Work

Section 8.1 Work

Work has a different meaning in physics than it has in everyday life. In physics we say that a force has done work on an object if it has caused the object to move. A weightlifter lifting a barbell above his head, a lift pulling passengers up to higher floors, a tugboat pulling a ship, a student lifting a backpack full of books upon his shoulders, etc, in each case, a work is done when a force is exerted on an object to cause its displacement.

As we have seen in previous chapters, the displacement of an object may or may not in the same direction as the force. The concept of work tries to capture the influence of the force in the direction of the displacement. Mathematically, the projection of the force vector in the direction of displacement captures this physical reality. As shown in Figure 8.1, the projection can be written in terms of magnitude of the force and the angle \(\theta\) that force makes with the displacement vector.

\begin{equation*} F_\parallel = F \cos\theta. \end{equation*}

Therefore, quantitatively, we define the work of a force \(\vec F\) when object has the infinitesimal displacement \(d \vec r\) by the product of the projection of the force \((F_{\parallel})\) on the displacement vector and the magnitude of the displacement as shown in Figure 8.1.

\begin{equation*} \text{Work} =F_{\parallel}\times |d\vec r| =F\cos\theta\, dr. \end{equation*}

Figure 8.1. \(W = F_{\parallel}\times |d\vec r|\text{.}\)

Zero Displacement: Clearly, no work is done when there is no displacement. For instance, when you lift a weight from ground to above your head, you do work only during the time you are moving the weight, but when you hold the weight steady, you do not do any work! But, your muscles do get tired because muscles are using energy, converting them into heat. And no work on the weight, i.e., your muscles are nt transfering any energy to the weight while you are just holding the weight.

Perpendicular Force: Also, note that work will be zero by any force that acts perpendicularly to the displacement. In this case, projection of the force is zero. You can also see that \(\theta = 90^\circ \implies \cos\theta = 0\text{,}\) and hence work is zero. For example, Normal force will not do any work because, object always moves perperndicularly to a Normal force.

Subsection 8.1.1 Work by a Constant Force on a Straight Displacement

Suppose you pull on a suitcase with a constant force \(\vec F \) at all instants of an interval that begins at \(t_i \) and ends at \(t_f\text{.}\) During this time, the suitcase has a displacement \(\vec r_{if} \) as shown in Figure 8.2. Other forces also may be acting on the suitcase, but here, we want work by this force alone.

Now, we can break up the displacement \(\vec r_{if}\) into infinitesimal parts. Suppose we divide the displacement into \(N\) parts of size \(\Delta r = r_{if}/N\text{.}\) On each part, say the \(j^\text{th}\) part, we will have same work by the constant force,

\begin{equation*} W_j = F\cos\theta\,\, \frac{r_{if}}{N}. \end{equation*}

Adding up work on all the parts we will get

\begin{equation*} W = N\times W_j = F\,r_{if}\cos\theta \end{equation*}

We will denote the total work by \(W_{if}\text{.}\)

\begin{equation} W_{if} = F\,r_{if}\cos\theta.\tag{8.1} \end{equation}

Actually, a more succinct way of stating this is by way of dot product of vectors \(\vec F\) and \(\vec r_{if}\text{.}\) Thus, the following dot product gives work done by a constant force on a straight-line dispacement \(\vec F\text{.}\)

\begin{equation} W_{if} = \vec F \cdot \vec r_{if}.\tag{8.2} \end{equation}

Note the caveat that the displacement has to be straight line. If not, the angle \(\theta\) between the force and displacement will change at different paces on the path. In that case your sum over various parts of the path wil become and integral. We will see this more general case below.

From the product of force and displacment, we find that the unit of work will be \(\text{N.m}\text{.}\) Now, if we expand this into base units we find that work can be written in various equivalent units, including the unit of energy.

\begin{equation*} \text{N.m} = \frac{\text{kg.m}}{\text{s}^2}.\text{m} = \text{kg}.\left(\text{m}/\text{s} \right)^2 = \text{J}. \end{equation*}

That is \(\text{N.m}\) is same as unit of kinetic energy, i.e., Joule. Just because the unit of work is same as the unit of energy, work is not energy. We will show by work-energy theorem below that work is amount of energy that gets transfered between bodies that interact via a force between them. When we study thermodynamics, we will encounter another mechanism by which energy is transferred between bodies - that mechanism will be called heat.

