Physics: Principles and Problems
Physics: Principles and Problems
9th Edition
Elliott, Haase, Harper, Herzog, Margaret Zorn, Nelson, Schuler, Zitzewitz
ISBN: 9780078458132
Textbook solutions

All Solutions

Page 128: Practice Problems

Exercise 17
Solution 1
Solution 2
Step 1
1 of 7
force = mass x acceleration
but there is a resistance opposing the force so the law is going to be
Step 2
2 of 7
force – resistance = mass x acceleration
and since the problem describes motion as (constant speed) so that means that the acceleration = zero
Step 3
3 of 7
force – resistance = zero
by adding +resistance to both sides
Step 4
4 of 7
force = resistance
substitute force with 36N
and acceleration with $mu_k$ $times$ 52
Step 5
5 of 7
36 = $52 times mu_k$
Step 6
6 of 7
$$
mu_k = dfrac{36}{52}
$$
Result
7 of 7
$mu_k$ = 0.69
Step 1
1 of 2
In this task we will calculate the friction factor for the given data.

Known:

$$
begin{align*}
F_g&=52 mathrm{N} \
F_p&=36 mathrm{N} tag{force exerted on sled by girl} \
end{align*}
$$

Unknown:

$$
begin{align*}
{mu } &=? \
\
F_f&={mu}cdot F_N tag{$F_N$ normal force} \
mu&=dfrac{F_f}{F_N} \
\
F_f&=F_p \
F_N&=F_g \
\
mu&=dfrac{F_p}{F_g} \
&=dfrac{36 mathrm{N} } {52 mathrm{N}} \
&=boxed{0.692}
end{align*}
$$

Notice that $mu$ is ratio and that it has no unit of measurement.

Result
2 of 2
$$
mu=0.692
$$
Exercise 18
Step 1
1 of 6
weight = mass x acceleration due to gravity
105 x 9.8 = 1029 N
the mass is given in KG but we work in Newton
so to convert it we multiply the mass with 9.8
Step 2
2 of 6
acceleration = zero
when it says (constant speed,uniform velocity,maximum velocity) that means that the acceleration = zero
a.k.a left = right
up = down
Step 3
3 of 6
force = resistance
substitute force with 102 N and resistance with 1029 x $mu_s$
Step 4
4 of 6
$$
102 = 1029 times mu_s
$$
Step 5
5 of 6
$$
mu_s = dfrac{102}{1029}
$$
Result
6 of 6
$mu_s$ = 0.099
Exercise 19
Step 1
1 of 4
In this problem we are calculating the force needed tu push a box.

Known:

$$
begin{align*}
F_g&=134 mathrm{N} \
mu&=0.55
end{align*}
$$

Unknown:

$$
begin{align*}
F_p&=?
end{align*}
$$

Step 2
2 of 4
Force needed to push a box is equal to the force of friction because that is the force that keeps the box at rest.

$$
begin{align*}
F_p&=F_f \
F_g&=F_N
end{align*}
$$

Step 3
3 of 4
Calculating the force needed.

$$
begin{align*}
F_p&=mucdot F_g \
&=0.55cdot 134 mathrm{N} \
&=boxed{73.98 mathrm{N}}
end{align*}
$$

Result
4 of 4
$$
F_p=73.94 mathrm{N}
$$
Exercise 20
Step 1
1 of 5
We are tasked to find the force $F_f$ required to pull the sled at a constant speed across the snow.
Step 2
2 of 5
Let us review the given values. We have the coefficient of kinetic friction $mu_k$, the weight $W_{person}$ of the person on the sled, and the weight $W_{sled}$ of the sled.

$$
begin{align*}
mu_k &= 0.12\
W_{person} &= 650 ,mathrm{N}\
W_{sled} &= 52 , mathrm{N}
end{align*}
$$

Step 3
3 of 5
begin{align}
intertext{Given the formula}
F_f &= mu_kF_N
intertext{such that}
F_N &= W_{person} + W_{sled}
end{align}

where $F_f$ is the force needed to needed to move the sled, $F_N$ is the normal force for the system, and $mu_k$ is the coefficient of kinetic friction acting against the system.

Step 4
4 of 5
We can now solve for the force by plugging in the given values.

$$
begin{align*}
F_f &= mu_kF_N\
&= mu_k(W_{person} + W_{sled})\
&= (0.12)(650.,mathrm{N} + 52.0,mathrm{N})\
&= boxed{84.0 ,mathrm{N}}
end{align*}
$$

Result
5 of 5
$$
F_f = 84.0 ,mathrm{N}
$$
Exercise 21
Solution 1
Solution 2
Step 1
1 of 2
By applying oil, the coefficient of friction $mu$ is reduced and thus the friction force is reduced so that the required force is also lower.

$$
begin{align*}
F_f&=5.8 mathrm{N} tag{Force when oil is not applied.} \
mu_{d}&=0.58 \
mu_{wo}&=0.06\
F’_f&=?tag{Force after oil application.}\
\
F_f&=mucdot F_N \
F_N&=dfrac{F_f}{mu_{d}} \
&=dfrac{5.8 mathrm{N}}{0.58} \
&=10 mathrm{N} \
\
F’_f&=mu_{wo}cdot F_N \
&=0.06cdot 10 mathrm{N} \
&=0.6 mathrm{N}
end{align*}
$$

Result
2 of 2
$$
F’_f=0.6 mathrm {N}
$$
Step 1
1 of 3
In the book we are given the coefficients of friction on a table. Steel on steel (dry) is .58 for kinetic, therefore we can assume that the rest of our (m)(g) is equal to 10 because of the 5.8N we start with.
Step 2
2 of 3
Using the same table we can see Steel on Steel (with oil) as a .06 so we multiply it by 10 to get 0.6N
Result
3 of 3
0.6N
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