Tag Archives: College Physics

College Physics by Openstax Chapter 4 Problem 3

A cleaner pushes a 4.50-kg laundry cart in such a way that the net external force on it is 60.0 N. Calculate its acceleration.

Solution:

From Newton’s Second Law of Motion, $\sum F =ma$ . Substituting the given values, we have

\begin{align*} \sum F & = ma \\ 60.0 \ \text{N} & = \left( 4.50 \ \text{kg} \right) \ a \\ a & = \frac{60.0 \ \text{N}}{4.50 \ \text{kg}} \\ a & = 13.3 \ \text{m/s}^{2} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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College Physics by Openstax Chapter 4 Problem 2

If the sprinter from the previous problem accelerates at that rate for 20 m, and then maintains that velocity for the remainder of the 100-m dash, what will be his time for the race?

Solution:

Solving for the time it takes to reach the first 20 meters.

\begin{align*} \Delta x & =v_{0}t+\frac{1}{2}at^{2} \\ 20 \ \text{m} & = \left( 0 \ \text{m/s} \right)t + \frac{1}{2}\left( 4.20 \ \text{m/s}^{2} \right)t^{2} \\ 20 & = 2.1 t^{2} \\ \frac{20}{2.1}& = \frac{\cancel{2.1} \ t^{2}}{\cancel{2.1}} \\ t^{2} & = 9.5238 \\ \sqrt{t^{2}} & = \sqrt{9.5238} \\ t_{1} & = 3.09 \ \text{s} \end{align*}

We can compute the velocity of the sprinter at the end of the first 20 meters.

\begin{align*} v^2 & = v_{0}^2 + 2ax \\ v^2 & = \left( 0 \ \text{m/s} \right)^2 + 2\left( 4.20 \ \text{m/s}^2 \right) \left( 20 \ \text{m} \right) \\ v & = \sqrt{2\left( 4.20 \right)\left( 20 \right)} \ \text{m/s} \\ v & = 12.96 \ \text{m/s} \end{align*}

For the remaining 80 meters, the sprinter has a constant velocity of 12.96 m/s. The sprinter’s time to run the last 80 meters can be computed as follows.

\begin{align*} \Delta x & = vt \\ 80 \ \text{m} & = \left( 12.96 \ \text{m/s} \right)\ t_{2} \\ \frac{80 \ \text{m}}{12.96 \ \text{m/s}} & =\frac{\cancel{12.96 \ \text{m/s} } \ \ t_{2}}{\cancel{12.96 \ \text{m/s}}} \\ t_{2} & = \frac{80}{12.96} \ \text{s} \\ t_{2} & = 6.17 \ \text{s} \\ \end{align*}

The sprinter’s total time, $t_{T}$ , to finish the 100-m race is the sum of the two times.

\begin{align*} t_{T} & = t_{1} + t_{2} \\ t_{T} & = 3.09 \ \text{s} + 6.17 \ \text{s} \\ t_{T} & = 9.26 \ \text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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College Physics by Openstax Chapter 3 Problem 22

A farmer wants to fence off his four-sided plot of flat land. He measures the first three sides, shown as A, B, and C in Figure 3.60, and then correctly calculates the length and orientation of the fourth side D. What is his result?

Solution:

For the four-sided plot to be closed, the resultant displacement of the four sides should be zero. The sum of the horizontal components should be zero, and the sum of the vertical components should also be equal to zero.

We need to solve for the components of each vector. Take into consideration that rightward and upward components are positive, while the reverse is negative.

