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College Physics by Openstax Chapter 2 Problem 52


An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b) Determine the final velocity at which the object hits the ground. (c) Determine the distance traveled during the last second of motion before hitting the ground.


Solution:

Consider Figure 1. The object was dropped from a height of 75.0 m. At the start of motion, the velocity is zero, v_{oy}=0.

The object traveled for a period of time t for the whole 75.0 m distance to the ground.

Part A

We are solving for the distance traveled by the object for the first 1 second. So, we have

\begin{align*}
\Delta y & = v_{oy} t + \frac{1}{2}at^2\\
\Delta y & = \left( 0 \ \text{m/s} \right)\left( 1 \ \text{s} \right)+\frac{1}{2}\left( -9.81 \ \text{m/s}^2 \right)\left( 1 \ \text{s} \right)^2 \\
\Delta y & = 0 -4.905 \ \text{m} \\
\Delta y & = -4.91 \ \text{m} \\
|\Delta y| & =4.91 \ \text{m}
\end{align*}

The negative sign of Δy indicates that the direction of the displacement is downward. Since we are looking for the scalar value of the distance, the answer is 4.91 m.

Part B

So we now consider the two positions of the object as shown in the figure to the right. The initial height of the object is 75.0 m above the ground, and the initial velocity is 0.

At the ground, we know that the position of the object is 0 m above the ground, but we do not know the time and velocity. Therefore, to determine the velocity of the object at this point, we proceed as follows:

\begin{align*}
\left( v_2 \right)^2 & =\left( v_1 \right)^2+2a \Delta y \\
\left( v_2 \right)^2 & =\left( v_1 \right)^2+2a \left( y_2-y_1 \right) \\
v_2 & = \pm \sqrt{\left( v_1 \right)^2+2a \left( y_2-y_1 \right)}\\
v_2 & = \pm \sqrt{\left( 0\ \text{m/s} \right)^2+2\left( -9.81\ \text{m/s}^2 \right)\left( 0 \ \text{m}-75\ \text{m} \right)} \\
v_2 & = \pm \ 38.4\ \text{m/s}\\
v_2 & =- 38.4\ \text{m/s}\\
\end{align*}

Since the object is directing downwards when it hit the ground, the velocity is negative.

Part C

First, we calculate the total time of the object’s motion from the beginning to the ground.

\begin{align*}
\Delta y & =\bcancel{v_{oy}t}+ \frac{1}{2}at^2 \\
\Delta y & = \frac{1}{2}at^2 \\
0 \text{m}-75\ \text{m} & = \frac{1}{2}\left( -9.81\ \text{m/s}^2 \right)t^2 \\
t^2& =\frac{-75\ \text{m}}{-4.905 \text{m/s}^2}\\
t&=\sqrt{\frac{75\ \text{m}}{4.905 \text{m/s}^2}}\\
t&=3.91 \ \text{s}
\end{align*}

Second, determine the total distance traveled from 0 s to 2.91 s, leaving out the last 1 s of the motion.

\begin{align*}
\Delta y & = v_{oy} t + \frac{1}{2}at^2\\
\Delta y & = \left( 0 \ \text{m/s} \right)\left( 1 \ \text{s} \right)+\frac{1}{2}\left( -9.81 \ \text{m/s}^2 \right)\left( 2.91 \ \text{s} \right)^2 \\
\Delta y & = 0 -41.5 \ \text{m} \\
\Delta y & = -41.5 \ \text{m} \\
|\Delta y| & =41.5 \ \text{m}
\end{align*}

Finally, subtract this distance from the total distance traveled to get the distance traveled in the last 1 second.

\begin{align*}
y_{_{\text{last 1 sec}}} & = 75.0 \ \text{m}-41.5 \ \text{m} \\
y_{_{\text{last 1 sec}}} & = 33.5\ \text{m}
\end{align*}

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Skidding on a Curve: Unbanked Curves – Uniform Circular Motion Example Problem

A 1000-kg car rounds a curve on a flat road of radius 50 m at a speed of 15 m/s Will the car follow the curve, or will it skid? Assume: (a) the pavement is dry and the coefficient of static friction is \mu _s=0.60; (b) the pavement is icy and \mu _s=0.25.


Solution:

The forces on the car are gravity mg downward, the normal force FN exerted upward by the road, and a horizontal friction force due to the road. They are shown in the free-body diagram of the car below. The car will follow the curve if the maximum static friction force is greater than the mass times the centripetal acceleration.

