). We also see that pushing parallelly *against* the motion of the object means an angle of \(180^\circ\) so the work done on that object is negative. This is logical because we are taking energy out of the object by pushing against it!

Suppose you decide to put all your books and magazines in one wooden box. You place the box on a table and you pull it using a rope attached to the box, as shown in the figure above. This pull generates a motion of the box that is exactly in the direction of the pull, namely precisely to the right. This means you are doing work on the box! Let us do an example calculation on this setup.

Suppose that you are exerting a constant force of \(250\,\mathrm{N}\) and you manage to drag the box towards you over a distance of \(2\,\mathrm{m}\). The work you exerted on the box doing this is

\[W=Fs=250\,\mathrm{N}\times2\,\mathrm{m}=500\,\mathrm{Nm}=500\,\mathrm{J}.\]

This means that the work done on the box is \(W=500\,\mathrm{J}\).

Now suppose that after this first pull you are tired, and your second pull is done with only half the force and the box only moves half the distance. In this case, the work done on the box in the second pull is

\[W=Fs=125\,\mathrm{N}\times1\,\mathrm{m}=125\,\mathrm{J}.\]

In the last situation, we suppose that the box is sliding towards you over ice and you try to stop it. You end up exerting a small force of \(F=10\,\mathrm{N}\) on the box because you don't have a lot of traction yourself on the ice, and the box comes to a stop after \(s=8\,\mathrm{m}\). The important thing to note in this situation is that the work done on the box by you is negative because the force you exerted on the box was opposite the direction of movement of the box. You did

\[W=-10\,\mathrm{N}\times8\,\mathrm{m}=-80\,\mathrm{J}\]

of work on the box.

We return to the case in which we are pulling the box on a table.

The surface of the table will resist the motion of the box by applying a force that opposes the direction of motion.

The force of Friction will always be directed against the motion of an object, so Friction always does negative work on objects.** **

If we want to calculate the work done by the friction force, we will need to know how much force was applied to the box by friction.

Suppose that on the first pull, the magnitude of the force of friction was equal to that of the force you exerted on the box. As the force and displacement are the same as in the example we already treated, we conclude that the force of friction did \(-500\,\mathrm{J}\) of work on the box. Note that we incorporate the fact that the friction was in the direction opposite to the movement of the box by including the minus sign!

In the example of us pulling the box, Gravity does no work because the movement of the box is horizontal while gravity acts vertically.

In general, the gravitational force on an object is its weight given in terms of its mass \(m\) and the gravitational acceleration \(g\) by \(-mg\). Here, the minus sign is there because Gravity acts downwards. Thus, the work that Gravity does on objects is calculated by

\[W=Fs=-mg\Delta h,\]

where \(\Delta h\) is the height difference the object undergoes.

You may recognise this quantity as the difference in Gravitational Potential Energy. This is exactly what it is: the work done by gravity on an object changes its Gravitational Potential energy accordingly.

A spring is always defined by how stiff it is, which is characterised by its **spring constant** \(k\), which we measure in \(\mathrm{N}/\mathrm{m}\). The potential energy \(E_\text{p}\) contained in a spring is determined by this spring constant and how much we squeeze or stretch it, called the **extension** \(x\), in the following manner:

\[E_\text{p}=\frac{1}{2}kx^2.\]

This potential energy defines how much work the spring can do on an object: with no extension, the potential energy is \(0\,\mathrm{J}\), so the work done on an object that is shot by a spring is equal to the Potential Energy of the spring just before releasing the spring:

\[W=E_\text{p}.\]

Q: A spring with spring constant \(k=6.0\,\mathrm{MN}/\mathrm{m}\) is squeezed until it has an extension of \(2.0\,\mathrm{cm}\). How much does it do on an object with mass \(m=4.3\,\mathrm{kg}\) if this object is being shot by this spring from its given squeezed configuration?

A: The work done on any object is completely determined by the Potential Energy of the spring, so the mass of the object is not relevant to answer this question. The work done can be calculated as follows:

\[W=\frac{1}{2}kx^2=\frac{1}{2}\times6.0\times10^6\,\mathrm{N}/\mathrm{m}\times\left(2.0\times10^{-2}\,\mathrm{m}\right)^2=1200\,\mathrm{J}.\]

**Work**is**t**he amount of energy transferred to an object by an external force when it is moved over a certain distance by that force.- The
**work done**on an object is the amount of energy transferred to an object through work. - The equation that describes the work \(W\) done on an object that moves a distance \(s\) while a force \(F\) is acting on it in the same direction as the object's movement is given by \(W=Fs\).
- \(1\,\mathrm{Nm}=1\,\mathrm{J}\).
- The direction of the force compared to that of the movement of the object is important: if they are opposite, negative work is done by the force on the object.
- Friction always does negative work.
- The work done by gravity is \(W=-mg\Delta h\).
- The work done by a spring when it goes from its extension \(x\) to no extension \(x_0=0\) is \(W=\frac{1}{2}kx^2\).

*W *done on an object by a force *F* that is moved over a distance *x* is calculated by *W=Fs*. If the force is opposite the direction of movement of the object, we introduce a minus-sign.

The **work done** on an object is the amount of energy transferred to an object through work.

Work done is measured in joules.

*W *done on an object by a force *F* that is moved over a distance *x* is calculated by *W=Fs*. If the force is opposite the direction of movement of the object, we introduce a minus-sign.

What units does work have?

J (joules).

A force applied on an object can perform work on it even if this object doesn't change its position.

False.

For the work to be done, the force must have the same direction as the displacement. If only part of the force applied is aligned with the displacement, then no work is done.

Is this statement true or false?

This statement is false.

Work can be done even if part of the force is aligned with the direction of displacement.

Does gravity do any work on an object moving horizontally?

No, never.

When does friction do positive work?

Never.

When does gravity do positive work?

Never.

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