Home Page Textbooks Physics: Principles and Problems Page 130: Section Review

Physics: Principles and Problems

9th Edition

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

ISBN: 9780078458132

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Page 130: Section Review

Exercise 27

Step 1

1 of 2

Friction, either static or kinetic, causes a friction force. The force of friction in both cases acts in the opposite direction from the external force that wants to move the body. Both frictions can be calculated by multiplying the normal force by the friction factor. The difference is primarily in the amount of friction factor. The kinetic friction factor is usually smaller than the static one, so the kinetic friction force is smaller too.

Result

2 of 2

Their similarities are the direction of action and both are the product of normal force and friction factors, the difference is in the amount and moment of action.

Exercise 28

Step 1

1 of 2

In this task we will calculate the friction force for the given data.

Known:

$$
begin{align*}
m&=25 mathrm{kg} \
mu &=0.15 \
end{align*}
$$

Unknown:

$$
begin{align*}
F_f&=? \
\
F_N&=F_g \
&=mcdot g \
&=25 mathrm{kg}cdot 9.8 mathrm{m/s^2} \
&=24.5 mathrm{N} \
\
F_f&=mucdot F_N \
&=0.15cdot 245 mathrm{N} \
&=boxed{36.75 mathrm{N}}
end{align*}
$$

Result

2 of 2

$$
F_f=36.75 mathrm{N}
$$

Exercise 29

Step 1

1 of 2

In this task we will calculate the initial velocity for the given data.

Known:

$$
begin{align*}
m&=2.3 mathrm{g} \
s&=0.35 mathrm{m} \
mu&=0.24 \
end{align*}
$$

Unknown:

$$
begin{align*}
v_i&=? \
\
F&=F_f \
mcdot a&=mucdot F_g \
mcdot a&=mu cdot mcdot g \
a&=0.24cdot cdot 9.8 mathrm{m/s^2} \
&=2.352 mathrm{m/s^2} \
\
v_i^2&=2cdot acdot s \
v_i&=sqrt{2cdot 2.352 mathrm{m/s^2}cdot 0.35 mathrm{m}} \
&=boxed{1.283 mathrm{m/s}}
end{align*}
$$

Result

2 of 2

$$
v_i=1.283 mathrm{m/s}
$$

Exercise 30

Step 1

1 of 2

In this task we will calculate the maximum force for the given data.

Known:

$$
begin{align*}
m&=40 mathrm{kg} \
mu&=0.43 \
end{align*}
$$

Unknown:

$$
begin{align*}
F_{max}&=? \
\
F_{max}&=F_f \
&=mucdot cdot mcdot g \
&=0.43cdot 40 mathrm{kg}cdot 9.8 mathrm{m/s^2} \
&=boxed{168.6 mathrm{N}}
end{align*}
$$

Result

2 of 2

$$
F_{max}=168.6 mathrm{N}
$$

Exercise 31

Step 1

1 of 2

Friction between the truck and the dresser will accelerate the dresser forward.

Step 2

2 of 2

However, if the force accelerating the dresser is greater than $(mu)_{s}mg$, then the dresser will slide backward.

Exercise 32

Step 1

1 of 2

In this task we will calculate the friction factors for the given data.

Known:

$$
begin{align*}
m&=13 mathrm{kg} \
F_1&=20 mathrm{N} \
F_2&=25 mathrm{N} \
a_1&=0 mathrm{m/s^2} \
a_2&=0.26 mathrm{m/s^2} \
end{align*}
$$

Unknown:

$$
begin{align*}
mu_s&=? \
mu_k&=? \
\
a_1&=0 mathrm{m/s^2} Rightarrow F_{fs}>F_1 \
F_{fs}&>F_1 \
mu_scdot mcdot g&> 20 mathrm{N} \
mu_s&>dfrac{20 mathrm{N}}{mcdot g} \
mu_s&>dfrac{20 mathrm{N}}{13 mathrm{kg}cdot 9.8 mathrm{m/s^2}} \
mu_s&>boxed{0.157} \
\
F_2&=F+F_{fk} \
F_2&=mcdot a+mu_kcdot mcdot g \
mu_k&=dfrac{F_2-mcdot a}{mcdot g} \
&=dfrac{25 mathrm{N}-13 mathrm{kg}cdot 0.26 mathrm{m/s^2}}{13 mathrm{kg}cdot 9.8 mathrm{m/s^2}} \
&=boxed{0.1697}
end{align*}
$$

Result

2 of 2

$$
begin{align*}
mu_s&>0.157 \
mu_k&=0.1697
end{align*}
$$

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Page 130: Section Review

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