Example 8.3. Work of a Constant Force Pulling on a Suitcase at an Angle.

A suitcase is pulled by a tension force of magnitude \(40\text{ N}\) in a strap acting at angle \(30^\circ\) above the horizontal direction as shown in Figure 8.4. Other forces on the suitcase are gravity, normal force, and rolling friction. As a result of these forces, the suitcase moves a distance of \(20\text{ m}\text{.}\) What is the work done on the suitcase by the tension force?

Answer.

\(693\text{ N.m}\text{.}\)

Solution.

The situation gives us all the quantities needed to compute the force by the tension force.

\begin{equation*} W_{if} = F_T\, \Delta r\, \cos\theta = 40\times 20\times\cos\,30^\circ = 693\text{ N.m}. \end{equation*}

Example 8.5. Different Work by the Same Force but Acting at Different Angles.

A boy pulls a \(5 \text{-kg}\) cart with a \(20 \text{ N}\) force at an angle of \(30^{\circ}\) with the horizon for some time. Over this time the cart moves a distance of \(12 \text{ m}\) on a horizontal floor.

(a) Find the work done on the cart by the boy.

(b) What will be the work done by the boy if he pulled with the same force horizontally instead of at an angle \(30^{\circ}\) with the horizon over the same distance?

Answer.

(a) \(208\text{ N.m}\) (b) \(240\text{ N.m}\text{.}\)

Solution 1. a

We use the definitition of work with angle \(\theta = 30^\circ\text{.}\)

\begin{equation*} W = 20 \text{ N} \times 12 \text{ m} \times \cos\, 30^\circ = 208\text{ N.m}. \end{equation*}

Solution 2. b

Now, the angle is \(\theta=0\text{,}\) which means \(\cos\, \theta=1\text{.}\) This gives

\begin{equation*} W = 20 \text{ N} \times 12 \text{ m} \times 1 = 240\text{ N.m}. \end{equation*}

More work is done now. Is this a waste of work? No, as you will see by work-energy theorem that cart in (b) be going faster in the end than in (a).

Example 8.6. Moving a Crate on a Rough Surface on Different Paths to Same Destination.

Workers want to move a crate of mass \(200\) kg from a site on the ground floor to a third floor apartment. They know that they can either use the elevator first, then slide it on the third floor to the apartment or first slide the crate to another location marked C in the figure, take another elevator to the third floor, and then slide it on the third floor a shorter distance.

The trouble is that the third floor is very rough compared to the ground floor. The coefficient of kinetic friction between the crate and the ground floor is \(0.1\) and between the crate and the third floor surface is \(0.3\text{.}\)

Find work needed by the workers for each path, i.e., for ABCDE and for AFE paths. Assume that the force workers need to do is just enough to slide the crate at constant velocity at zero acceleration. Note: The work by the elevator against the force of gravity is not done by the workers.

Answer.

\(W_\text{ABCDE}=13,730\text{ J}\text{;}\) \(W_\text{AFE}=17,660\text{ J}\text{.}\)

Solution 1. Setting up

The workers will need to apply a horizontal force equal in magnitude to the kinetic friction force. Since acceleration is assumed to be zero, the balancing of forces on the crate gives us the following for the magnitude of the kinetic friction force.

\begin{equation*} F_k = \mu_k\, m \, g. \end{equation*}

That means the force by workers has this magnitude also. Let us denote this by \(F\text{.}\)

\begin{equation*} F = \mu_k\, m \, g. \end{equation*}

Since the force by the workers is in the same direction as the displacements, the work by them for a displacement \(\Delta r \) is

\begin{equation*} W = F \Delta r = \mu_k\, m \, g\, \Delta r. \end{equation*}

Now, we use this expression on each path, making sure to use the appropriate \(\mu_k\) and \(\Delta r\text{.}\) Since \(mg \) is common factor, we can just work out the number for this once.

