Tag Archives: Physics

College Physics by Openstax Chapter 7 Problem 2

A 75.0-kg person climbs stairs, gaining 2.50 meters in height. Find the work done to accomplish this task. (Neglect friction in your calculations.)

Solution:

Work done against gravity in lifting an object becomes potential energy of the object-Earth system. The change in gravitational potential energy is \Delta PE_{g} = mgh, with h being the increase in height and g the acceleration due to gravity.

W=mgh

We are given the following values: m=75.0\ \text{kg}, g=9.80\ \text{m/s}^2, and h=2.50\ \text{m}.

Substitute the given in the formula.

\begin{align*}
W & = mgh \\
W & = \left( 75.0\ \text{kg} \right)\left( 9.80\ \text{m/s}^2 \right)\left( 2.50\ \text{m} \right)\\
W & = 1837.5\ \text{Nm} \\
W & = 1837.5\ \text{J} \\
W & = 1.84 \times 10 ^{3} \ \text{J} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The work done is about 1.84 \times 10 ^ {3}\ \text{Joules} .

College Physics by Openstax Chapter 7 Problem 1

How much work does a supermarket checkout attendant do on a can of soup he pushes 0.600 m horizontally with a force of 5.00 N? Express your answer in joules and kilocalories.

Solution:

The work W that a force F does on an object is the product of the magnitude F of the force, times the magnitude d of the displacement, times the cosine of the angle \theta between them. In symbols,

W=Fd \cos \theta

We are given the following values: F=5.00\ \text{N}, d=0.600\ \text{m}, and \theta=0^\circ.

Substitute the given values in the formula for work.

\begin{align*}
W & = Fd \cos \theta \\
W & = \left( 5.00\ \text{N} \right)\left( 0.600\ \text{m} \right) \cos 0^\circ \\
W & = 3.00\ \text{Nm} \\
W & = 3.00\ \text{J} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The work done is 3.00 Joules. Now, we can convert this in unit of kilocalories knowing that 1\ \text{kcal} = 4186\ \text{J}. 

\begin{align*}
W & = 3.00\ \text{J} \\
W & = 3.00\ \text{J}\ \times \ \frac{1\ \text{kcal}}{4186\ \text{J}} \\
W & = 0.000717\ \text{kcal} \\
W & = 7.17 \times 10 ^{-4} \ \text{kcal} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The work done in kilocalories is about 7.17 \times 10 ^{-4}.

Problem 6-19: The angular velocity of an “artificial gravity”

A rotating space station is said to create “artificial gravity”—a loosely-defined term used for an acceleration that would be crudely similar to gravity. The outer wall of the rotating space station would become a floor for the astronauts, and centripetal acceleration supplied by the floor would allow astronauts to exercise and maintain muscle and bone strength more naturally than in non-rotating space environments. If the space station is 200 m in diameter, what angular velocity would produce an “artificial gravity” of 9.80 m/s2 at the rim?

Solution:

We are given the following quantities:

\text{radius} = \frac{\text{diameter}}{2} = \frac{200\ \text{m}}{2} = 100 \ \text{m}
\text{centripetal acceleration}, a_c = 9.80 \ \text{m/s}^2

Centripetal acceleration is the acceleration experienced while in uniform circular motion. It always points toward the center of rotation. The formula for centripetal acceleration is

a_{c} = r \omega ^2

If we solve for the angular velocity in terms of the other quantities, we have

\omega = \sqrt{\frac{a_c}{r}}

Substituting the given quantities,

\begin{align*}
\omega & = \sqrt{\frac{a_c}{r}}  \\ \\
\omega & = \sqrt{\frac{9.80 \ \text{m/s}^2}{100\ \text{m}}} \\ \\
\omega & = 0.313 \ \text{rad/sec} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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College Physics by Openstax Chapter 3 Problem 4

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 \vec{A} and \vec{B} , as in Figure 3.53, then this problem asks you to find their sum \vec{R}=\vec{A}+\vec{B} .)

Figure 3.53

Solution:

Figure 3.4A

Consider Figure 3.54A.

The resultant of the two vectors \vec{A} and \vec{B} is labeled \vec{R}. This \vec{R} is directed \theta ^{\circ} from the x-axis.

