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In which of these scenarios are you doing more work: pulling your ridiculously full trash can out to the side of the road with all your might and not moving it even an inch…or lifting a single pencil up into the air? Now, your mom might force you to take the trash out regardless, but it turns out that by…

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In which of these scenarios are you doing more work: pulling your ridiculously full trash can out to the side of the road with all your might and not moving it even an inch…or lifting a single pencil up into the air? Now, your mom might force you to take the trash out regardless, but it turns out that by using the physics definition of work, we can prove that you do more work in the pencil scenario than in the trash can one. Crazy, right? This article will describe the relationship between force and position and what their deal is with work.

And what’s the most fun way to show the relationship between two physical principles of nature, such as force and position? Yep, you guessed it: graphs! By the end of this article, you will not have had enough of the wonders that force vs. position graphs offer.

Before we dive right into all the fun with graphs (I can just sense your excitement), we have to go over a key definition: directly proportional relationships.

For two things to have a directly proportional relationship, their ratio must be equal to a constant value.

This definition will come in handy as you continue reading, especially when you get to all the talk about spring constants. When you get there, think about what it means for force and position to have a directly proportional relationship. If \(k\) is the slope of our force vs. position graph and also a constant, what does that mean for the relationship between force and position?

One of the essential skills in physics to develop is finding relationships. Physics is all about how one thing is related to another. That is how we get all of our equations; it is why modeling relationships through graphs is meaningful. Becoming relationship-minded and asking how the concepts you learn relate to each other will accelerate your understanding of physics.

The area under a force vs. position graph is equal to the work done by the force on whatever object is getting displaced. Recall the equation

$$W = F \Delta x$$

that describes the work done that you learned in AP Physics 1.

Notice how work is simply the product of force and position. Recognizing this relationship makes understanding the area of a force vs. position graph easier. The area under the curve is equal to the work because the integral of the graph will relate the force to the object’s position. Therefore, a more advanced equation for work can be obtained by applying integration like so:

$$W = \int_{\vec a}^{\vec b} \vec{F}\cdot \mathrm{d}\vec{r}$$

where \(W\) is the work done, \(\vec a\) and \(\vec b\) are your initial and final positions, and \(\vec F\) is the force as a function of position \(\vec r\).

Fig. 1 – Here, the area under the curve of a force vs. position graph is shaded and labeled as work.

Notice that the work equation involves evaluating the dot product of two vectors.

The dot product is an operation used for vectors that equals the product of the vector’s magnitudes multiplied by the cosine of the angle between them.

Expressed mathematically, a dot product is

$$\vec{A} \cdot \vec{B} = A B \cos{\theta},$$

where \(\vec A\) and \(\vec B\) are our two vectors, \(A\) and \(B\) are their magnitudes, and \(\theta\) is the angle between them.

Fig. 1 above shows a force vs. displacement graph with a constant slope. So, what happens when our force is not constant, and we have a variable slope? Let’s illustrate this possibility as well as incorporate our trash can and pencil scenario.

Your mom asks you to take out the trash. After seconds of trying to move the massive tin bin, you give up without moving it at all. You then go to your mom’s room and confess to her that you could not accomplish your chore.

“You always slack off when it comes to doing work,” she says.

“Actually,” you respond while lifting a pencil, “I am doing more work on this pencil now than I ever did on that trash can.”

Are you correct in your statement?

Remember that the integral of a force vs. displacement curve equals the work done by that force. Therefore, since you did not move the trash bin at all, though you pushed it with all your might, the total displacement of the bin is \(0\,\mathrm{m}\). No displacement means no force vs. displacement curve, which means you did no work on the trash can. The pencil, however, is a different story. Let’s say you lifted the pencil with a varying force which can be described by the equation

$$F(x)=ax^2+\frac{1}{2}bx+1.00\,\mathrm{N},$$

where \(a=1.00\,\mathrm{\tfrac{N}{m^2}}\) and \(b=1.00\,\mathrm{\tfrac{N}{m}}\). How much work do you do on the pencil if you raise the pencil \(0.750\,\mathrm{m}\) from where it lay at rest compared to the amount of work you did on the trash can?