\begin{equation*} mg = 200\times 9.81 = 1962 \text{ N}. \end{equation*}

Solution 2. AFE

(Path AFE) Therefore, we will get the following for the work on path AFE.

\begin{align*} W_\text{AFE} = \amp W_\text{AF} + W_\text{FE}\\ \amp = 0 + 0.3\times 1962\times 30 = 17,660 \text{ N.m}. \end{align*}

Solution 3. ABCDE

(Path ABCDE) On path ABCDE we will get the following work.

\begin{align*} W_\text{ABCDE} = \amp W_\text{AB} + W_\text{BC} + W_\text{CD} + W_\text{DE}\\ \amp = 0.1\times 1962\times 30 + 0.1\times 1962\times 10 + 0 + 0.3\times 1962\times 10\\ \amp = 13,730 \text{ N.m}. \end{align*}

Example 8.8. Work By Various Forces in Pulling a Cart.

A cart of mass \(M\) is pulled a distance \(D\) on a flat horizontal surface by a constant force \(F\) that acts at an angle of \(\theta\) with the horizontal direction. The other forces on the object during this time are gravity \((F_w)\text{,}\) Normal forces \((F_{N1})\) and \((F_{N2})\text{,}\) and rolling frictions \(F_{r1}\) and \(F_{r2}\) as shown in Figure 8.9. What is the work done by each force?

Solution.

We can find the work done by each force by using the definition of wwork for constant force on straight line displacement. Let us organize the information of magnitude and directions for each force with respect to the displacement vector in a table, and also present the answer for the work by the force in this table.

Force	Magnitude	Angle with displacement	Work
\(\vec F\)	\(F\)	\(\theta\)	\(FD\cos\theta\)
\(\vec F_w\)	\(Mg\)	\(90^{\circ}\)	\(0\)
\(\vec F_{N1}\)	\(F_{N1}\)	\(90^{\circ}\)	\(0\)
\(\vec F_{N2}\)	\(F_{N2}\)	\(90^{\circ}\)	\(0\)
\(\vec F_{r1}\)	\(F_{r1}\)	\(180^{\circ}\)	\(-F_{r1}D\)
\(\vec F_{r2}\)	\(F_{r2}\)	\(180^{\circ}\)	\(-F_{r2}D\)

Subsection 8.1.2 (Calculus) Fundamental Definition of Work

To define work by any force, not just work by a constant force, and on any path of the motion of the object, we need to use Calculus. Once again, we want work (or total work) by a force \(\vec F \) during an interval \(t_i \) to \(t_f \text{.}\) Suppose the body follows the path shown in the Figure 8.10, under the influence of this force as well as other forces, not shown, but we will find the formula for this force only.

To define work by force \(\vec F \) we first divide the full path into small straight segments. Then, we define the work on each segment by a dot product between the force vector and the displacement of that segment. For instance, we obtain the following work \(\delta W \) on the segment \(\Delta \vec r \) in the figure.

\begin{equation} \delta W = \vec F\, \cdot\, \Delta \vec r,\tag{8.3} \end{equation}

which is obviously \(F_{\parallel} |\Delta \vec r|\text{.}\)

Summing \(\delta W\)’s over the entire path gives us the total work \(W_{if} \) during the interval from \(t_i \) to \(t_f \text{.}\) When the segments become infinitesimally small we write the sum as a line integral, which we decorate with \(\text{path} \) to indicate that the value may depend on the path taken between positions \(\vec r_i\) and \(\vec r_f\text{.}\)

\begin{equation} W_{if} = \int_{\text{path}}\, \vec F\, \cdot\, d\vec r.\tag{8.4} \end{equation}

The integral is an example of a line integral, sometimes also called the work integral or path integral . This integral needs to be evaluated over the path given by the position vector \(\vec r\) at times in the interval \(t_i \) to \(t_f\text{.}\) It is best to think of integral in (8.4) as “conceptual”, which needs to be applied to a given situation to turn into an integral that you can perform. See the exercises below.