For vector A, the components are

\begin{align*} A_x & = \left( 4.70 \ \text{km} \right) \cos 7.5^\circ \\ A_x & = 4.6598 \ \text{km} \end{align*}

\begin{align*} A_y & = -\left( 4.70 \ \text{km} \right) \sin 7.5^\circ \\ A_y & = -0.6135 \ \text{km} \end{align*}

The components of vector B are

\begin{align*} B_x & =-\left( 2.48 \ \text{km} \right) \sin 16^\circ \\ B_x & = -0.6836 \ \text{km} \end{align*}

\begin{align*} B_y & =\left( 2.48 \ \text{km} \right) \cos 16^\circ \\ B_y & =2.3839 \ \text{km} \end{align*}

For vector C, the components are

\begin{align*} C_x & = -\left( 3.02 \ \text{km} \right) \cos 19^\circ \\ C_x & = -2.8555 \ \text{km} \end{align*}

\begin{align*} C_y & = \left( 3.02 \ \text{km} \right) \sin 19^\circ \\ C_y & = 0.9832 \ \text{km} \end{align*}

Now, we need to take the sum of the x-components and equate it to zero. The x-component of D is unknown.

\begin{align*} A_x+B_x+C_x+D_x & =0 \\ 4.6598 \ \text{km}-0.6836 \ \text{km}-2.8555 \ \text{km}+ D_x & =0 \\ 1.1207 \ \text{km} +D_x & =0 \\ D_x & = -1.1207 \ \text{km} \end{align*}

We also need to take the sum of the y-component and equate it to zero to solve for the y-component of D.

\begin{align*} A_y +B_y+C_y+D_y & =0 \\ -0.6135 \ \text{km}+2.3839 \ \text{km}+0.9832 \ \text{km}+ D_y & =0 \\ 2.7536 \ \text{km} +D_y & =0 \\ D_y & = -2.7536 \ \text{km} \end{align*}

To solve for the distance of D, we shall use the Pythagorean Theorem.

\begin{align*} D & = \sqrt{\left( D_x \right)^2+\left( D_y \right)^2} \\ D & = \sqrt{\left( -1.1207 \ \text{km} \right)^2+\left( -2.7536 \ \text{km} \right)^2} \\ D & = 2.97 \ \text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

Then we can solve for θ using the tangent function. Since it is taken from the vertical axis, it can be solved by:

\begin{align*} \theta & = \tan^{-1} \left| \frac{D_x}{D_y} \right| \\ \theta & = \tan^{-1} \left| \frac{-1.1207 \ \text{km}}{-2.7536 \ \text{km}} \right| \\ \theta & = 22.1 ^ \circ \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

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College Physics by Openstax Chapter 3 Problem 21

You fly 32.0 km in a straight line in still air in the direction 35.0º south of west. (a) Find the distances you would have to fly straight south and then straight west to arrive at the same point. (This determination is equivalent to finding the components of the displacement along the south and west directions.) (b) Find the distances you would have to fly first in a direction 45.0º south of west and then in a direction 45.0º west of north. These are the components of the displacement along a different set of axes—one rotated 45º.

Solution:

Part A

Consider the illustration shown.

The south and west components of the 32.0 km distance are denoted by D_S and D_W, respectively. The values of these components are solved below:

\begin{align*} D_S & = \left( 32.0\ \text{km} \right) \sin 35.0 ^\circ \\ D_S & = 18.4^\circ \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

\begin{align*} D_W & = \left( 32.0\ \text{km} \right) \cos 35.0 ^\circ \\ D_W & = 26.2^\circ \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

Part B

Consider the new set of axes (X-Y) as shown below. This new set of axes is rotated 45° from the original axes. Thus, axis X is 45° south of west, and axis Y is 45° west of north. First, we can obviously see that θ has a value of 10°.

Therefore, the components of the 32.0 km distance along X and Y axes are:

\begin{align*} D_X & = \left( 32.0 \ \text{km} \right) \cos 10^\circ \\ D_X & = 31.5^\circ \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

\begin{align*} D_Y & = \left( 32.0 \ \text{km} \right) \sin 10^\circ \\ D_Y & = 5.56^\circ \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

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College Physics by Openstax Chapter 3 Problem 20

A new landowner has a triangular piece of flat land she wishes to fence. Starting at the west corner, she measures the first side to be 80.0 m long and the next to be 105 m. These sides are represented as displacement vectors A from B in Figure 3.59. She then correctly calculates the length and orientation of the third side C. What is her result?

Solution:

Consider the illustration shown.