Let’s start our solution off by plotting a force vs. position graph of our pencil scenario.

Fig. 2 – The area (blue) under this curve from \(0.000\,\mathrm{m}\) to \(0.750\,\mathrm{m}\) gives us the total work you do on the pencil.

Ok, now that we have a better visual for what is happening in this problem, we will dive into our solution.

Recall our formula for work:

$$W = \int_{\vec a}^{\vec b} \vec{F}\cdot \mathrm{d}\vec{r}.$$

In this scenario, we are only looking at one dimension so we can ditch the vectors, and we know that \(a=0.000\,\mathrm{m}\) and \(b=0.750\,\mathrm{m}\) because that is how much we displace the pencil (our bounds of integration). We then have to integrate our function \(F(x)\) over those bounds with respect to our displacement \(x\):

$$\begin{align*} W &= \int_{0.000\,\mathrm{m}}^{0.750\,\mathrm{m}} F(x)\cdot \mathrm{d}x \\ &= \int_{0.000\,\mathrm{m}}^{0.750\,\mathrm{m}} \left(ax^2+\frac{1}{2}bx+1.00\,\mathrm{N}\right)\, \mathrm{d}x \\ &= \left.\left(\frac{1}{3}ax^3 + \frac{1}{4}bx^2 +(1.00\,\mathrm{N})x\right) \right|_{0.000\,\mathrm{m}}^{0.750\,\mathrm{m}} \\ &= \frac{1}{3}\left(1.00\,\mathrm{\tfrac{N}{m^2}}\right)(0.750\,\mathrm{m})^3 + \frac{1}{4}\left(1.00\,\mathrm{\tfrac{N}{m}}\right) (0.750\,\mathrm{m})^2 \\ &+1.00\,\mathrm{N}(0.750\,\mathrm{m}) – 0\,\mathrm{J} \\ &= 1.03\,\mathrm{J}. \\ \end{align*}$$

Therefore, we do \(1.03\,\mathrm{J}\) more work on the pencil than on the trash can. We also conclude that the area under the curve in the figure above is \(1.03\,\mathrm{J}\).

Remember how we said that physics is all about finding relationships? Let’s see if we can find a relationship between force and position that will explain the slope of a force vs. position graph.

The slope is equal to the rise over the run; therefore, the slope of our graph would look something like this:

$$\frac{F}{\Delta x}\\\mathrm{.}$$

Does this equation ring any bells? It is the formula for the spring constant \(k\) that shows up in Hooke’s Law.

Hooke’s Law relates the force exerted by a spring to its displacement times a constant that quantifies the stretchiness of that particular spring \(k\).

Written mathematically, Hooke’s Law looks something like this

$$F_\text{spring}=-kx,$$

where \(F_\text{spring}\) is the magnitude of the force of the spring and \(x\) is the spring’s distance from equilibrium.

This means that a force vs. position graph that was referring to the displacement’s effect on the force of a spring would have a spring constant that could be calculated using the formula

$$k = \frac{F}{\Delta x}\\\mathrm{.}$$

Fig. 3 – Notice that the variable \(k\) is being multiplied by \(x\) to show that \(k\) is the slope. Since \(k\) is a constant value, the force and the displacement above have a directly proportional relationship.

The slope of a force vs. position graph is constant, as seen in Fig. 2 above when we refer to springs and Hooke’s Law: in that specific case, the slope of our force-displacement graph is constant and equals the spring constant \(k\). However, when generalizing the slope of a force vs. displacement graph, the slope does not have to be constant. For example, \(F(x)\) could, perhaps, be a variable force described by a higher-order equation, meaning that the slope of our graph could follow a more parabolic or cubic trajectory. Therefore, in a general sense, we have to say that the slope of a force vs. position graph is its derivative, which can have applications for many other physical scenarios: the spring constant \(k\) is only one example in case the force is directly proportional to the displacement.