Example 8.11. (Calculus) Work Integral for a Path on Axis.

Suppose we have a force, \(\vec F = - 20\, x\, \hat i + 10\, \hat j\text{,}\) which is in units of \(\text{N}\) when \(x\) is in \(\text{m}\text{.}\) What is the work done by this force for a motion on the \(x\) axis, on a direct path from \(x = 2\text{ m}\) to \(x = 5\text{ m}\text{?}\)

Answer.

\(-210\text{ N.m}.\)

Solution.

First, we notice that the infinitesimal displacement vector, \(d\vec r = \hat i\, dx + \hat j\, dy + \hat k\, dz\) on the path is simply along the \(x \) axis.

\begin{equation*} d\vec r = \hat i\, dx, \end{equation*}

obtained by simply setting \(dy = 0\) and \(dz = 0\text{.}\) Therefore, the dot product on the path takes a simple form as well.

\begin{equation*} \vec F\cdot d\vec r = -20\, x \, dx. \end{equation*}

This is now restricted on the path, and we can do the integral over the variable \(x \) as a regular integral.

\begin{equation*} \int_{\text{path}}\,\vec F\cdot d\vec r = \int_2^5\, -20\, x \, dx = -10\left( 5^2-2^2\right) = -210\text{ N.m}, \end{equation*}

where I have put the unit of work in the last step.

Example 8.12. Work by Area Under \(F_x\) vs \(x\).

For simplicity, let path of motion be along \(x\) axis from \(x=x_i\) to \(x=x_f\) with no turn around in the path. Then, work will be only by \(F_x (x_f - x_i)\text{,}\) where \(F_x\)m> is \(x\) component of the force. In this simple setting, work in Eq. (8.4) can also be evaluated by area under the ciurve.

\begin{equation} \text{Work } = \text{ Area under }F_x\text{ versus }x.\tag{8.5} \end{equation}

An example of this calculation is illustrated in Figure 8.13. During the displacement from \(x=0\) to \(x=40\text{ m}\text{,}\) work done is positive, \(+500\text{ N.m}\text{,}\) and during displacement from \(x=40\text{ m}\)to \(x=60\text{ m}\text{,}\) work done is negative, \(-300\text{ N.m}\text{.}\) The net work will be \(200\text{ N.m}\text{.}\)

Figure 8.13. Work by area-under-the-curve method for a motion on \(x\) axis that does not have a turn around in the motion on the axis.

Now, if you have a turn around in the motion, e.g., a ball that is shot up and returns back, you should be careful and find work for each direction separately. Figure 8.14 illustrates an example of a motion that has a turnaround on \(x\)-axis; for comparing purposes, this is same as Figure 8.13 up to \(x=40\text{ m}\text{,}\) but then, the force as well as displacement turns around. During the displacement from \(x=0\) to \(x=40\text{ m}\text{,}\) work done is positive, \(+500\text{ N.m}\text{,}\) and during displacement from \(x=40\text{ m}\)to \(x=20\text{ m}\text{,}\) work done is, again, positive, \(+300\text{ N.m}\text{.}\) The net work will now be \(800\text{ N.m}\text{.}\)

Figure 8.14. Work by area-under-the-curve method for a motion on \(x\) axis that has a turn around in the motion on the axis.

Example 8.15. (Calculus) Work Integral for a Two Dimensional Case.

A friction force acts opposite to the displacement direction and depends on the location. Suppose a kinetic friction force of magnitude, \(F = 10\, x\, y^2\text{,}\) acts on the box when the box is at the coordinates \((x,y)\text{.}\) The unit of the force in \(\text{N}\) when \(x\) and \(y\) are in \(\text{m}\text{.}\)

We want to compare work done by this force on two paths between A and C: A-B-C versus A-C direct. Let the coordinates of these points be \((x_A, y_A) = (0, 2\text{ m})\text{,}\) \((x_B, y_B) = (5\text{ m}, 2\text{ m})\text{,}\) and \((x_C, y_C) = (5\text{ m}, 3\text{ m})\text{.}\)

(a) Find work done by \(F\) on the path A-to-B-to-C.

