Chapter 7 - Work and energy - Modeling with Physics

7.1Overview¶

In this chapter, we introduce a new way to build models derived from Newton’s theory of classical physics. We will introduce the concepts of work and energy, which will allow us to model situations using scalar quantities, such as energy, instead of vector quantities, such as forces. It is important to remember that even when we are using energy and work, these tools are derived from Newton’s Laws; that is, we may not be using Newton’s Second Law explicitly, but the models that we develop are still based on the same theory of classical physics.

7.2Work¶

We introduce the concept of work as the starting point for building models using energy instead of forces. Work is a scalar quantity that is meant to represent how a force exerted on an object over a given distance results in a change in speed of that object. We will first introduce the concept of work done by a force on an object, and then look at how work can change the kinematics of the object. This is analogous to how we first defined the concept of force, and then looked at how force affects motion (by using Newton’s Second Law, which connected the concept of force to the acceleration of the object).

The work done by a force, $\vec F$ , on an object over a displacement, $\vec d$ , is defined to be:

\boxed{W = \vec F \cdot \vec d = Fd\cos\theta = F_xd_x+F_yd_y+F_zd_z}

(1)

where θ is the angle between the vectors when they are placed tail to tail, as in Figure 1. The dimension of work, force times displacement, is also called “energy”. The S.I. unit for energy is the Joule (abbreviated $\text{J}$ ) which is equivalent to $\text{Nm}$ or ${\rm kg m^2/s^2}$ in base units.

When determining the scalar product $\vec F\cdot \vec d = Fd\cos\theta$, $\theta$ is the angle between the vectors when they are placed tail to tail. — Figure 1:When determining the scalar product $\vec F\cdot \vec d = Fd\cos\theta$ , θ is the angle between the vectors when they are placed tail to tail.

The work “done” by the force is the scalar product of the force vector and the displacement vector of the object. We say that the force “does work” if it is exerted while the object moves (has a displacement vector) and in such a way that the scalar product of the force and displacement vectors is non-zero. A force that is perpendicular to the displacement vector of an object does no work (since the scalar product of two perpendicular vectors is zero). A force exerted in the same direction as the displacement will do positive work ( $\cos\theta$ positive), and a force in the opposite direction of the displacement will do negative work ( $\cos\theta$ negative). As we will see, positive work corresponds to increasing the speed of the object, whereas negative work corresponds to decreasing its speed. No work corresponds to no change in speed (but could corresponds to a change in velocity).

You may be tempted to ask, “Why work? Why not something else? Why that scalar product in particular? How could we possibly have thought of that?”. In general, it seems arbitrary that we introduce the quantity “work” and then find that it leads to a convenient way of building models. However, we did not just pull this quantity out of thin air! Many theorists, over many years, tried all sorts of quantities and ways to rephrase Newton’s Theory that were not helpful. The quantities that make it into textbooks are the ones that turned out to be useful. You should also keep in mind that, just like force, work is a “made-up” mathematical tool that is helpful in describing the world around us. There is no such thing as work or energy; they are just useful mathematical tools.

7.2.1Work in one dimension.¶

Work involves vectors, so we can first examine the concept in one dimension, before extending this to two and three dimensions. We can choose $x$ as the coordinate in one dimension, so that all vectors only have an $x$ component. We can write a force vector as $\vec F=F\hat x$ , where $F$ is the $x$ component of the force (which could be positive or negative). A displacement vector can be written as $\vec d = d \hat x$ , where again, $d$ is the $x$ component of the displacement, and can be positive or negative. In one dimension, work is thus:

W = \vec F \cdot \vec d = (F\hat x) \cdot ( d\hat x ) = Fd (\hat x\cdot\hat x)=Fd

(2)

where $\hat x \cdot \hat x = 1$ . Consider, for example, the work done by a force, $\vec F$ , on a box, as the box moves along the $x$ axis from position $x=x_0$ to position $x=x_1$ , as shown in Figure 3.

A force, $\vec F$, exerted on an object as it moves from position $x=x_0$ to position $x=x_1$. — Figure 3:A force, $\vec F$ , exerted on an object as it moves from position $x=x_0$ to position $x=x_1$ .

We can write the length of the displacement vector as $||\vec d|| =d= \Delta x = x_1-x_0$ . The work done by the force is given by:

W = \vec F \cdot \vec d = F\hat x\cdot \Delta x\hat x =F\Delta x =F(x_1-x_0)

(3)

which is a positive quantity, since $x_1 > x_0$ , with our choice of coordinate system.

7.2.2Work in one dimension - varying force¶

Suppose that instead of a constant force, $\vec F$ , we have a force that changes with position, $\vec F(x)$ , and can take on three different values between $x=x_0$ and $x=x_3$ :

\vec F (x)= \begin{cases} F_1\hat x & x<\Delta x \\ F_2\hat x & \Delta x \leq x< 2\Delta x \\ F_3\hat x & 2\Delta x \leq x \end{cases}

(4)

as illustrated in Figure 3, which shows the force on an object as it moves from position $x=x_0$ to position $x=x_3$ , along three (equal) displacement vectors, $\vec d_1=\vec d_2=\vec d_3=\Delta x \hat x$ .

A varying force, $\vec F(x)$, exerted on an object as it moves from position $x=x_0$ to position $x=x_3$. — Figure 4:A varying force, $\vec F(x)$ , exerted on an object as it moves from position $x=x_0$ to position $x=x_3$ .

The total work done by the force over the three separate displacements is the sum of the work done over each displacement:

\begin{align*} W^{tot}&=W_1+W_2+W_3\\ &=\vec F_1\cdot \vec d_1+\vec F_2\cdot \vec d_2+\vec F_3\cdot \vec d_3\\ &= F_1\Delta x +F_2\Delta x + F_3\Delta x \end{align*}

(5)

If instead of 3 segments we had $N$ segments and the $x$ component of the force had the $N$ corresponding values $F_i$ in the $N$ segments, the total work done by the force would be:

W^{tot} = \sum_{i=0}^N\vec F_i \cdot \Delta \vec x

(6)

where we introduced a vector $\Delta \vec x$ to be the vector of length $\Delta x$ pointing in the positive $x$ direction. In the limit where $\vec F(x)$ changes continuously as a function of position, we take the limit of an infinite number of infinitely small segments of length $dx$ , and the sum becomes an integral:

\boxed{W^{tot} = \int_{x_0}^{x_f}\vec F(x) \cdot d\vec x}

(7)

where the work was calculated in going from $x=x_0$ to $x=x_f$ , and $d\vec x=dx\hat x$ is an infinitely small displacement vector (of length $dx$ ) in the positive $x$ direction.