We need to solve for an interior angle of the triangle. So, we need to solve for the value of α first. This can be done by simply subtracting the sum of 21 and 11 degrees from 90 degrees.

\begin{align*} \alpha & = 90 ^ \circ -\left( 21^\circ +11^\circ \right) \\ \alpha & = 58^\circ \end{align*}

Then, using the cosine law, we can now solve for the magnitude of vector C. That is

\begin{align*} C^2 & = A^2 + B^2 - 2AB \cos \alpha \\ C^2 & = \left( 80\ \text{m} \right)^2+\left( 105\ \text{m} \right)^2-2\left( 80\ \text{m} \right)\left( 105\ \text{m} \right) \cos 58^\circ \\ C^2 & = 8522.3564 \\ C & = \sqrt{8522.3564} \\ C & = 92.3 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

Before we can solve for the value of θ, we need to know the value of β first. This can be done by using the sine law.

\begin{align*} \frac{\sin \beta}{80\ \text{m}} & = \frac{\sin 58^\circ }{92.3 \ \text{m}} \\ \sin \beta & = \frac{\left( 80 \ \text{m} \right)\sin 58^\circ }{92.3 \ \text{m}} \\ \beta & = \arcsin \left[ \frac{\left( 80 \ \text{m} \right)\sin 58^\circ }{92.3 \ \text{m}} \right] \\ \beta & = 47.3^\circ \end{align*}

Finally, we can solve for θ.

\begin{align*} \theta & = \left( 90 ^\circ +11^\circ \right) - \beta \\ \theta & = \left( 90 ^\circ +11^\circ \right) - 47.3^\circ \\ \theta & = 53.7^\circ \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

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College Physics by Openstax Chapter 3 Problem 18

You drive 7.50 km in a straight line in a direction 15º east of north. (a) Find the distances you would have to drive straight east and then straight north to arrive at the same point. (This determination is equivalent to find the components of the displacement along the east and north directions.) (b) Show that you still arrive at the same point if the east and north legs are reversed in order.

Solution:

Part A

Consider the illustration shown.

Let D_Ebe the east component of the distance, and D_N be the north component of the distance.

\begin{align*} D_E & = 7.50 \ \sin 15^\circ \\ D_E & = 1.9411 \ \text{km} \\ D_E & = 1.94 \ \text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

\begin{align*} D_N & = 7.50 \ \cos 15^\circ \\ D_N & = 7.2444\ \text{km} \\ D_N & = 7.24 \ \text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

Part B

It can be obviously seen from the figure below that you still arrive at the same point if the east and north legs are reversed in order.

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College Physics by Openstax Chapter 3 Problem 17

Repeat Exercise 3.16 using analytical techniques, but reverse the order of the two legs of the walk and show that you get the same final result. (This problem shows that adding them in reverse order gives the same result—that is, B + A = A + B .) Discuss how taking another path to reach the same point might help to overcome an obstacle blocking your other path.

**Figure 3.58** The two displacements A and B add to give a total displacement R having magnitude R and direction θ.

Solution:

Considering the right triangle formed by the vectors A, B, and R. We can solve for the magnitude of R using the Pythagorean Theorem. That is

\begin{align*} R & = \sqrt{A^2+B^2} \\ & = \sqrt{\left( 18.0 \text{m} \right)^2+\left( 25.0 \text{m} \right)^2} \\ & =30.806 \text{m} \\ & \approx 30.8 \text{m} \qquad {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

Then we solve for the compass direction by solving the value θ using the same right triangle.

\begin{align*} \theta & = \arctan \left( \frac{B}{A} \right) \\ & = \arctan \left( \frac{25.0 \text{m}}{18.0 \text{m}} \right) \\ & = 54.246 ^\circ \\ & \approx 54.2 ^ \circ \\ \end{align*}

Therefore, the compass direction of the resultant is 54.2° North of West.

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College Physics by Openstax Chapter 3 Problem 16

Solve the following problem using analytical techniques: Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements A and B, as in Figure 3.58, then this problem asks you to find their sum R=A+B.)