(b) Find work done on by \(F\) the direct path A-to-C.

Answer.

(a) \(- 817\text{ N.m}\) (b) \(-914\text{ N.m}\)

Solution 1. a

We can break up the calculation into two parts: (i) A-to-B and (ii) B-to-C. We do this because the path descriptions of these two segments are different.

\begin{align*} \amp \text{On AB we have } y = 2,\\ \amp \text{On BC we have } x = 5. \end{align*}

The force and displacement vectors on these paths are:

\begin{align*} \text{On AB:}\amp \\ \amp d\vec r = \hat i\, dx, \\ \amp \vec F = -\left( 10\, x\, y^2\right)_{y=2}\,\hat i = -40\,x\,\hat i .\\ \text{On BC:}\amp \\ \amp d\vec r = \hat j\, dy,\\ \amp \vec F = -\left( 10\, x\, y^2 \right)_{x=5}\hat j = -50\,y^2\,\hat j. \end{align*}

Therefore, we have the following integral on the path AB.

\begin{equation*} W_{AB} = \int_0^5\, -40\,x\, dx = -500\text{ N.m}, \end{equation*}

and on path BC

\begin{equation*} W_{BC} = \int_2^3\, -50\,y^2\, dy = -\dfrac{50}{3} \left(3^3 - 2^3\right) = -317\text{ N.m}. \end{equation*}

Adding the two we get the total work on A-B-C path.

\begin{equation*} W_{ABC} = W_{AB} + W_{BC} = - 817\text{ N.m}. \end{equation*}

Solution 2. (b) and (c)

(b) On path AC, \(x \) and \(y \) are related by

\begin{equation*} y = mx + b, \end{equation*}

with

\begin{equation*} m = \dfrac{3-2}{5-0} = 0.2,\ \ b = 2. \end{equation*}

That is,

\begin{equation*} y = 0.2\, x + 2. \end{equation*}

This can be used to eliminate \(y \) and \(dy\) from equations below.

We use the unit vector for the direction from A to C to write the force vector in an analytic form. This unit vector is

\begin{equation*} \hat u = \dfrac{ (5-0)\hat i + (3-2)\hat j}{ \sqrt{(5-0)^2 + (3-2)^2}} = \dfrac{5}{\sqrt{26}}\, \hat i + \dfrac{1}{\sqrt{26}}\, \hat j. \end{equation*}

Since kinetic friction is pointed in the opposite direction to the displacement, we get the following analytic expression for the force vector, since the magnitude \(10\, x\, y^2\) is positive.

\begin{align*} \vec F \amp = -10\, x\, y^2\, \hat u\\ \amp = -10\, x\, y^2 \left( \dfrac{5}{\sqrt{26}}\, \hat i + \dfrac{1}{\sqrt{26}}\, \hat j \right). \end{align*}

The force and displacement vectors on this path are:

\begin{align*} \amp d\vec r = \hat i\, dx + \hat j\, dy = (\hat i\, + 0.2\, \hat j)\, dx, \\ \amp \vec F = -10\, x\, (0.2x + 2)^2 \left( \dfrac{5}{\sqrt{26}}\, \hat i + \dfrac{1}{\sqrt{26}}\, \hat j \right), \end{align*}

where I used \(dy = 0.2 dx \) on this path so that we can write the work integral as an integral over \(dx\text{.}\)

\begin{equation*} W_{AC} = \int_{AC}\, \vec F\cdot d\vec r = \int_{0}^5\, \left( \text{ something here }\right) dx. \end{equation*}

The dot product is ugly to work out. You can do that on a separate piece of paper. I found the following.

\begin{equation*} W_{AC} = - \dfrac{52}{\sqrt{26}}\, \int_0^5\left(0.04\,x^3 + 0.8\, x^2 + 4\, x \right)dx = -914\text{ N.m}. \end{equation*}

\begin{equation*} \left| W_{AC} \right| \gt \left| W_{ABC} \right|. \end{equation*}

This says that it would cost less energy to go on path A-B-C than to go directly.