Therefore, we can rewrite our equation for a ‘variable spring constant’ \(k_\text{variable}\) in a new light, involving our knowledge of derivatives:

$$k_\text{variable}=\frac{\mathrm{d}}{\mathrm{d}x}F(x).$$

The above equation translates to “\(k_\text{variable}\) equals the derivative of our force doing work taken with respect to our displacement \(x\).”

For ideal springs, \(k\) is not variable and is a constant depending on the inherent characteristics of the spring.

Fig. 4 – As the spring expands or contracts (displaces some amount \(x\) from equilibrium), the force exerted changes; the constant \(k\) quantifies the stretchiness of the spring.

The graph in Fig. 4 takes a closer look at the meaning of the spring constant \(k\). As the spring displaces some amount \(x\) from equilibrium, the restoring force tries to pull it back (represented by the green arrow \(F_\text{s}\)). The force exerted on the spring, causing it to displace is represented by the purple arrow \(F\). The spring constant \(k\) is the stretchiness of the spring, it quantifies how difficult it is for our two forces to displace the spring, or in other words, to change \(x\).

Through a similar process, the units for the slope of a force vs. position graph can be found. Start with our work equation,

$$\frac{F}{\Delta x}\\\mathrm{,}$$

and recognize that the units for force are newtons, and for position, meters:

$$\mathrm{\frac{N}{m}\\}\mathrm{.}$$

These units make a lot of logical sense. Seeing as how \(k\) is the spring constant, the fact that its units are newtons over meters shows that it can be found by doing the force per unit length. This gives the stretchiness of the spring. Recall that newtons are given as

$$\mathrm{N=\frac{m\,kg}{s^2}\\}\mathrm{,}$$

therefore, we can relate our initial understanding of the units, force over meters, to a new understanding:

$$\mathrm{\frac{N}{m}\\=\frac{kg}{s^2}\\}\mathrm{.}$$

Going even further allows us to see how the spring constant relates to surface tension. Surface tension is found by doing the force over a length, and it has units of kilograms over seconds squared, just like the spring constant.

Force vs. position graphs give us enough information to find the velocity of an object. By calculating the work from the graph by finding the area under the curve, one can find the velocity using the Work-Energy Theorem.

Remember that the Work-Energy Theorem states that the work done by a force on an object is equal to that object’s change in kinetic energy.

A horizontal force is exerted on an object whose mass is \(10.0\,\mathrm{kg}\). That object starts at rest with a displacement of \(x = 0.00\,\mathrm{m}\). It then continues horizontally until reaching a position \(5.00\,\mathrm{m}\) away from where it began and stops. The force exerted on the object as a function of position is given below.

Fig. 5 – \(c\) equals \(1\,\mathrm{\frac{N}{m}}\). This image represents the force on the object as a function of the object’s displacement.

What is the object’s velocity?

By calculating the integral of the above equation, we can find the work done by the force on the object:

$$W = \int_{0.00\,\mathrm{m}}^{5.00\,\mathrm{m}} F(x)\, \mathrm{d}x.$$

In this case, we represent our displacement by \(x\), not \(\vec{r}\), because the force exerted is in one dimension, \(x\).

We know that \(F(x)=cx\) where \(c\) equals \(1\,\mathrm{\frac{N}{m}}\), therefore, we can replace \(F(x)\) in our equation for \(x\) and then calculate the integral:

$$\begin{align*} W &= \int_{0.00\,\mathrm{m}}^{5.00\,\mathrm{m}} cx \,\mathrm{d}x \\ &= \frac{cx^2}{2} \Big |_{0.00\,\mathrm{m}}^{5.00\,\mathrm{m}} \\ &= \left(1\,\mathrm{\frac{N}{m}}\right) \frac{(5.00\,\mathrm{m})^2}{2} – 0.00\,\mathrm{N\,m} \\ W &= 12.5\,\mathrm{J}.\end{align*}$$

Note that \(1\) joule is equivalent to \(1\) newton meter.