Example 7.1

A block is pressed against the free end of a horizontal spring with spring constant, $k$ , so as to compress the spring by a distance $D$ relative to its rest length, as shown in Figure 5. The other end of the spring is fixed to a wall. What is the work done by the spring force on the block in going from $x=-D$ to $x=0$ ? What is the work done by the block on the spring over the same displacement?

Figure 5:A block is pressed against a horizontal spring so as to compress the spring by a distance $D$ relative to its rest length.

7.2.3Work in multiple dimensions¶

First, consider the work done by a force $\vec F$ in pulling a crate over a displacement $\vec d$ , in the case where the force is directed at an angle θ above the horizontal, as shown in Figure 6, and the displacement is along the $x$ axis (or rather, we chose the $x$ axis to be parallel to the displacement).

The work done by the force is given by:

\begin{align*} W = \vec F \cdot \vec d &= Fd\cos\theta\\ &= F_{\parallel}d\\ &= Fd_{\parallel}\\ \end{align*}

(12)

where we highlighted the fact that the scalar product “picks out” components of vectors that are parallel to each other. $F_{\parallel} = F\cos\theta$ is the component of $\vec F$ that is parallel to $\vec d$ , and $d_{\parallel}=d\cos\theta$ is the component of $\vec d$ that is parallel to $\vec F$ . These are also shown in Figure 6.

In general, if an object is moving along an arbitrary path, we cannot choose the $x$ axis to be parallel to the displacement or to the force. If the path can be sub-divided into straight segments over which the force is constant, as in Figure 7, we can calculate the work done by the force over each segment and add the work done in each segment together to obtain the total work done by the force. Note that, in general, the work done by a force as an object moves from one position to another depends on the particular path that was taken between the two positions, since different paths will have difference lengths.

Figure 7:An arbitrary two dimensional path of an object from $A$ to $B$ broken into three straight segments.

Example 7.2

Compare the work done by the force of kinetic friction in sliding a crate along a horizontal surface from position $A$ (coordinates $x_A, y_A$ ) to position $B$ (coordinates $x_B, y_B$ ) using the two different paths depicted in Figure 8. Assume that the mass of the crate is $m$ and that the coefficient of kinetic friction between the crate and the ground is $\mu_k$ .

Figure 8:Two possible paths to slide a crate from position $A$ to position $B$ , as seen from above.

Solution

The force of kinetic friction is always in the direction opposite to that of motion. Thus, regardless of the path taken, the force of friction will do negative work.

Let us first calculate the work done by the force of kinetic friction along the first path (the straight line). The force of kinetic friction will have a magnitude:

f_k = \mu_k N = \mu_k mg

(13)

The normal force will have the same magnitude as the weight because the crate is not moving (accelerating) in the direction perpendicular to the $xy$ plane. The displacement vector from $A$ to $B$ can be written as:

\begin{align*} \vec d &= (x_B-x_A)\hat x + (y_B-y_A) \hat y\\ \therefore ||\vec d|| &=d= \sqrt{(x_B-x_A)^2 + (y_B-y_A)^2} \end{align*}

(14)

The force of kinetic friction will be in the opposite direction of the displacement vector, so the angle between the two vectors is $180{\rm \degree}$ ( $\cos\theta=-1$ ). The work done by the force of kinetic friction is thus:

W = \vec f_k \cdot\vec d = f_k d \cos\theta = -\mu_k mg\sqrt{(x_B-x_A)^2 + (y_B-y_A)^2}

(15)

and is negative, as expected.

For path 2, we break up the motion into two segments, with displacements vectors $\vec d_1$ (along $y$ ) and $\vec d_2$ (along $x$ ). We can write the two displacement vectors as:

\begin{align*} \vec d_1 &= 0\hat x + (y_B-y_A) \hat y\\ \therefore ||\vec d_1||&=d_1=(y_B-y_A)\\ \vec d_2 &= (x_B-x_A)\hat x + 0 \hat y\\ \therefore ||\vec d_2||&=d_2=(x_B-x_A)\\ \end{align*}

(16)

Along each segment, the force of kinetic friction is anti-parallel to the displacement (note that the force of friction changes direction over the two segments), but the magnitude is $f_k=\mu_kmg$ . The work done along the first segment is thus:

W_1 = \vec f_k \cdot \vec d_1 = f_k d_1 \cos\theta = -\mu_k mg(y_B-y_A)

(17)

The work done along the second segment is:

W_2 = \vec f_k \cdot \vec d_2 = f_k d_2 \cos\theta = -\mu_k mg(x_B-x_A)

(18)

And the total work done by the force of kinetic friction over the second path is:

W^{tot} = W_1 + W_2 = -\mu_k mg \left((x_B-x_A) + (y_B-y_A)\right)

(19)

which is more work than was done along path 1. This makes sense because for both paths, the force of friction has the same magnitude and is always in the opposite direction of motion; thus, the longer the path, the more work will be done by the force.

Example 7.3

A box of mass $m$ is moved from the floor onto a table using two different paths, as shown in Figure 9. The table is a horizontal distance $L$ away from where the box starts and a height $H$ above the floor. Compare the work done by the weight of the box along the two possible paths.

Figure 9:Two possible paths to move a box from the floor onto a table.

The most general case for which we can calculate the work done by a force is the case when the force changes continuously along a path where the displacement also changes direction continuously. This is illustrated in Figure 11 which shows an arbitrary path between two points $A$ and $B$ , and a force, $\vec F(\vec r)$ , that depends on position ( $\vec r$ ). In general, the work done by the force on an object that goes from $A$ to $B$ will depend on the actual path that was taken.

An arbitrary path between two points $A$ and $B$ with a force that depends on position, $\vec F(\vec r)$. — Figure 11:An arbitrary path between two points $A$ and $B$ with a force that depends on position, $\vec F(\vec r)$ .

The strategy for calculating the work in the general case is the same: we break up the path into small straight segments with displacement vectors $d\vec l$ (Figure 12) where we assume that the force is constant over the segment. The total work is the sum of the work over each segment:

\boxed{W = \int_A^B \vec F(\vec r) \cdot d\vec l}

(27)

As usual, we use the integral symbol to indicate that you need to take an infinite number of infinitely small segments $d\vec l$ in order to calculate the sum.