Exercises 8.1.3 Exercises

1. Calculating Work by Force of Gravity.

A baseball of mass \(0.3 \text{ kg} \) is lobbed vertically up. It reaches a height of \(20.0\text{ m}\text{.}\)

(a) How much work does the weight of the ball do on the way up?

(b) How much work does the weight of the ball do on the way down?

Hint.

(a) \(\theta = 180^{\circ}\text{,}\) (b) \(\theta = 0^{\circ} \text{.}\)

Answer.

(a) \(-59\text{ N.m} \) (b) \(+59\text{ N.m}\)

Solution 1. a

To compute the work, we need magnitudes of force and displacement vectors, and angle between them.

\begin{align*} \amp F = mg = 0.3\times 9.81 = 2.943\text{ N},\\ \amp r_{if} = h = 20.0\text{ m},\\ \amp\theta = 180^{\circ}. \end{align*}

Now, we just multiply them together to obtain work,

\begin{equation*} W_{if} = 2.943 \times 20.0 \times \cos\, 180^{\circ} = -59\text{ N.m}. \end{equation*}

Solution 2. b

We do not need to do an entirely new calculation here. We just need to notice that the angle between force and displacement now will be \(0 ^{\circ} \) rather than \(180^{\circ}\text{.}\) That would make work positive.

\begin{equation*} W_{if} = +59\text{ N.m}. \end{equation*}

2. Calculating Work of Gravity when No Change in Height.

A baseball of mass \(0.3 \text{ kg} \) is lobbed at \(45^{\circ}\) above the horizontal. It reaches some height on its parabolic path and returns before it is caught at the same height as it was launched from at a horizontal distance of \(40 \text{ m}\text{.}\) Find the work done by the force of weight during the flight.

Hint.

Work by weight on way up is negative and work on way down in positive.

Answer.

Solution.

As shown in Exercise 8.1.3.1, the work when the ball is going up will be negative and equal amount of positive work will be done when the ball is in its way down. The sum of the two works will be zero.

3. Work by Spring Force - Work by Area-Under-the-Curve.

A block attached to a spring is moving on a horizontal surface along \(x\) axis. The \(x\) component of the spring force is given by \(F_x = -k\, x\text{,}\) where \(k\) is the spring constant and \(x\) is the \(x\) coordinate of the block.

What is the work done by the spring force when the block’s position has moved from \(x=x_i\) to \(x=x_f\text{?}\) The plot of \(F_x\) versus \(x\) is given in figure.

Hint.

Use area under the curve. (triangle + rectangle)

Answer.

\(-\dfrac{1}{2} k x_f^2 + \dfrac{1}{2} k x_i^2.\)

Solution.

We need to find the shaded area in the figure. The area is made up of a triangle and a rectangle.

\begin{align*} W \amp = \dfrac{1}{2}\left( -kx_f + k x_i \right) \left(x_f - x_i \right) + k x_i\left( x_f - x_i\right), \\ \amp = -\dfrac{1}{2} k \left(x_f - x_i \right)^2 -kx_ix_f + kx_i^2, \\ \amp = -\dfrac{1}{2} k x_f^2 + kx_fx_i -\dfrac{1}{2}kx_i^2 -kx_ix_f + kx_i^2,\\ \amp = -\dfrac{1}{2} k x_f^2 + \dfrac{1}{2} k x_i^2. \end{align*}

4. (Calculus) Hockey Puck on a Rough Surface with Varying Friction.

