Home Page Textbooks Physics: Principles and Problems Page 156: Practice Problems

Physics: Principles and Problems

9th Edition

Elliott, Haase, Harper, Herzog, Margaret Zorn, Nelson, Schuler, Zitzewitz

ISBN: 9780078458132

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Page 156: Practice Problems

Exercise 12

Step 1

1 of 2

begin{align*}
intertext{Data given in the problem:}
&v=8.8hspace{1mm}frac{m}{s}\
&r=25hspace{1mm}m
intertext{We are looking for centripetal acceleration, and we can find it using formula:}
&a_c=frac{v^2}{r}\
&a_c=frac{8.8^2}{25}\
&a_c=3.09hspace{1mm}frac{m}{s^2}
intertext{Friction force acting on the runner, because it is acting on his shoes.}
end{align*}

Result

2 of 2

$$
a_c=3.09hspace{1mm}dfrac{m}{s^2}
$$

Friction force

Exercise 13

Step 1

1 of 2

begin{align*}
intertext{Data given in the problem:}
&v=22hspace{1mm}frac{m}{s}\
&r=56hspace{1mm}m
intertext{First we are looking for centripetal acceleration, and we can find it using formula:}
&a_c=frac{v^2}{r}\
&a_c=frac{22^2}{56}\
&a_c=8.64hspace{1mm}frac{m}{s^2}
intertext{Now we are looking for minimum
coefficient of static friction if we want car to avoid slipping. }
intertext{We know that friction is:}
&F_f=- mu mgrightarrow\
&mu=frac{F_f}{mg}
intertext{We can put expression for friction:}
&F_f=m a_c\
rightarrow\
&mu=frac{ma_c}{mg}\
&mu=frac{a_c}{g}\
&mu=frac{8.64}{9.8}\
&mu=0.88\
end{align*}

Result

2 of 2

$$
begin{align*}
&a_c=8.64hspace{1mm}frac{m}{s^2}\
&mu=0.88\
end{align*}
$$

Exercise 14

Solution 1

Solution 2

Step 1

1 of 2

begin{align*}
intertext{Data given in the problem:}
&v=201hspace{1mm}frac{m}{s}\
&a_c=5hspace{1mm}frac{m}{s^2}
intertext{First we are looking for centripetal acceleration, and we can find it using formula:}
&a_c=frac{v^2}{r}\
rightarrow\
&r=frac{v^2}{a_c}\
&r=frac{207^2}{5}\
&r=8569.8hspace{1mm}m
intertext{Let’s convert that into km. We know that $1km=1000m$ or $1m=frac{1}{1000}km$}
&r=8569.8frac{1}{1000}hspace{1mm}km\
&r=8.57hspace{1mm}km\
end{align*}

Result

2 of 2

$$
begin{align*}
&r=8.57hspace{1mm}km\
end{align*}
$$

Step 1

1 of 4

First, let us review the given values,

$$
begin{align*}
v &= 201 ,mathrm{frac{m}{s}}\
a_c &= 5.0 ,mathrm{frac{m}{s^2}}
end{align*}
$$

such that $v$ is the velocity of the airplane and $a_c$ is the maximum centripetal acceleration allowed for the plane.

Step 2

2 of 4

We know that the formula for centripetal acceleration $a_c$ is

$$
begin{align}
a_c = frac{v^2}{r}
end{align}
$$

Step 3

3 of 4

We can isolate the radius $r$ and replace the equal sign $(=)$ with an inequality, less than $(<)$, that suggests $r$ is less than the value obtained in the right-hand-side of the equation.

begin{align*}
a_c &= frac{v^2}{r}\
r &< frac{v^2}{a_c}
intertext{We can now plug in the values.}
r &< frac{(201 ,mathrm{frac{m}{s}})^2}{5.0 ,mathrm{frac{m}{s^2}}}\
r &< 8080.2 ,mathrm{m} \
r &< boxed{8.1 ,mathrm{km}}
end{align*}

Result

4 of 4

$$
r < 8.1 ,mathrm{km}
$$

Exercise 15

Step 1

1 of 2

The centripetal acceleration is:

$a_c = dfrac{v^2}{r} = dfrac{(4.1)^2}{6.3} = 2.668 m/s^2$

The force of friction is:

$$
F = m a = (45)*(2.668) = 120 N
$$

Result

2 of 2

$$
120 N
$$

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Page 156: Practice Problems

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