We divide the path into infinitesimally small segments with displacement vectors $d\vec l$. — Figure 12:We divide the path into infinitesimally small segments with displacement vectors $d\vec l$ .

You should note that this is not an integral like any other that we have seen so far: the integral is not over a single integration variable (usually we use $x$ ), but it is the integral (the sum!) over the specific path that we have chosen in going from $A$ to $B$ . This is called a “path integral”, and is generally difficult to evaluate.

Example 7.4

Figure 13:A parabolic path between $A$ and $B$ .

A force, $\vec F(\vec r) = \vec F(x,y) = F_x\hat x + F_y \hat y$ , is exerted on an object. The object starts at position $A$ and ends at position $B$ , along a parabolic path, $y(x) = a+bx^2$ , as depicted in Figure 13. What is the work done by the force, $\vec F$ , along this trajectory?

Solution

In this case, the force can change with position (if $F_x$ and $F_y$ are not constant), and the direction of the path changes continuously. When we break up the path into small segments $d\vec l$ , we need to incorporate the equation of the parabola to include the fact that $d\vec l$ must always be tangent to the parabola. Consider one small segment along the trajectory and the infinitesimal displacement vector $d\vec l$ at that point, as in Figure 14.

$The infinitesimal displacement vector, $d\vec l$.$

Figure 14:The infinitesimal displacement vector, $d\vec l$ .

We can write the $x$ and $y$ components of the vector as infinitesimal distances, $dx$ and $dy$ , along the $x$ and $y$ axes, respectively. The vector $d\vec l$ can thus be written:

d\vec l = dx \hat x + dy \hat y

(28)

The total work done by the force is then:

\begin{align*} W &= \int_A^B \vec F(\vec r) \cdot d\vec l\\ &=\int_A^B (F_x\hat x + F_y \hat y) \cdot (dx \hat x + dy \hat y)\\ &=\int_A^B (F_x dx + F_ydy)\\ \therefore W&= \int_A^B F_x dx + \int_A^B F_ydy \end{align*}

(29)

where in the last line, we simply used the property that the integral of a sum is the sum of the corresponding integrals. At this point, we have two integrals over integration variables ( $x$ and $y$ ) that are meaningful. However, we have not yet used the fact that our path is a parabola, and in general, we expect that the shape of the path is important. By saying that we are integrating (or calculating the work) over a specific path, we are really saying that $x$ and $y$ are not independent; that is, if we know the value of $x$ at some point on the path, we know the corresponding value of $y$ ( $y = a+bx^2$ ).

Since $x$ and $y$ are not independent, we can use a “substitution of variables” in order to express $y$ in terms of $x$ , and $dy$ in terms of $dx$ :

\begin{align*} y(x) &= a + bx^2\\ \frac{dy}{dx} &= 2bx\\ \therefore dy &= 2bxdx \end{align*}

(30)

This allows us to convert the integral over $y$ to an integral over $x$ , which also allows us to be explicit for the limits of the integral (in our example, the integral goes from $x=0$ to $x=x_1$ ):

\begin{align*} W&= \int_A^B F_x dx + \int_A^B F_ydy\\ &=\int_0^{x_1} F_x dx + \int_0^{x_1} F_y(2bxdx)\\ &=\int_0^{x_1} (F_x + 2bxF_y)dx \end{align*}

(31)

where we would need to know how $F_x$ and $F_y$ depends on $x$ and $y$ in order to actually evaluate the integral.

For example, if the force were constant ( $F_x$ and $F_y$ constant), then the work done along the parabolic path would be:

\begin{align*} W &= \int_0^{x_1} (F_x + 2bxF_y)dx\\ &=\left[F_x x + bF_yx^2 \right]_0^{x_1}\\ &=F_x x_1 + bF_yx_1^2 \end{align*}

(32)

As we mentioned earlier, if the force is constant in magnitude and direction, then the work done is independent of path. We can easily check this, using the displacement vector $\vec d = x_1\hat x + bx_1^2 \hat y$ :

\begin{align*} W &= \vec F \cdot \vec d = (F_x\hat x+ F_y\hat y) \cdot (x_1\hat x + bx_1^2 \hat y)\\ &=F_x x_1 + bF_yx_1^2 \end{align*}

(33)

as we found above.

7.2.4Net work done¶

So far, we have considered the work done on an object by a single force. If more than one force is exerted on an object, then each force can do work on the object, and we can calculate the “net work” done on the object by adding together the work done by each force. We will show that this is equivalent to first calculating the net force on the object, $F^{net}$ (i.e. the vector sum of the forces on the object), and then calculating the work done by the net force.

Suppose that three forces, $\vec F_1$ , $\vec F_2$ , and $\vec F_3$ are exerted on an object as it moves such that its displacement vector is $\vec d$ . The net work done on the object is easily shown to be equivalent to the work done by the net force::

\begin{align*} W^{net} &= W_1 + W_2 + W_3 \\ &= \vec F_1 \cdot \vec d + \vec F_2 \cdot \vec d + \vec F_3 \cdot \vec d \\ &=(F_{1x}d_x+F_{1y}d_y+F_{1z}d_z)+ (F_{2x}d_x+F_{2y}d_y+F_{2z}d_z) + (F_{3x}d_x+F_{3y}d_y+F_{3z}d_z)\\ &=(F_{1x} + F_{2x} + F_{3x})d_x+(F_{1y} + F_{2y} + F_{3y})d_y+(F_{1z} + F_{2z} + F_{3z})d_z\\ &=\vec F^{net} \cdot \vec d \end{align*}

(34)

where $\vec F^{net} = \vec F_1 + \vec F_2 + \vec F_3$ is the net force. The result is easily generalized to any number of forces, including if those forces change as a function of position:

W^{net} = \int_A^B F^{net}(\vec r) \cdot d\vec l

(35)

Example 7.5

You push with an unknown horizontal force, $\vec F$ , against a crate of mass $m$ that is located on an inclined plane that makes an angle θ with respect to the horizontal, as shown in Figure 15. The coefficient of kinetic friction between the crate and the incline is $\mu_k$ . You push in such a way that that crates moves at a constant speed up the incline. What is the net work done on the crate if it moves up the incline by a distance $d$ ?

Figure 15:A crate being pushed up an incline.