A hockey puck of mass \(300\text{ grams}\) is shot across a rough floor with the roughness different at different places, which can be described by a position-dependent coefficient of kinetic friction. For a puck moving along the \(x\)-axis, the coefficient of kinetic friction is the following function of \(x\text{,}\) \(\mu(x) = 0.1 + 0.05\,x\text{,}\) where \(x \) is in meter.

Find work done by the kinetic frictional force on the hockey puck when it has moved (a) from \(x = 0\) to \(x = 2\text{ m}\text{,}\) and (b) from \(x = 2\text{ m}\) to \(x = 4\text{ m}\text{.}\)

Hint.

Use \(F_k(x) = \mu_k(x) F_N\) and integrate \(\int F_k(x) dx \text{.}\)

Answer.

(a) \(-0.883 \text{ N.m}\text{,}\) (b) \(-1.47 \text{ N.m}\text{.}\)

Solution.

From the vertical direction of the free-body diagram of the puck, we have \(N = mg\text{.}\) This gives the following for the kinetic friction force.

\begin{equation*} F_k = \mu_k\, N = \mu_k(x) m g. \end{equation*}

Since \(\vec F_k\) is in the opposite direction of the displacement, we get the followinf for the work integtral.

\begin{align*} W_{if} \amp = -\int_{x_i}^{x_f} F_k(x) dx = -mg\int_{x_i}^{x_f} (0.1 + 0.05\, x) dx, \\ \amp = -mg( 0.1\, x_f - 0.1\, x_i + 0.025\, x_f^2 - 0.025\, x_i^2). \end{align*}

We now use the values of \(x_i\) and \(x_f\) to find the values of \(W_{iuf}\text{.}\)

(a) Here \(x_i=0\) and \(x_f=2\text{.}\) Therefore,

\begin{equation*} W_{if} = -0.3\times 9.81 ( 0.1\times 2 + 0.025 \times 2^2 ) = -0.883\text{ N.m}. \end{equation*}

(b) Here \(x_i=2\) and \(x_f=4\text{.}\) Therefore,

\begin{align*} W_{if} \amp = -0.3\times 9.81 ( 0.1\times 4 - 0.1\times 2 \\ \amp \ \ \ \ + 0.025 \times 4^2 - 0.025 \times 2^2 ) = -1.47\text{ N.m}. \end{align*}

5. Work Done by a Position-Dependent Force in a Plane.

Consider a particle on which several forces act, one of which is known to be a force whose magnitude and direction depends on the position of the particle. In a particular coordinate system, the force is given by \(\vec F_2(x,y) = 2y\hat u_x + 3x\hat u_y\text{,}\) where \(x\) and \(y\) are the coordinates of the position of the particle.

(a) Find the work done by this force when the particle moves from the origin to a point A\((5\ \text{m},0)\) on the \(x\)-axis.

(b) Find the work done by this force, if the particle first moves on the \(y\)-axis from origin to a point B\((0,3\ \text{m})\text{,}\) turns \(90^{\circ}\text{,}\) moves parallel to the \(x\)-axis from point B to point C\((5\ \text{m}, 3\ \text{m})\text{,}\) moves parallel to the \(y\)-axis from C to A, and finally ends up at A.

Solution 1. a

A position-dependent force has values varuing with position \((x,y,z)\) on the path. For instance, here \(F_{2x}(x,y,z) = 2y\text{,}\) which says that \(F_{2x}=0\) on a path on the \(x\)-axis since on that path \(y=0\text{.}\) Let us work out the work on each segment.

On path OA the displacement of the particle is on the \(x\)-axis. Therefore, the displacement vector element will be \(d\vec r = dx\ \hat u_x\text{,}\) with \(y=0\text{.}\) We now take the dot product of the force \(\vec F_{2}\) with \(d\vec r\text{,}\) and set \(y=0\) in the result to restrict the integration to this path.

\begin{equation*} \vec F_2 \cdot d\vec r = 2ydx\Big{\vert}_{y=0} = 0. \end{equation*}

The work done by \(\vec F_2\) when the particle moves from origin to \(x=5\ m\) on the \(x\)-axis is zero.