We shall use the right triangle formed to solve for the unknowns.

Solve for the magnitude of the resultant.

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

Solve for the value of \theta .

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

We need the complementary angle for the compass angle. 

\begin{align*}
90^\circ -54.2^\circ =35.8^\circ 
\end{align*}

Therefore, the compass angle reading is

\begin{align*}
35.8^\circ , \text{W of N} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}
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College Physics by Openstax Chapter 3 Problem 2

Find the following for path B in Figure 3.52:
(a) The total distance traveled, and
(b) The magnitude and direction of the displacement from start to finish.

Figure 3.54 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side
Figure 3.52 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side

Solution:

Part A

The total distance traveled is 

\begin{align*}

\text{d} & = \left(4 \times 120 \text{m} \right) + \left(3 \times 120\ \text{m} \right) + \left(3 \times 120\ \text{m} \right) \\
\text{d} & = 1 200\ \text{m}  \ \qquad \  {\color{DarkOrange} \left( \text{Answer} \right)}\\

\end{align*}

Part B

The magnitude of the displacement is 

\begin{align*}
\text{s} & = \sqrt{\left( s_x \right)^2+\left( s_y \right)^2} \\
\text{s} & = \sqrt{\left( 1 \times 120\ \text{m} \right)^2+ \left( 3 \times 120 \ \text{m} \right)^2} \\
\text{s} & = \sqrt{\left( 120\ \text{m} \right)^2+ \left( 360 \ \text{m} \right)^2} \\
\text{s} & = 379 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

The direction is

\begin{align*}
\theta & = \arctan \left( \frac{s_y}{s_x} \right) \\
\theta & = \arctan \left( \frac{360\ \text{m}}{120\ \text{m}} \right) \\
\theta & = 71.6^\circ , \ \text{N of E} \ \qquad {\color{DarkOrange} \left( \text{Answer} \right)}\\
\end{align*}
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College Physics 2.51 – Time of the hiker to move out from a falling rock

Standing at the base of one of the cliffs of Mt. Arapiles in Victoria, Australia, a hiker hears a rock break loose from a height of 105 m. He can’t see the rock right away but then does, 1.50 s later. (a) How far above the hiker is the rock when he can see it? (b) How much time does he have to move before the rock hits his head?

Solution:

Part A

We know that the initial height, y_0 of the rock is 105 meters, and the initial velocity, v_0 is zero. We shall solve for the distance traveled by the rock for 1.5 seconds from the initial position first to find the height at detection.

The change in height is

\displaystyle \begin{aligned}
\Delta \text{y}&=\text{v}_0\text{t}+\frac{1}{2}\text{at}^2 \\
&=\left( 0  \right)\left( 1.50 \ \text{s} \right)+\frac{1}{2}\left( 9.81\ \text{m/s}^{2} \right)\left( 1.50\ \text{s} \right)^{2}\\
&=0+11.036\ \text{m} \\
&=11.04 \ \text{m} 
\end{aligned}

So, the rock falls about 11.04 m from the initial height for 1.50 seconds. Therefore, the height of the rock above his head at this point is

\displaystyle \begin{aligned}
\text{y}&=\text{y}_{0}-\Delta \text{y} \\
&=105\ \text{m}-11.04\ \text{m} \\
&=93.96 \ \text{m}
\end{aligned}

Part B

We shall solve for the total time of travel, that is, from the initial position to his head. Then we shall subtract 1.50 s from that to solve for the unknown time of moving out. The total time of travel is

\begin{aligned}
\text{y} & =\frac{1}{2}\text{at}^{2} \\
&\text {Solving for t, we have}\\
\text{t}&=\sqrt{\frac{\text{2y}}{\text{a}}} \\
&=\sqrt{\frac{2\left( 105\ \text{m} \right)}{9.81 \ \text{m/s}^{2}}} \\
&=4.63 \ \text{s}

\end{aligned}

Therefore, to move out the hiker has about

\begin{aligned}
\text{t}&=4.63 \ \text{s}-1.50\ \text{s}\\
&=3.13\ \text{s}
\end{aligned}

College Physics 2.50 – Motion of a Jumping Kangaroo

A kangaroo can jump over an object 2.50 m high. (a) Calculate its vertical speed when it leaves the ground. (b) How long is it in the air?