Solution

Although the answer may be obvious, let’s go the long way about it and calculate the work done by each force, and then sum them together to get the total work done. We start by identifying the forces exerted on the crate:

$\vec F$ , the applied force, of unknown magnitude, $\vec F$ .
$\vec F_g$ , the weight of the crate, with magnitude $mg$ .
$\vec N$ , a normal force exerted by the incline.
$\vec f_k$ , a force of kinetic friction, with magnitude $\mu_k N$ , that points in the direction opposite of $\vec d$ .

These are shown in the free-body diagram in Figure 16, along with our choice of coordinate system, and the displacement vector.

Figure 16:Free-body diagram for the crate on the incline.

With our choice of coordinate system, the displacement vector is given by:

\vec d = d (\cos\theta \hat x + \sin\theta \hat y)

(36)

Before calculating the work done by each force, we need to determine the magnitude of the normal force (and thus of the force of kinetic friction). Since the crate is moving at a constant velocity, its acceleration is zero, so the sum of the forces must be zero. Writing out the $y$ component of Newton’s Second Law allows us to find the magnitude of the normal force:

\begin{align*} \sum F_y &= N\cos\theta -F_g - f_k\sin\theta = 0\\ \therefore mg &= N\cos\theta-\mu_kN\sin\theta = N(\cos\theta-\mu_k\sin\theta)\\ \therefore N &= \frac{mg}{\cos\theta-\mu_k\sin\theta} \end{align*}

(37)

Writing out the $x$ component of Newton’s Second Law allows us to find the magnitude of the unknown force $F$ :

\begin{align*} \sum F_x &= F - N\sin\theta - f_k\cos\theta = 0\\ \therefore F &= N\sin\theta+\mu_kN\cos\theta = N(\sin\theta+\mu_k\cos\theta)\\ &=mg\frac{\sin\theta+\mu_k\cos\theta}{\cos\theta-\mu_k\sin\theta} \end{align*}

(38)

We now proceed to calculate the work done by each force. The work done by the normal force is identically zero, since it is perpendicular to the displacement vector. The work done by the applied force, $\vec F = F\hat x$ , is:

\begin{align*} W_F &= \vec F \cdot \vec d = (F\hat x)\cdot(d (\cos\theta \hat x + \sin\theta \hat y))\\ &=Fd\cos\theta=mg\frac{\sin\theta+\mu_k\cos\theta}{\cos\theta-\mu_k\sin\theta}d\cos\theta \end{align*}

(39)

The work done by the force of gravity, $\vec F_g = -mg \hat y$ , is:

\begin{align*} W_g &= \vec F_g \cdot \vec d = (-mg \hat y)\cdot(d (\cos\theta \hat x + \sin\theta \hat y))\\ &=-mgd\sin\theta \end{align*}

(40)

The work done by the force of friction, $\vec f_k$ , noting that $\vec f_k$ and $\vec d$ are antiparallel:

\begin{align*} W_f &= \vec f_k \cdot \vec d = -f_kd = -\mu_kNd\\ &=-\mu_k\frac{mg}{\cos\theta-\mu_k\sin\theta}d \end{align*}

(41)

The net work done on the crate is thus:

\begin{align*} W^{net}&=W_F + W_g + W_f\\ &=mg\frac{\sin\theta+\mu_k\cos\theta}{\cos\theta-\mu_k\sin\theta}d\cos\theta-mgd\sin\theta -\mu_k\frac{mg}{\cos\theta-\mu_k\sin\theta}d\\ &=mgd \left( \frac{\sin\theta+\mu_k\cos\theta}{\cos\theta-\mu_k\sin\theta}\cos\theta - \sin\theta - \mu_k\frac{1}{\cos\theta-\mu_k\sin\theta} \right)\\ &=mgd \left( \frac{(\sin\theta+\mu_k\cos\theta)\cos\theta - \sin\theta(\cos\theta-\mu_k\sin\theta) - \mu_k}{\cos\theta-\mu_k\sin\theta} \right)\\ &=mgd \left( \frac{\sin\theta\cos\theta+\mu_k\cos^2\theta - \sin\theta\cos\theta+\mu_k\sin^2\theta - \mu_k}{\cos\theta-\mu_k\sin\theta} \right)\\ &=mgd \left( \frac{\mu_k(\cos^2\theta+\sin^2\theta) - \mu_k}{\cos\theta-\mu_k\sin\theta} \right)\\ &=0 \end{align*}

(42)

where we used the fact that $\cos^2\theta+\sin^2\theta=1$ . Thus we find that the net work done on the crate is zero!

Discussion: Of course, this makes sense, because the net force on the crate is zero, since it is not accelerating, so the net work done is also zero. As a consequence, or rather, by construction, we have the condition that if the net work done on an object is zero, then that object does not accelerate. We thus have a scalar quantity (work) that can tell us something about whether an object is changing speed. In the next section, we introduce a new quantity, “kinetic energy”, to describe how an object’s speed changes when the net work done is not zero.

Olivia’s Thoughts

Pay close attention to the words “on” and “by.” There are a few things about this that can be tricky:

In Example 7.5, we were asked to find the net work done on the crate. Sometimes, the question won’t specify that it wants you to find the net work, and will just say “What is the work done on the crate?” When you are just asked for the work done “on” an object, the question is implicitly asking for the net work done on the object.
Just because the net work done on an object is zero doesn’t mean that the work done by each of the forces is zero. This may seem obvious, but it’s easy to get tripped up on a test or exam. If you are reading a question about work and it says that the object is moving at a constant speed, it’s tempting to just jump ahead and say that the work must be equal to zero. However, you can only say this if it’s asking you for the net work done on the object. For instance, in Example 7.5, we concluded that since the crate was moving at a constant speed, the net work was equal to zero. But if the question asked you to find the work done on the crate by gravity, that would mean something different. The work done by gravity in this case is not equal to zero (it’s actually negative).
The work done “on” an object is not the same as the net work done “by” that object. For example, say you are in a tug-of-war and you pull the other team towards you, but you yourself do not move. The net work done on you is zero, but the work done by you is not zero. So, when you are talking about work, you should always state explicitly whether the work is being done “on” the object or “by” the object.

Note: The wording won’t always be like this - sometimes it will say “How much work do you do on the box?” instead of “How much work is done by you on the box,” so always be careful. Still, looking for key words like “by” and “on” is a good place to start.