Solution:

Considering the right triangle formed by the vectors A, B, and R. We can solve for the magnitude of R using the Pythagorean Theorem. That is

\begin{align*} R & = \sqrt{A^2+B^2} \\ & = \sqrt{\left( 18.0\ \text{m} \right)^2+\left( 25.0\ \text{m} \right)^2} \\ & =30.806 \ \text{m} \\ & \approx 30.8 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

Then we solve for the compass direction by solving the value θ using the same right triangle.

\begin{align*} \theta & = \arctan\left( \frac{B}{A} \right) \\ & = \arctan\left( \frac{25.0\ \text{m}}{18.0\ \text{m}} \right) \\ & = 54.246 ^\circ \\ & \approx 54.2 ^ \circ \end{align*}

Therefore, the compass direction of the resultant is 54.2° North of West.

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College Physics by Openstax Chapter 3 Problem 15

Find the north and east components of the displacement from San Francisco to Sacramento shown in Figure 3.57.

Solution:

Consider the following figure.

Using the right triangle formed, we can solve for the east component and the north component. Let S_E be the east component and S_N be the north component of S.

\begin{align*} S_E & = S \cos 45^\circ \\ & = \left( 123 \ \text{km} \right) \cos45^\circ \\ & = 86.974 \ \text{km} \\ & \approx 87.0 \ \text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

\begin{align*} S_N & = S \sin 45^\circ \\ & = \left( 123 \ \text{km} \right) \sin 45^\circ \\ & = 86.974 \ \text{km} \\ & \approx 87.0 \ \text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

Therefore, the east and north components are equal at about 87.0 km.

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College Physics by Openstax Chapter 3 Problem 14

Find the following for path D in Figure 3.56: (a) the total distance traveled and (b) the magnitude and direction of the displacement from start to finish. In this part of the problem, explicitly show how you follow the steps of the analytical method of vector addition.

**Figure 3.56** The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side.

Solution:

Part A

Looking at path D, we can see that it moves 2 blocks downward, 6 blocks to the right, 4 blocks upward, and 1 block to the left. Thus, the total distance of path D is

\begin{align*} \text{distance} & = \left( 2\times 120\ \text{m} \right)+\left( 6\times 120\ \text{m} \right)+\left( 4\times 120\ \text{m} \right)+\left( 1\times 120\ \text{m} \right) \\ & = 1\ 560 \ \text{m} \\ & = 1.56 \times 10^{3} \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

Part B

Looking at the initial and final position of path D, the final position is 5 blocks to the right or 600 meters to the right of the initial position, and 2 blocks or 240 meters upward from the initial position. Refer to the figure below.

Using the right triangle, we can solve for the displacement using the Pythagorean Theorem.

\begin{align*} \text{displacement} & = \sqrt{\left( 600\ \text{m} \right)^2+\left( 240\ \text{m} \right)^2} \\ & = 646.2198 \ \text{m} \\ & \approx 646 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

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Tag Archives: College Physics

College Physics by Openstax Chapter 4 Problem 3

A cleaner pushes a 4.50-kg laundry cart in such a way that the net external force on it is 60.0 N. Calculate its acceleration.

College Physics by Openstax Chapter 4 Problem 2

If the sprinter from the previous problem accelerates at that rate for 20 m, and then maintains that velocity for the remainder of the 100-m dash, what will be his time for the race?

College Physics by Openstax Chapter 3 Problem 22

A farmer wants to fence off his four-sided plot of flat land. He measures the first three sides, shown as A, B, and C in Figure 3.60, and then correctly calculates the length and orientation of the fourth side D. What is his result?

College Physics by Openstax Chapter 3 Problem 21

College Physics by Openstax Chapter 3 Problem 20

College Physics by Openstax Chapter 3 Problem 18

College Physics by Openstax Chapter 3 Problem 17

College Physics by Openstax Chapter 3 Problem 16

College Physics by Openstax Chapter 3 Problem 15

Find the north and east components of the displacement from San Francisco to Sacramento shown in Figure 3.57.

College Physics by Openstax Chapter 3 Problem 14

Find the following for path D in Figure 3.56: (a) the total distance traveled and (b) the magnitude and direction of the displacement from start to finish. In this part of the problem, explicitly show how you follow the steps of the analytical method of vector addition.