Part A

The motion of the kangaroo is under free-fall. We are looking for the initial velocity, and we know that the velocity in the highest position is zero.

From 

\begin{aligned}
\text{v}^2 &=\left (\text{v}_0 \right )^2+2\text{ay},\\
\end{aligned}

we have

\begin{aligned}
\text{v}^2 &=\left (\text{v}_0 \right )^2+2\text{ay}\\
\text{v}^2-2\text{ay} &= \left ( \text{v}_0\right)^2\\
\text{v}_0&=\sqrt{\text{v}^2-2\text{ay}}
\end{aligned}

Substituting the known values, 

\begin{aligned}
\text{v}_0&=\sqrt{\text{v}^2-2\text{ay}} \\
\text{v}_0&=\sqrt{0^2-2\left(-9.81 \text{m/s}^2\right)\left(2.50 \text{m}\right)}\\
\text{v}_0&= {\color{green}7.00 \  \text{m/s}}
\end{aligned}

Therefore, the vertical speed of the kangaroo when it leaves the ground is 7.00 m/s.

Part B

Since the motion of the kangaroo has uniform acceleration, we can use the formula

\text{y}=\text{v}_o\text{t}+\frac{1}{2}\text{a}\text{t}^2

The initial and final position of the kangaroo will be the same, so y is equal to zero. The initial velocity is 7.00 m/s, and the acceleration is -9.81 m/s2.

\begin{aligned}
\text{y} & =\text{v}_0\text{t}+\frac{1}{2}\text{a}\text{t}^2\\
0 & = \left( 7.00\ \text{m/s} \right)\text{t}+\frac{1}{2}\left( -9.81\ \text{m/s}^{2} \right)\text{t}^2\\
0 & =7\text{t}-4.905\text{t}^{2}\\
7\text{t}-4.905\text{t}^{2}&=0 \\
\text{t}\left( 7-4.905\text{t} \right) & =0 \\
\text{t}=0 \qquad &\text{or} \qquad 7-4.905\text{t}=0 \\

\end{aligned}

Discard the time 0 since this refers to the beginning of motion. Therefore, we have

\begin{aligned}
7-4.905\text{t} &=0 \\
4.905\text{t} & = 7 \\
\text{t} & =\frac{7}{4.905} \\
 \text{t}&={\color{green}1.43 \  \text{s}} 
\end{aligned}

The kangaroo is about 1.43 seconds long in the air.

Solution Guides to College Physics by Openstax Chapter 13 Banner

Chapter 13: Temperature, Kinetic Theory, and the Gas Laws

Temperature

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

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Thermal Expansion of Solids and Liquids

Problem 9

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Problem 15

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The Ideal Gas Law

Problem 22

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Kinetic Theory: Atomic and Molecular Explanation of Pressure and Temperature

Problem 39

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Problem 44

Problem 45

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Humidity, Evaporation, and Boiling

Problem 49

Problem 50

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Solution Guides to College Physics by Openstax Chapter 12 Banner

Chapter 12: Fluid Dynamics and Its Biological and Medical Applications

Flow Rate and Its Relation to Velocity

Problem 1

Problem 2

Problem 3

Problem 4

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Problem 7

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Bernoulli’s Equation

Problem 17

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The Most General Applications of Bernoulli’s Equation

Problem 25

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Viscosity and Laminar Flow; Poiseuille’s Law

Problem 29

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The Onset of Turbulence

Problem 51

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Molecular Transport Phenomena: Diffusion, Osmosis, and Related Processes

Problem 62

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Solution Guides to College Physics by Openstax Chapter 7 Banner

Chapter 7: Work, Energy, and Energy Resources

Work: The Scientific Definition

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Kinetic Energy and the Work-Energy Theorem

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Gravitational Potential Energy

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Conservative Forces and Potential Energy

Problem 22

Problem 23

Nonconservative Forces

Problem 24

Problem 25

Conservation of Energy

Problem 26

Problem 27

Problem 28

Problem 29

Power

Problem 30

Problem 31

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Work, Energy, and Power in Humans

Problem 44

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World Energy Use

Problem 60

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