Essential Physics Chapter 5 Answers

Check Your Understanding

5.i

14 N, $56\text{°}$ measured from the positive x-centrality

5.2

a. His weight acts downward, and the strength of air resistance with the parachute acts up. b. neither; the forces are equal in magnitude

v.5

a. $159.0\widehat{i}+770.0\widehat{j}\text{N}$ ; b. $0.1590\widehat{i}+0.7700\widehat{j}\text{Northward}$

v.6

$a=\mathrm{two.78}{\text{m/south}}^2$

5.7

a. $\mathrm{iii.0}\text{m}\text{/}{\text{due south}}^2$ ; b. eighteen N

5.8

a. $\mathrm{1.vii}{\text{m/south}}^2;$ b. $\mathrm{ane.3}{\text{thousand/s}}^2$

Conceptual Questions

ane.

Forces are directional and have magnitude.

3.

The cupcake velocity earlier the braking action was the aforementioned as that of the car. Therefore, the cupcakes were unrestricted bodies in motion, and when the automobile suddenly stopped, the cupcakes kept moving forward co-ordinate to Newton's showtime police force.

five.

No. If the force were cypher at this point, then there would exist null to alter the object'southward momentary zero velocity. Since nosotros do non discover the object hanging motionless in the air, the force could non be null.

7.

The astronaut is truly weightless in the location described, because there is no large torso (planet or star) nearby to exert a gravitational force. Her mass is lxx kg regardless of where she is located.

9.

The force you exert (a contact force equal in magnitude to your weight) is small. Globe is extremely massive by comparison. Thus, the acceleration of World would be incredibly small. To see this, employ Newton's second law to calculate the acceleration you would crusade if your weight is 600.0 Due north and the mass of Earth is $6.00\times {10}^{24}\text{kg}$ .

11.

a. action: World pulls on the Moon, reaction: Moon pulls on Earth; b. activity: foot applies force to ball, reaction: brawl applies force to foot; c. action: rocket pushes on gas, reaction: gas pushes dorsum on rocket; d. activity: automobile tires button backward on road, reaction: road pushes forward on tires; e. action: jumper pushes downwardly on footing, reaction: footing pushes upwards on jumper; f. action: gun pushes forward on bullet, reaction: bullet pushes backward on gun.

13.

a. The rifle (the crush supported by the burglarize) exerts a force to expel the bullet; the reaction to this force is the force that the bullet exerts on the rifle (shell) in contrary direction. b. In a recoilless rifle, the beat out is not secured in the burglarize; hence, every bit the bullet is pushed to move forward, the shell is pushed to eject from the opposite finish of the barrel. c. Information technology is non safety to stand up behind a recoilless rifle.

fifteen.

a. Aye, the force tin can exist acting to the left; the particle would feel dispatch contrary to the motion and lose speed. B. Yes, the force tin exist acting downward because its weight acts down fifty-fifty as it moves to the correct.

17.

two forces of different types: weight acting downwards and normal force acting upward

Problems

19.

a. ${\overrightarrow{F}}_{\text{cyberspace}}=5.0\widehat{i}+10.0\widehat{j}\text{N}$ ; b. the magnitude is $F_{\text{internet}}=\mathrm{eleven}\text{N}$ , and the direction is $\theta =63\text{°}$

21.

a. ${\overrightarrow{F}}_{\text{net}}=660.0\widehat{i}+150.0\widehat{j}\text{N}$ ; b. $F_{\text{net}}=676.6\text{Due north}$ at $\theta =\mathrm{12.eight}\text{°}$ from David'southward rope

23.

a. ${\overrightarrow{F}}_{\text{cyberspace}}=95.0\widehat{i}+283\widehat{j}\text{N}$ ; b. 299 N at $71\text{°}$ north of east; c. ${\overrightarrow{F}}_{\text{DS}}=\text{−}\left(95.0\widehat{i}+283\widehat{j}\right)\text{N}$

25.