7.3Kinetic energy and the work energy theorem¶

At this point, you should be comfortable calculating the net work done on an object upon which several forces are exerted. As we saw in the previous section, the net work done on an object is connected to the object’s acceleration; if the net force on the object is zero, then the net work done and acceleration are also zero. In this section, we derive a new quantity, kinetic energy, which allows us to connect the work done on an object with its change in speed. This will allow us to describe motion using only scalar quantities. Like the definition of work, the following derivation appears to “come out of thin air”. Remember, though, that theorists have tried all sorts of mathematical tricks to reformulate Newton’s Theory, and this is the one that worked.

Consider the most general case of an object of mass $m$ acted upon by a net force, $\vec F^{net}(\vec r)$ , which can vary in magnitude and direction. We wish to calculate the net work done on the object as it moves along an arbitrary path between two points, $A$ and $B$ , in space, as shown in Figure 17. The instantaneous acceleration of the object, $\vec a$ , is shown along with an “element of the path”, $d\vec l$ .

An object moving along an arbitrary path between points $A$ and $B$ that is acted upon by a net force $\vec F^{net}$. — Figure 17:An object moving along an arbitrary path between points $A$ and $B$ that is acted upon by a net force $\vec F^{net}$ .

The net work done on the object can be written:

W^{net} = \int_A^B F^{net}(\vec r) \cdot d\vec l

(43)

and is in general a difficult integral to evaluate for an arbitrary path. Our goal is to find a way to evaluate this integral by finding a function, $K$ , with the property that:

\int_A^B F^{net}(\vec r) \cdot d\vec l =K_B - K_A

(44)

That is, we will only have to evaluate $K$ at the end points of the path in order to determine the value of the integral. In this way, the function $K$ is akin to an anti-derivative.

In order to determine the form for the function $K$ , we start by noting that, by using Newton’s Second Law, we can write the integral for work in terms of the acceleration of the object:

\begin{align*} \sum \vec F &= \vec F^{net} = m\vec a\\ \therefore \int_A^B F^{net}(\vec r) \cdot d\vec l &= \int_A^B m\vec a\cdot d\vec l =m\int_A^B \vec a\cdot d\vec l \end{align*}

(45)

where we assumed that the mass of the object does not change along the path and can thus be factored out of the integral. Consider the scalar product of the acceleration, $\vec a$ , and the path element, $d\vec l=dx\hat x +dy\hat y + dz\hat z$ , written in terms of the velocity vector:

\begin{align*} \vec a & = \frac{d\vec v}{dt}\\ \therefore \vec a\cdot d\vec l &= \frac{d\vec v}{dt}\cdot d\vec l\\ &=\left(\frac{dv_x}{dt}\hat x+ \frac{dv_y}{dt}\hat y + \frac{dv_z}{dt}\hat z\right) \cdot (dx\hat x +dy\hat y + dz\hat z)\\ &=\frac{dv_x}{dt}dx+\frac{dv_y}{dt}dy+\frac{dv_z}{dt}dz \end{align*}

(46)

Any of the terms in the sum can be re-arranged so that the time derivative acts on the element of path ( $dx$ , $dy$ , or $dz$ ) instead of the velocity, for example:

\frac{dv_x}{dt}dx = \frac{dx}{dt}dv_x

(47)

where we recognize that $\frac{dx}{dt} = v_x$ . We can thus write the scalar product between the acceleration vector and the path element as:

\begin{align*} \vec a\cdot d\vec l&= \frac{dv_x}{dt}dx+\frac{dv_y}{dt}dy+\frac{dv_z}{dt}dz\\ &=\frac{dx}{dt}dv_x + \frac{dy}{dt}dv_y+\frac{dz}{dt}dv_z\\ &=v_xdv_x + v_ydv_y + v_zdv_z \end{align*}

(48)

The integral for the net work done can be written as:

\begin{align*} W^{net} &= \int_A^B F^{net}(\vec r) \cdot d\vec l =m \int_A^B (v_xdv_x + v_ydv_y + v_zdv_z)\\ &=m\int_A^B v_xdv_x +m\int_A^B v_ydv_y + m\int_A^B v_zdv_z \end{align*}

(49)

which corresponds to the sum of three integrals over the three independent components of the velocity vector. The components of the velocity vector are functions that change over the path and have fixed values at either end of the path. Let the velocity vector of the object at point $A$ be $\vec v_A=(v_{Ax}, v_{Ay}, v_{Az})$ and the velocity vector at point $B$ be $\vec v_B=(v_{Bx}, v_{By}, v_{Bz})$ . The integral over, say, the $x$ component of velocity is then:

\begin{align*} m\int_A^B v_xdv_x &= m\int_{v_{Ax}}^{v_{Bx}} v_xdv_x= m\left[\frac{1}{2}v_x^2 \right]_{v_{Ax}}^{v_{Bx}}\\ &=\frac{1}{2}m(v_{Bx}^2-v_{Ax}^2) \end{align*}

(50)

We can thus write the net work integral as:

\begin{align*} W^{net} &=m\int_A^B v_xdv_x +m\int_A^B v_ydv_y + m\int_A^B v_zdv_z\\ &=\frac{1}{2}m(v_{Bx}^2-v_{Ax}^2) + \frac{1}{2}m(v_{By}^2-v_{Ay}^2) +\frac{1}{2}m(v_{Bz}^2-v_{Az}^2)\\ &=\frac{1}{2}m(v_{Bx}^2+v_{By}^2+v_{Bz}^2)-\frac{1}{2}m(v_{Ax}^2+v_{Ay}^2+v_{Az}^2)\\ &=\frac{1}{2}mv_B^2 - \frac{1}{2}mv_A^2 \end{align*}

(51)

where we recognized that the magnitude (squared) of the velocity is given by $v_A^2 = v_{Ax}^2+v_{Ay}^2+v_{Az}^2$ . We have thus arrived at our desired result; namely, we have found a function of speed, $K(v)$ , that when evaluated at the endpoints of the path allows us to calculate the net work done on the object over that path:

\boxed{K(v) = \frac{1}{2}mv^2}

(52)

That is, if you know the speed at the start of the path, $v_A$ , and the speed at the end of the path, $v_B$ , then the net work done on the object along the path between $A$ and $B$ is given by:

\boxed{W^{net} = \Delta K = K(v_B) - K(v_a)}

(53)

We call $K(v)$ the “kinetic energy” of the object. We can say that the net work done on an object in going from $A$ to $B$ is equal to its change in kinetic energy (final kinetic energy minus initial kinetic energy). It is important to note that we defined kinetic energy in a way that it is equal to the net work done. You may have already seen kinetic energy from past introductions to physics as a quantity that is just given; here, we instead derived a function that has the desired property of being equal to the net work done and called it “kinetic energy”.