Part A

In the vertical direction (y) there is no acceleration. Newton’s second law tells us that the normal force on the car is equal to the weight mg since the road is flat:

\begin{align*}
\sum_{}^{}F_y & =m\ a_c\\
\\
F_N\ -\ mg & =0\\
\\
F_N&=mg\\
\\
F_N &=\left( 1000\ \text{kg} \right)\left( 9.81\ \text{m/s}^2 \right)\\
\\
F_N &=9810\  \text{N}
\end{align*}

In the horizontal direction the only force is friction, and we must compare it to the force needed to produce the centripetal acceleration to see if it is sufficient. The net horizontal force required to keep the car moving in a circle around the curve is

\begin{align*}
\sum_{}^{}F_c & =m\ a_c\\
\\
 & =m\cdot \frac{v^2}{r}\\
\\
& =\left( 1000\ \text{kg} \right)\cdot \frac{\left( 15\ \text{m/s} \right)^2}{50\ \text{m}}\\
\\
&=4500\ \text{N}
\end{align*}

Now we compute the maximum total static friction force (the sum of the friction forces acting on each of the four tires) to see if it can be large enough to provide a safe centripetal acceleration. For (a), \mu _s=0.60, and the maximum friction force attainable is

\begin{align*}
\sum_{}^{}F_{fr_{max}}& =\mu _s \ F_N\\
\\
&=\left( 0.60 \right)\left( 9810\ \text{N} \right)\\
\\
&=5886\ \text{N}
\end{align*}

Since a force of only 4500 N is needed, and that is, in fact, how much will be exerted by the road as a static friction force, the car can follow the curve.

Part B

The maximum static friction force possible is

\begin{align*}
\sum_{}^{}F_{fr_{max}}& =\mu _s \ F_N\\
\\
&=\left( 0.25 \right)\left( 9810\ \text{N} \right)\\
\\
&=2452.5\ \text{N}
\end{align*}

The car will skid because the ground cannot exert sufficient force (4500 N is needed) to keep it moving in a curve of radius 50 m at a speed of 54 km/h.


Acceleration of a Revolving Ball – Uniform Circular Motion Example

A 150-g ball at the end of a string is revolving uniformly in a horizontal circle of radius 0.600 m, as in the Figure 1 below. The ball makes 2.00 revolutions in a second. What is its centripetal acceleration?

Figure 1: A small object moving in a circle, showing how the velocity changes. At each point, the instantaneous velocity is in a direction tangent to the circular path.

Solution:

The linear velocity of the ball can be computed by dividing the total arc length traveled by the total time of travel. That is, the ball traveled 2 revolutions (twice the circumference of the circle) for 1 second. Thus,

\begin{align*}
\text{v} &= \frac{2\cdot2 \pi \text{r}}{\text{t}} \\
\\
& = \frac{4\pi \text{r}}{\text{t}} \\
\\
& = \frac{4\pi\left( 0.600\ \text{m} \right)}{1 \ \text{s}} \\
\\
& = 7.54 \ \text{m/s}
\end{align*}

Since the linear velocity has already been computed, we can now compute for the centripetal acceleration, ac.

\begin{align*}
\text{a}_\text{c} & = \frac{\text{v}^{2}}{\text{r}} \\
\\
& = \frac{\left( 7.54\ \text{m/s} \right)^{2}}{0.600\ \text{m}}\\
\\
& =94.8 \ \text{m/s}^{2}
\end{align*}

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

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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

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Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

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Kinetic Theory: Atomic and Molecular Explanation of Pressure and Temperature

Problem 39

Problem 40

Problem 41

Problem 42

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Problem 48

Humidity, Evaporation, and Boiling

Problem 49

Problem 50

Problem 51

Problem 52

Problem 53

Problem 54

Problem 55

Problem 56

Problem 57

Problem 58

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Problem 60

Problem 61

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Problem 72


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

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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

Problem 52

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Molecular Transport Phenomena: Diffusion, Osmosis, and Related Processes

Problem 62

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Problem 66


Solution Guides to College Physics by Openstax Chapter 11 Banner

Chapter 11: Fluid Statics

Density

Problem 1

Problem 2

Problem 3

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Pressure

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Variation of Pressure with Depth in a Fluid

Problem 14

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Pascal’s Principle

Problem 24

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Gauge Pressure, Absolute Pressure, and Pressure Measurement

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Archimedes’ Principle

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Cohesion and Adhesion in Liquids: Surface Tension and Capillary Action

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Pressures in the Body

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Solution Guides to College Physics by Openstax Chapter 10 Banner

Chapter 10: Rotational Motion and Angular Momentum

Angular Acceleration

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Kinematics of Rotational Motion

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Dynamics of Rotational Motion: Rotational Inertia

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Rotational Kinetic Energy: Work and Energy Revisited

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Angular Momentum and Its Conservation

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Collisions of Extended Bodies in Two Dimensions

Problem 43

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Problem 45

Problem 46

Problem 47

Gyroscopic Effects: Vector Aspects of Angular Momentum

Problem 48


Solution Guides to College Physics by Openstax Chapter 9 Banner

Chapter 9: Statics and Torque

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The Second Condition for Equilibrium

Problem 1

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Stability

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Applications of Statics, Including Problem-Solving Strategies

Problem 17

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Simple Machines

Problem 19

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Forces and Torques in Muscles and Joints

Problem 26

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Problem 28

Problem 29

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