Running from rest, the sprinter attains a velocity of $v=12.96\text{m/s}$ , at end of acceleration. We find the time for acceleration using $x=\mathrm{xx.00}\text{m}=0+0.5a{t}_1{}^2$ , or $t_{\mathrm{i}}=3.086\text{s.}$ For maintained velocity, ${\mathrm{ten}}_2=5t_2$ , or $t_2=x_{\mathrm{two}}\text{/}5=80.00\text{m}\text{/}12.96\text{m}\text{/}\text{s}=6.173\text{s}$ . $\text{Total time}=\mathrm{nine.259}\text{s}$ .

27.

a. $\mathrm{yard}=56.0\text{kg}$ ; b. $a_{\text{meas}}=a_{\text{astro}}+a_{\text{ship}},\text{where}{a}_{\text{ship}}=\frac{{\mathrm{chiliad}}_{\text{astro}}{a}_{\text{astro}}}{m_{\text{send}}}$ ; c. If the force could be exerted on the astronaut past another source (other than the spaceship), then the spaceship would non feel a recoil.

29.

$F_{\text{net}}=4.12\times {\mathrm{ten}}^5\text{Northward}$

31.

$a={253\text{m/s}}^2$

33.

$F_{\text{net}}=F-f=ma\Rightarrow F=1.26\times {\mathrm{ten}}^{\mathrm{three}}\text{North}$

35.

$\begin{array}{}\\ \\ {\mathrm{five}}^2=v_0^{\mathrm{two}}+2ax\Rightarrow a=\text{−}\mathrm{7.fourscore}{\text{m/s}}^2\hfill \\ F_{\text{net}}=\mathrm{-7.80}\times {\mathrm{x}}^3\text{N}\hfill \end{array}$

37.

a. ${\overrightarrow{F}}_{\text{net}}=\mathrm{thousand}\overrightarrow{a}\Rightarrow \overrightarrow{a}=9.0\widehat{i}{\text{m/s}}^2$ ; b. The dispatch has magnitude $9.0{\text{m/south}}^{\mathrm{two}}$ , so $x=110\text{one thousand}$ .

39.

$\mathrm{ane.6}\widehat{i}-0.8\widehat{j}{\text{grand/s}}^2$

41.

a. $\begin{array}{ccc}\hfill {\mathrm{due\; west}}_{\text{Moon}}& =\hfill & m{\mathrm{chiliad}}_{\text{Moon}}\hfill \\ \hfill g& =\hfill & 150\text{kg}\hfill \\ \hfill w_{\text{Earth}}& =\hfill & 1.5\times {10}^3\text{Due north}\hfill \end{array}$ ; b. Mass does non change, and so the suited astronaut'south mass on both Globe and the Moon is $150\text{kg.}$

43.

a. $\begin{array}{ccc}\hfill F_{\text{h}}& =\hfill & 3.68\times {10}^3\text{N and}\hfill \\ \hfill w& =\hfill & \mathrm{seven.35}\times {10}^2\text{North}\hfill \\ \hfill \frac{F_{\text{h}}}{w}& =\hfill & 5.00\text{times greater than weight}\hfill \end{array}$ ;
b. $\begin{array}{ccc}\hfill F_{\text{net}}& =\hfill & 3750\text{N}\hfill \\ \hfill \theta & =\hfill & 11.3\text{°}\text{from horizontal}\hfill \end{array}$

45.

$\begin{array}{ccc}\hfill \mathrm{westward}& =\hfill & 19.6\text{Northward}\hfill \\ \hfill F_{\text{net}}& =\hfill & 5.40\text{North}\hfill \\ \hfill F_{\text{net}}& =\hfill & \mathrm{one\; thousand}a\Rightarrow a=\mathrm{ii.seventy}{\text{grand/s}}^2\hfill \end{array}$

51.

a. $F_{\text{net}}=\mathrm{two.64}\times {10}^{\mathrm{vii}}\text{N;}$ b. The forcefulness exerted on the transport is likewise $\mathrm{ii.64}\times {\mathrm{x}}^7\text{N}$ because it is contrary the shell's direction of motion.

53.