Solution 2. b

There are three path segments: OB, BC, CA.

SEGMENT OB:

Work done on OB, \(W_\text{OB} = 0\) since \(F_{2y} (x=0) = 0\text{.}\)

SEGMENT BC:

Displacement vector infinitesimal element, \(d\vec r = \hat u_x\ dx\text{.}\) Other condition on the path: \(y = 3\ \text{m}\text{.}\) Dot product of the displacement vector element with force: \(\vec F_2\cdot d\vec r = 2ydx\text{.}\) Evaluating this for \(y = 3\ \text{m}\) we obtain \(\vec F_2\cdot d\vec r = 6dx\text{.}\)

Now, we integrate \(\vec F_2\cdot d\vec r\) for the segment. The integration variable is \(x\text{,}\) which changes from 0 to \(5\ \text{m}\) over the segment. Therefore, the work on BC segment is

\begin{equation*} W_\text{BC} = \int_0^{5\ \text{m}} 6 dx = 30\ \text{N.m}. \end{equation*}

SEGMENT CA:

We repeat the process for this segment. Displacement vector infinitesimal element, \(d\vec r = dy\ \hat u_y\text{.}\) Other condition on the path: \(x = 5\ \text{m}\text{.}\) Dot product of displacement vector element with force: \(\vec F_2\cdot d\vec r = 3xdy\text{.}\) Evaluating this for \(x = 5\ \text{m}\) we obtain \(\vec F_2\cdot d\vec r = 15\ dy\text{.}\)

Now, we integrate this for the segment. The integration variable is \(y\text{,}\) which changes from \(3\ m\) to \(0\) over the segment. Therefore, work on BC segment is

\begin{equation*} W_\text{CA} = \int_{3\text{m}}^0 15dy = -45\ \text{N.m}. \end{equation*}

The net work done on the path O-B-C-A path is

\begin{equation*} W_\text{OBCA} = W_\text{OB} + W_\text{BC} + W_\text{CA} = -15\ \text{N.m}. \end{equation*}

We find that work done on path OA is different from the work on path OBCA. If the work of a force depends on the path we say that the force is non-conservative. Therefore, the force \(\vec F_2\) given here is an example of a non-conservative force.

6. Work by Gravity on Pendulum Bob.

In the motion of a pendulum the direction between the weight and the displacement of pendulum changes along the path of the pendulum. Consider a pendulum of mass \(m\) and length \(L\text{.}\) Find the work done by the force of gravity (i.e. weight) of the pendulum for a displacement of the pendulum from a position where the suspending thread makes an angle \(\theta_0\) with respect to the vertical axis to the point where the pendulum is hanging vertically.

Solution.

Notice that, although weight \(mg\) of the pendulum is constant, it does different amount of work on different segments \(d\vec r\) along the path of the pendulum since the angle between weight vector and \(d\vec r\) changes along the path. Therefore, we must use the fundamental definition of work in terms of the line integral.

To set up the work integral, we look at an infinitesimal displacement at an arbitrary displacement in its motion, say between the suspension angles \(\theta\) and \(\theta+d\theta\text{,}\) where \(d\theta\) is negative for the motion under study. The displacement vector is in the direction of velocity and has the magnitude of the corresponding arc, \(-Ld\theta\text{,}\) which is positive since \(d\theta \lt 0\text{.}\) Since the direction of the displacement is tangent to the circle of motion, the angle between the weight and the displacement vector is equal to \(\left( 90^{\circ}-\theta \right)\) as shown in Figure 8.22.

Therefore, the work done by the weight during the infinitesimal displacement under consideration is

\begin{equation*} dW = (mg)(-Ld\theta) \cos(90^{\circ}-\theta) = -mgL\sin\theta\ d\theta. \end{equation*}

Now, we integrate from \(\theta = \theta_0\) to \(\theta = 0\) to obtain the required work.

\begin{equation*} W = \int_{\theta_0}^{0} [-mgL\sin\theta]\ d\theta = mgL(1-\cos\theta_0). \end{equation*}

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