The relation between the net work done and the change in kinetic energy is called the “Work-Energy Theorem” (or Work-Energy Principle). It is the connection that we were looking for between the dynamics (the forces from which we calculate work) and the kinematics (the change in kinetic energy). Unlike Newton’s Second Law, which relates two vector quantities (the vector sum of the forces and the acceleration vector), the Work-Energy Theorem relates two scalar quantities to each other (work and kinetic energy). Although we introduced the kinetic energy as a way to calculate the integral for the net work, if you know the value of the net work done on an object, then the Work-Energy Theorem can be used to calculate the change in speed of the object.

Most importantly, the Work-Energy theorem introduces the concept of “energy”. As we will see in later chapters, there are other forms of energy in addition to work and kinetic energy. The Work-Energy Theorem is the starting point for the idea that you can convert one form of energy into another. The Work-Energy Theorem tells us how a force, by doing work, can provide kinetic energy to an object or remove kinetic energy from an object.

Example 7.7

Figure 18:A block is pressed against a horizontal spring so as to compress the spring by a distance $D$ relative to its rest length.

If the block is released from rest and there is no friction between the block and the horizontal surface, what is the speed of the block when it leaves the spring?

Example 7.8

Figure 19:A block is pressed against a horizontal spring so as to compress the spring by a distance $D$ relative to its rest length.

If the block is released from rest and the coefficient of kinetic friction between the block and the horizontal surface is $\mu_k$ , what is the speed of the block when it leaves the spring?

Solution

This is the same example as the previous one, but with kinetic friction. The forces on the block are:

$\vec F_g$ , its weight, with magnitude $mg$ .
$\vec N$ , the normal force exerted by the ground on the block.
$\vec F(x)$ , the force from the spring, with magnitude $kx$ .
$\vec f_k$ , the force of kinetic friction, with magnitude $\mu_kN$ .

Both the normal force and weight are perpendicular to the displacement, so they will do no work. Furthermore, since the acceleration in the vertical direction is zero, the normal force will have the same magnitude as the weight ( $N=mg$ ). The magnitude of the force of kinetic friction is thus $f_k = \mu_k mg$ . The net work done will be the sum of the work done by the spring, $W_F$ , and the work done by friction, $W_f$ :

W^{net} = W_F + W_f

(58)

We have already determined the work done by the spring:

W_F = \frac{1}{2}kD^2

(59)

The work done by the force of kinetic friction will be negative (since it is in the direction opposite of the motion) and is given by:

W_f = \vec f_k \cdot \vec d = -f_kD = -\mu_kmgD

(60)

Applying the work energy theorem, and noting that the block started at rest ( $v_i=0$ ), the final speed $v_f$ is found to be:

\begin{align*} W^{net} =W_F + W_f&= \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 \\ \frac{1}{2}kD^2-\mu_kmgD &=\frac{1}{2}mv_f^2\\ \therefore v_f &=\sqrt{\frac{kD^2}{m}-2\mu_kgD} \end{align*}

(61)

Discussion: We can think of this in terms of the concept of energy. The spring does positive work on the block, and so it increases its kinetic energy. Friction does negative work on the block, decreasing its kinetic energy. Only the spring is “introducing” energy into the block, as friction is removing that energy by doing negative work. Another way to think about it is that the spring is inputting energy; some of that energy goes into increasing the kinetic energy of the block, and some of it is lost by friction.

The energy that is lost to friction can be thought of as “thermal energy” (heat) that goes up into heating the block and the surface. Indeed, if you rub your hand against the table, you will notice that it gets warmer; you are losing some of the energy introduced to your hand by the work done by your arm into heating up the table and your hand! This shows that we can think about modelling friction using thermal energy rather than a force.

7.4Using PhET to Study the Work-Energy Theorem¶

In the PhET simulation below, click on the “Measure” tab. Leave all of the default settings as they are. Place the skater at the top of the ramp and release them. They should continue to skate up and down the ramp endlessly since there is no friction. Place the measurement sensor at the top of the ramp. Attempt to adjust the sensor until the kinetic energy reads zero as the skater passes the sensor.

Does the sensor’s measurement of potential energy agree with the amount of work you calculated above? If the kinetic energy is not zero, does the sum of potential and kinetic energy sum to the calculated energy above?

Figure 20:A PhET simulation to explore work-energy.

Move the measurement sensor to the bottom of the ramp. Try to place it so that the potential energy is zero. How does the kinetic energy compare to the calculation of gravitational work from above?

Olivia’s Thoughts

The skater begins at the top of the ramp with zero kinetic energy and $3528 {\rm J}$ of potential energy. As the skater goes to the bottom of the ramp, gravity is continuously doing work on the skater. As the potential energy drops, it is converted as work by the force of gravity to kinetic energy. At the bottom of the ramp, all of the potential energy has been used to do work on the skater. The skater’s kinetic energy is now equal to the original potential energy. Since gravity is a conservative force, the conversion of energy from potential to kinetic is complete, without losses. As a check of the relationship

W=\Delta K

(63)

you can calculate that the skater’s speed at the bottom should be

\begin{align*} mgh &= \frac{1}{2}mv^2_f−\frac{1}{2}mv^2_i\\ mgh &= \frac{1}{2}mv^2_f − 0 \end{align*}

(64)

Therefore,

gh = \frac{1}{2}v^2_f

(65)

and

\begin{align*} v_f &= \sqrt{2gh}\\ v_f &= \sqrt{2\times 9.8\times 6}\\ v_f &= 10.8 {\rm m/s} \end{align*}

(66)

Next, increase and decrease the skater’s mass. Notice their motion does not change. This is seemingly a contradiction. Can you explain why this scenario has mass independent motion?

Now, set the friction to a small value. What do you observe happening? Place the measuring sensor somewhere on the ramp. You should notice the thermal energy has a value when it did not when there was no friction. Where is the skater’s energy going?