Because the weight of the history book is the force exerted by Earth on the history book, we stand for it every bit ${\overrightarrow{F}}_{\text{EH}}=\mathrm{-xiv}\widehat{j}\text{N}\text{.}$ Bated from this, the history book interacts only with the physics book. Because the acceleration of the history book is nada, the internet force on it is zero past Newton'south second constabulary: ${\overrightarrow{F}}_{\text{PH}}+{\overrightarrow{F}}_{\text{EH}}=\overrightarrow{0},$ where ${\overrightarrow{F}}_{\text{PH}}$ is the force exerted past the physics book on the history book. Thus, ${\overrightarrow{F}}_{\text{PH}}=\text{−}{\overrightarrow{F}}_{\text{EH}}=\text{−}\left(\mathrm{-fourteen}\widehat{j}\right)\text{N}=\mathrm{fourteen}\widehat{j}\text{N}\text{.}$ We discover that the physics book exerts an upward force of magnitude 14 N on the history book. The physics book has three forces exerted on it: ${\overrightarrow{F}}_{\text{EP}}$ due to Globe, ${\overrightarrow{F}}_{\text{HP}}$ due to the history book, and ${\overrightarrow{F}}_{\text{DP}}$ due to the desktop. Since the physics volume weighs xviii N, ${\overrightarrow{F}}_{\text{EP}}=\mathrm{-18}\widehat{j}\text{North}\text{.}$ From Newton's third law, ${\overrightarrow{F}}_{\text{HP}}=\text{−}{\overrightarrow{F}}_{\text{PH}},$ so ${\overrightarrow{F}}_{\text{HP}}=\mathrm{-14}\widehat{j}\text{North}\text{.}$ Newton'southward second law applied to the physics book gives $\sum \overrightarrow{F}=\overrightarrow{0},$ or ${\overrightarrow{F}}_{\text{DP}}+{\overrightarrow{F}}_{\text{EP}}+{\overrightarrow{F}}_{\text{HP}}=\overrightarrow{0},$ so ${\overrightarrow{F}}_{\text{DP}}=\text{−}\left(\mathrm{-18}\widehat{j}\right)-\left(\mathrm{-xiv}\widehat{j}\right)=32\widehat{j}\text{N}\text{.}$ The desk exerts an upward force of 32 N on the physics book. To make it at this solution, nosotros apply Newton'due south 2d law twice and Newton's third law once.

55.

a. The free-body diagram of caster 4:

b. $T=\mathrm{one\; thousand}\mathrm{one\; thousand},F=2T\text{cos}\theta =2m\mathrm{1000}\text{cos}\theta$

57.

a. $0.106{\text{m/s}}^{\mathrm{two}}$ in the direction of the 2nd team. The 2d team wins the tug-of-war.
b. $1.22\times {\mathrm{ten}}^{\mathrm{four}}{\text{m/south}}^2$

59.

a. $T=\mathrm{ane.96}\times {10}^{\mathrm{-4}}\text{N;}$
b. $\begin{array}{}\\ \\ T^{\prime }=4.71\times {\mathrm{ten}}^{\mathrm{-iv}}\text{N}\hfill \\ \frac{T^{\prime }}{T}=2.40\text{times the tension in the vertical strand}\hfill \end{array}$

61.

$$ $\begin{array}{}\\ \\ F_{y\text{net}}=F_{\perp }-2T\text{sin}\theta =0\hfill \\ F_{\perp }=\mathrm{two}T\text{sin}\theta \hfill \\ T=\frac{F_{\perp }}{\mathrm{two}\text{sin}\theta }\hfill \end{array}$

65.

a. 5.6 kg; b. 55 Due north; c. $T_2=60\text{N}$ ;
d.

67.

a. $4.9{\text{one thousand/s}}^2$ , 17 Due north; b. 9.viii Due north

Additional Problems

77.

a. $F_{\text{net}}=\frac{m({\mathrm{five}}^2-v_0{}^{\mathrm{two}})}{\mathrm{ii}\mathrm{10}}$ ; b. 2590 Northward

79.

$\begin{array}{}\\ \\ {\overrightarrow{F}}_{\text{net}}={\overrightarrow{F}}_1+{\overrightarrow{F}}_2+{\overrightarrow{F}}_3=(6.02\widehat{i}+14.0\widehat{j})\text{Northward}\hfill \\ {\overrightarrow{F}}_{\text{net}}=k\overrightarrow{a}\Rightarrow \overrightarrow{a}=\frac{{\overrightarrow{F}}_{\text{net}}}{k}=\frac{\mathrm{half-dozen.02}\widehat{i}+14.0\widehat{j}\text{North}}{\mathrm{x.0}\text{kg}}=(0.602\widehat{i}+1.40\widehat{j}){\text{m/south}}^{\mathrm{two}}\hfill \end{array}$

81.