Using the Work-Energy Theorem, we can relate the work to the object’s velocity. Remember that the Work-Energy Theorem states

$$W = \Delta K$$

as a mathematical relationship, therefore,

$$W = \frac{1}{2}\\mv^2\mathrm{.}$$

The change in the kinetic energy is the same as the regular formula for kinetic energy because the object started at rest, so the initial velocity was \(0\). Thus, the change in the speed is just the total speed.

Now that we have our equation

$$\frac{2W}{m}\\ = v^2$$

set up, let’s solve it symbolically,

$$v=\sqrt{\frac{2W}{m}\\}\mathrm{,}$$

before we plug in any numbers.

That looks good. Now we’ll plug and chug,

$$\sqrt{\frac{2\times 12.5\,\mathrm{N\,m}}{10.0\,\mathrm{kg}}\\} = 1.58\,\mathrm{\frac{m}{s}\\}\mathrm{,}$$

to get the answer:

$$v = 1.58\,\mathrm{\frac{m}{s}\\}\mathrm{.}$$

Our answer for the object’s velocity is \(1.58\) meters per second.

So now, when your mom tells you to take out the trash, tell her that you could do more work by simply lifting a pencil (that is if you fail to actually move the trash can).

A Force vs. Position Graph models the force on an object as a function of that object’s displacement. This is especially useful because the area under the curve of a Force vs. Position Graph is equal to the work done.

An object’s velocity can be found with a Force vs. Position Graph by calculating the area under the curve to find the work. Then, the work can be related to the velocity through the Work-Energy Theorem. After simplifying, the object’s velocity equals the square root of 2 times the work over the object’s mass.

The slope of a Force vs. Position Graph gives the force per unit length, which has units of newtons over meters.

The slope of a Force vs. Position graph is the surface tension. In reference to springs, this translates to the spring constant k.

Question

Which scenario describes a directly proportional relationship?

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Answer

The more ice cream I eat, the more pounds I gain.

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Question

Let’s say you lifted the pencil with a varying force. That force can be described by the equation

$$F(x)=x^3+\frac{1}{4}x+1.$$

How much work do you do on the pencil if you raise the pencil \(0.500\,\mathrm{m}\) from where it lay at rest?

(Note that the constants multiplied to each variable, \(1\) and \(\frac{1}{4}\), have the proper units to make the right hand side of the formula have units of \(\mathrm{N}\).

Show answer

Answer

\(0.547\,\mathrm{J}\).

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Question

Work is ___.

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Answer

calculated as an integral.

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Question

We use a dot product with___

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Answer

Vectors.

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Question

What is the formula for a dot product?

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Answer

$$\vec A \cdot \vec B = AB\cos{\theta}$$

where \(\vec A\) and \(\vec B\) are our two vectors, \(A\) and \(B\) are their magnitudes, and \(\theta\) is the angle between them.

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Question

What is the slope of a force vs. position graph?

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Answer

With springs, it is the spring constant \(k\).

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Question

What are the units for \(k\)?

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Answer

\(\frac{\mathrm{N}}{\mathrm{m}}\\\).

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Question

The velocity of an object can be found with a Force vs. Position graph.

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Answer

Yes, always.

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Question

$$W=\Delta K \mathrm{,}$$

where \(W\) is the work done, and \(K\) is the kinetic energy.

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Answer

True.

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Question

$$\sqrt{\frac{2W}{m}\\} = v $$

where \(W\) is the work done, \(m\) is the mass, and \(v\) is the velocity.

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Answer

True.

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Question

What is the definition of slope?

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Answer

The rise over the run.

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Question

What is the integral formula for work?

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Answer

\(W=\displaystyle \int_{\vec a}^{\vec b} \vec F \cdot \mathrm{d} \vec r\).

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Question

What is the area under a parallel force vs. position graph?

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Answer

Work.

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