7.5Calculating Work Numerically¶

It’s possible, many students in Physics 1 have not had Calculus 2, and integrals are not familiar. Let’s take a graphical look at the work. In Figure 21, a constant $10~ {\rm N}$ net force is moving an object at constant velocity over a distance of $0.1~ {\rm m}$ . The action is split into ten segments. Considering the relationship

W = \vec{F}\cdot\vec{d}

(67)

we can see that when the force and displacement are aligned, the work is $W=\lvert\vec{F}\rvert\lvert\vec{d}\rvert = Fd$ . This is the height times width, or area, of the graph. We could break the graph into segments and add the segments. As described in the previous section, the area of the segments can be written $\vec F_i \cdot \Delta \vec x$ and summed to obtain the full area of the graph of $F$ vs. $x$ .

W^{tot} = \sum_{i=0}^N\vec F_i \cdot \Delta \vec x_i

(68)

Example

A block is pushed with $10 {\rm N}$ of net force to move the block with constant velocity a distance of $0.1 {\rm m}$ . Split the work into intervals of $\Delta x = 0.01$ meters and calculate the total work done as the sum of the work over all of the intervals.

Figure 21:A constant net force of 10 Newtons applied to an object moving a of 0.1 meter.

Solution

Table 1:A table of all of the intervals.

$F_i ({\rm N})$	$\Delta~x_i ({\rm m})$	$W_i ({\rm J})$
10	$0.01 - 0.00 = 0.01$	$10 \times 0.01 = 0.1$
10	$0.02 - 0.01 = 0.01$	$10 \times 0.01 = 0.1$
10	$0.03 - 0.02 = 0.01$	$10 \times 0.01 = 0.1$
10	$0.04 - 0.03 = 0.01$	$10 \times 0.01 = 0.1$
10	$0.05 - 0.04 = 0.01$	$10 \times 0.01 = 0.1$
10	$0.06 - 0.05 = 0.01$	$10 \times 0.01 = 0.1$
10	$0.07 - 0.06 = 0.01$	$10 \times 0.01 = 0.1$
10	$0.08 - 0.07 = 0.01$	$10 \times 0.01 = 0.1$
10	$0.09 - 0.08 = 0.01$	$10 \times 0.01 = 0.1$
10	$0.10 - 0.09 = 0.01$	$10 \times 0.01 = 0.1$

Now, we can simply add the last column on the right.

\begin{align*} W^{tot} &= \sum_{i=0}^N\vec F_i \cdot \Delta \vec x_i \\ &= 0.1 + 0.1 + 0.1 + 0.1 + 0.1 +0.1 + 0.1 + 0.1 + 0.1 + 0.1\\ &= 1~{\rm J} \end{align*}

(69)

Notice that this is the same as

W^{tot} = Fd = 10\cdot 0.1 = 1~{\rm J}

(70)

Next, let’s look at the spring example from Section 7.3. Instead of compressing the spring we will consider stretching the spring in the $+\hat{x}$ -direction. The force applied to stretching the spring is $\vec{F}=kx \hat{x}$ . If we plot $F$ vs. $x$ for stretching a spring with spring constant $k=10~\text{N/m}$ from $x=0$ to $x=0.1$ meters, we get the graph shown in . We can see that the force increases linearly with displacement. When we divide the displacement into intervals, we no longer can fit rectangular areas to fully describe the area of the force vs. displacement. That is okay, we can approximate the area with these rectangular areas.

Force vs. displacement for stretching a spring with $k=10 {\rm N/m}$ a distance of $0.1 {\rm m}$. The graph is split into 10 segments. — Figure 22:Force vs. displacement for stretching a spring with $k=10 {\rm N/m}$ a distance of $0.1 {\rm m}$ . The graph is split into 10 segments.

Example

A spring with spring constant $k=10 {\rm N/m}$ is stretched from its equilibrium length $0.0 {\rm m}$ by an amount $0.1 {\rm m}$ . Use the numerical summation described above to find the work done stretching the spring. Use the value of the force at the left edge of the rectangular segments as shown in Figure 22

Solution

Table 1:A table of all of the intervals for a spring with $k=10 {\rm N/m}$ displaced $0.1 {\rm m}$ .

$F_i=kx_i ({\rm N})$	$\Delta~x_i ({\rm m})$	$W_i ({\rm J})$
$10 \times 0.00 = 0.00$	$0.01 - 0.00 = 0.01$	$0.00 \times 0.01 = 0.000$
$10 \times 0.01 = 0.10$	$0.02 - 0.01 = 0.01$	$0.10 \times 0.01 = 0.001$
$10 \times 0.02 = 0.20$	$0.03 - 0.02 = 0.01$	$0.20 \times 0.01 = 0.002$
$10 \times 0.03 = 0.30$	$0.04 - 0.03 = 0.01$	$0.30 \times 0.01 = 0.003$
$10 \times 0.04 = 0.40$	$0.05 - 0.04 = 0.01$	$0.40 \times 0.01 = 0.004$
$10 \times 0.05 = 0.50$	$0.06 - 0.05 = 0.01$	$0.50 \times 0.01 = 0.005$
$10 \times 0.06 = 0.60$	$0.07 - 0.06 = 0.01$	$0.60 \times 0.01 = 0.006$
$10 \times 0.07 = 0.70$	$0.08 - 0.07 = 0.01$	$0.70 \times 0.01 = 0.007$
$10 \times 0.08 = 0.80$	$0.09 - 0.08 = 0.01$	$0.80 \times 0.01 = 0.008$
$10 \times 0.09 = 0.90$	$0.10 - 0.09 = 0.01$	$0.90 \times 0.01 = 0.009$

Now, we can simply add the last column on the right.

\begin{align*} W^{tot} &= \sum_{i=0}^N\vec F_i \cdot \Delta \vec x_i \\ &= 0.000 + 0.001 + 0.002 + 0.003 + 0.004 + 0.005 + 0.006 + 0.007 + 0.008 + 0.009\\ &= 0.045~{\rm J} \end{align*}

(71)

Let’s compare to the work computed from integration in the previous section.

W^{tot} = \frac{1}{2}kx^2 = \frac{1}{2}10\cdot\left(0.1\right)^2 = 0.05~\text{J}

(72)

So, our approximation is close. You might notice that the area of this graph is a triangle, and we could calculate

\begin{align*} W &= A_{triangle}\\ &= \frac{1}{2}\text{base} \cdot~\text{height} \\ &= \frac{1}{2}Fx \\ &= \frac{1}{2}kx \cdot x \\ &= \frac{1}{2}kx^2 \end{align*}

(73)

We could improve our approximation by dividing the $F$ vs. $x$ graph into smaller intervals as shown in . The calculation using a table would be tedious. Let’s see how a computer can help us solve this problem.