$\begin{array}{}\\ \\ {\overrightarrow{F}}_{\text{internet}}={\overrightarrow{F}}_{\text{A}}+{\overrightarrow{F}}_{\text{B}}\hfill \\ {\overrightarrow{F}}_{\text{internet}}=A\widehat{i}+\left(\mathrm{-1.41}A\widehat{i}-1.41A\widehat{j}\right)\hfill \\ {\overrightarrow{F}}_{\text{net}}=A\left(\mathrm{-0.41}\widehat{i}-1.41\widehat{j}\right)\hfill \\ \theta =254\text{°}\hfill \end{array}$
(Nosotros add $180\text{°}$ , because the angle is in quadrant IV.)

83.

$F=2\mathrm{one\; thousand}{k}^{\mathrm{ii}}{x}^2$ ; First, take the derivative of the velocity function to obtain $a=\mathrm{two}\mathrm{thou}\mathrm{10}5=2\mathrm{thousand}x\left(k{x}^2\right)=\mathrm{ii}{k}^2{x}^3$ . Then employ Newton's second law $F=ma=2m{k}^2{x}^{\mathrm{two}}$ .

85.

a. For box A, $N_{\text{A}}=m\mathrm{thou}$ and $N_{\text{B}}=gg\text{cos}\theta$ ; b. $N_{\text{A}}>{\mathrm{Northward}}_{\text{B}}$ because for $\theta <90\text{°}$ , $\text{cos}\theta <\mathrm{ane}$ ; c. ${\mathrm{Due\; north}}_{\text{A}}>{\mathrm{North}}_{\text{B}}$ when $\theta =\mathrm{ten}\text{°}$

87.

a. 8.66 N; b. 0.433 grand

Challenge Problems

93.

a.

; b. No; ${\overrightarrow{F}}_{\text{R}}$ is non shown, because it would supplant ${\overrightarrow{F}}_{\mathrm{i}}$ and ${\overrightarrow{F}}_2$ . (If we desire to testify it, we could describe information technology and so place squiggly lines on ${\overrightarrow{F}}_{\mathrm{i}}$ and ${\overrightarrow{F}}_2$ to evidence that they are no longer considered.

95.

a. 14.1 m/southward; b. 601 N

101.

$$ $\overrightarrow{a}=\mathrm{-248}\widehat{i}-433\widehat{j}\text{m}\text{/}{\text{s}}^2$

103.

$0.548{\text{grand/s}}^{\mathrm{two}}$

105.

a. $T_{\mathrm{i}}=\frac{2\mathrm{thou}g}{\text{sin}\theta }$ , $T_{\mathrm{two}}=\frac{\mathrm{thousand}\mathrm{thou}}{\text{sin}\left(\text{arctan}\left(\frac{1}{\mathrm{ii}}\text{tan}\theta \right)\right)}$ , $T_3=\frac{2gg}{\text{tan}\theta };$ b. $\varphi =\text{arctan}\left(\frac{1}{2}\text{tan}\theta \right)$ ; c. $2.56\text{°}$ ; (d) $x=d\left(2\text{cos}\theta +\mathrm{two}\text{cos}\left(\text{arctan}\left(\frac{1}{2}\text{tan}\theta \right)\right)+1\right)$

107.

a. $\overrightarrow{a}=\left(\frac{5.00}{\mathrm{thou}}\widehat{i}+\frac{3.00}{m}\widehat{j}\right)\text{thousand}\text{/}{\text{south}}^2;$ b. 1.38 kg; c. 21.ii m/s; d. $\overrightarrow{\mathrm{five}}=\left(18.1\widehat{i}+\mathrm{10.nine}\widehat{j}\right)\text{g}\text{/}{\text{southward}}^{\mathrm{two}}$

109.

a. $0.900\widehat{i}+0.600\widehat{j}\text{N}$ ; b. 1.08 N

Essential Physics Chapter 5 Answers,

Source: https://openstax.org/books/university-physics-volume-1/pages/chapter-5

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