Force vs. displacement for stretching a spring with $k=10 {\rm N/m}$ a distance of $0.1 {\rm m}$. The graph is split into 50 segments. — Figure 23:Force vs. displacement for stretching a spring with $k=10 {\rm N/m}$ a distance of $0.1 {\rm m}$ . The graph is split into 50 segments.

In the trinket below we will compute the spring force as a function of stretch distance. First, we should define some variables such as the spring constant $k$ , the total stretch distance $x$ , the intervals of stretch $\Delta x$ , and the work $W$ . The total distance and the work are going to be summed in a loop. So, we initialize them to zero. We’ll begin by repeating the calculation above to make sure we get the expected results.

k = 10 #spring constant in N/m
x = 0 #total stretch distance
dx = 0.01 #stretch interval
W = 0

Next, we will create a loop that continues while the stretch is less than 0.1 meters. This should make the last left edge be $x=0.09$ and the last $\Delta x = 0.1-0.09$ . Inside that loop we want to calculate the force at the left side of the rectangle (this is $x$ . Then, add the work this force does over the increment of displacement $F\Delta x$ to the work that has been done in previous increments of displacement. We’ll print the results that are in the table above, and we can’t forget to increment $x$ for the next loop.

while x < 0.1:
    rate(2)
    F = k*x #force at left side of interval
    W = W + F*dx
    print(x, F, W)
    x = x + dx

Run this code to see that it works to produce the values in the table above. If you want to create more intervals, you only need to decrease the value of $\Delta x$ or dx in your code. Try changing this value to dx=0.005 (20 intervals). Then, try dx = 0.001. Does this make the summation closer to the exact value of $W=\frac{1}{2}kx^2=0.05~\text{J}$ ? How small does dx need to be to be within 1% of the exact value?

Figure 24:An empty trinket.

For fun, we can graph and simulate the scenario of stretching a spring. To create a graph add the following two lines after the Web VPython 3.2 line. This will create a graph and f1 is the data that is displayed on the graph.

spring_graph = graph(xtitle='x (m)', ytitle='W (J)')
f1 = gdots(color=color.red)

On the line after the calculation of work in the loop, add the following line to put position on the x-axis and work on the y-axis. This will add a single point to the graph each time the loop iterates.

f1.plot(x, W) #add this data point to a graph

Check that this works properly in your trinket. Once that works, you can add an object simulation. After the line with f1, create a spring with the following code.

spring = helix(pos=vec(-0.125, 0, 0), axis=vec(0.25, 0, 0), radius=0.1, coils = 3)

After the line where the work is calculated, add the following line to extend the spring from its equilibrium length to the equilibrium length plus $x$ .

spring.axis=vec(0.25+x, 0, 0)

7.6Power¶

We finish the chapter by introducing the concept of “power”, which is the rate at which work is done on an object, or more generally, the rate at which energy is being converted from one form to another. If an amount of work, $\Delta W$ , was done in a period of time $\Delta t$ , then the work was done at a rate of:

\boxed{P = \frac{\Delta W}{\Delta t}}

(74)

where $P$ is called the power. The SI unit for power is the “Watt”, abbreviated $\text{W}$ , which corresponds to ${\rm J/s}={\rm kg m^2/s^3}$ in base SI units. If the rate at which work is being done changes with time, then the instantaneous power is defined as:

\boxed{P = \frac{dW}{dt}}

(75)

You have probably already encountered power in your everyday life. For example, your $1000 {\rm W}$ hair dryer consumes “electrical energy” at a rate of $1000 {\rm J}$ per second and converts it into the kinetic energy of the fan as well as the thermal energy to heat up the air. Horsepower ( $\text{hp}$ ) is an imperial unit of power that is often used for vehicles, the conversion being $1 {\rm hp} = 746 {\rm W}$ . A $100 {\rm hp}$ car thus has an engine that consumes the chemical energy released by burning gasoline at a rate of $7.46e4 {\rm J}$ per second and converts it into work done on the car as well as into heat.

Example 7.9

If a car engine can do work on the car with a power of $P$ , what will be the speed of the car at some time $t$ if the car was at rest at time $t=0$ ?

Example 7.10

You are pushing a crate along a horizontal surface at constant speed, $v$ . You find that you need to exert a force of $\vec F$ on the crate in order to overcome the friction between the crate and the ground. How much power are you expending by pushing on the crate?

7.7Summary¶

The work, $W$ , done on an object by a force, $\vec F$ , while the object has moved through a displacement, $\vec d$ , is defined as the scalar product:

\begin{align*} W = \vec F \cdot \vec d &= Fd\cos\theta\\ &= F_xd_x+F_yd_y+F_zd_z \end{align*}

(81)

If the force changes with position and/or the object moves along an arbitrary path in space, the work done by that force over the path is given by:

W =\int_A^B \vec F(\vec r) \cdot d\vec l

(82)

If multiple forces are exerted on an object, then one can calculate the net force on the object (the vector sum of the forces), and the net work done on the object will be equal to the work done by the net force:

W^{net} = \int_A^B \vec F^{net}(\vec r) \cdot d\vec l

(83)

If the net work done on an object is zero, that object does not accelerate.

We can define the kinetic energy, $K(v)$ of an object of mass $m$ that has speed $v$ as:

K(v) = \frac{1}{2} mv^2

(84)

The Work-Energy Theorem states that the net work done on an object in going from position $A$ to position $B$ is equal to the object’s change in kinetic energy:

W^{net} = \Delta K = \frac{1}{2} mv_B^2 - \frac{1}{2} mv_A^2

(85)

where $v_A$ and $v_B$ are the speed of the object at positions $A$ and $B$ , respectively.

The rate at which work is being done is called power and is defined as:

P = \frac{dW}{dt}

(86)

If a constant force $\vec F$ is exerted on an object that has a constant velocity $\vec v$ , then the power that corresponds to the work being done by that force is:

\begin{align*} P &= \frac{d}{dt} W = \frac{d}{dt}(\vec F \cdot \vec d)\\ &= \vec F \cdot \frac{d}{dt}\vec d = \vec F \cdot \vec v \end{align*}

(87)

7.8Thinking about the material¶

7.9Sample problems and solutions¶

7.9.1Problems¶

7.9.2Solutions¶

Part 1 - Mechanics

Chapter 6 - Applying Newton's Laws

Part 1 - Mechanics

Chapter 8 - Potential Energy and Conservation of Energy