Review Questions and Answers for Chapters 4 and 5 Apes

vii.

(a) $12{\text{m/s}}^2$ .

(b) The dispatch is not one-4th of what information technology was with all rockets burning because the frictional forcefulness is nonetheless every bit large as it was with all rockets called-for.

9.

(a) The system is the child in the wagon plus the wagon.

(b

(c) $a=0\text{.}\text{130}{\text{1000/s}}^2$ in the management of the 2d kid's button.

(d) $a=0.00{\text{chiliad/southward}}^2$

11.

(a) $3.68\times {\mathrm{ten}}^{\mathrm{three}}\text{Northward}$ . This strength is 5.00 times greater than his weight.

(b) $\text{3750 North; xi.3º}\text{higher up horizontal}$

13.

$1.5\times {\mathrm{ten}}^3\text{N},\text{150 kg},\text{150 kg}$

15.

Strength on shell: $2\text{.}\text{64}\times {\text{10}}^7\text{N}$

Force exerted on transport = $-2\text{.}\text{64}\times {\text{10}}^7\text{Northward}$ , by Newton'south third police force

17.

1. $0.{\text{11 m/s}}^2$
2. $\mathrm{ane}\text{.}2\times {\text{10}}^4\text{N}$

19.

(a) $\mathrm{vii}\text{.}\text{84}\times {\text{ten}}^{\mathrm{-iv}}\text{N}$

(b) $1\text{.}\text{89}\times {\text{10}}^{\mathrm{–3}}\text{N}$ . This is 2.41 times the tension in the vertical strand.

21.

Newton's 2nd police force practical in vertical direction gives

$$F_y=F-2T\text{sin}\theta =0$$

$$F_{}=\mathrm{ii}T\text{sin}\theta$$

$$T=\frac{F_{}}{\text{2 sin}\theta }.$$

23.

Using the gratuitous-body diagram:

$F_{\text{net}}=T-f-mg=\text{ma}$ ,

so that

$a=\frac{T-f-\text{mg}}{\mathrm{1000}}=\frac{1\text{.}\text{250}\times {\text{10}}^7\text{Due north}-4.50\times {\text{ten}}^{\text{6}}N-(5.00\times {\text{ten}}^5\text{kg})(\mathrm{ix.}{\text{80 1000/south}}^2)}{5.00\times {\text{ten}}^5\text{kg}}=\text{half dozen.xx}{\text{yard/s}}^{\mathrm{ii}}$ .

25.

1. Use Newton's laws of movement.

   

2. Given : $a=4.00g=(\mathrm{four.00})(\mathrm{ix.}{\text{80 yard/south}}^{\mathrm{ii}})=\text{39.2}{\text{thousand/s}}^2\text{;}$ $\mathrm{one\; thousand}=\text{70}\text{.}\text{0 kg}$,

   Notice: $F$ .

3. $$ $$ $\sum F\text{=+}F-w=\text{ma}\text{,}$ and so that $F=\text{ma}+\mathrm{due\; west}=\text{ma}+\text{mg}=m(a+\mathrm{thousand})$.

   $F=(\text{70.0 kg})[(\text{39}\text{.}{\text{two m/s}}^2)+(\mathrm{ix}\text{.}{\text{80 1000/s}}^{\mathrm{two}})]$ $=3.\text{43}\times {\text{10}}^3\text{N}$ . The force exerted by the high-jumper is really down on the ground, simply $F$ is up from the basis and makes him jump.

4. This result is reasonable, since it is quite possible for a person to exert a force of the magnitude of ${\text{x}}^3\text{N}$.

27.

(a) $4\text{.}\text{41}\times {\text{10}}^5\text{Northward}$

(b) $\mathrm{i}\text{.}\text{50}\times {\text{x}}^{\mathrm{five}}\text{N}$

29.

(a) $\text{910 N}$

(b) $\mathrm{one}\text{.}\text{11}\times {\text{10}}^3\text{N}$

31.

$a=\text{0.139 chiliad/due south}$ , $\theta =\mathrm{12.4º}$ north of eastward

33.

1. Employ Newton's laws since we are looking for forces.
2. Depict a forcefulness diagram:

   

3. The tension is given every bit $T=\text{25.0 N.}$ Find $F_{\text{app}}\text{.}$ Using Newton's laws gives: ${\text{Σ F}}_x=\mathrm{0,}$ so that applied force is due to the y-components of the 2 tensions: $F_{\text{app}}=\mathrm{two}T\text{sin}\text{θ}=\mathrm{ii}(\text{25.0 N})\text{sin}\left(\text{15º}\right)=\text{12.9 North}$

   The ten-components of the tension cancel. $\sum F_{\mathrm{10}}=0$ .

4. This seems reasonable, since the applied tensions should be greater than the forcefulness practical to the molar.

40.

$\text{10.2}{\text{thousand/s}}^2\text{, 4.67º from vertical}$

42.

$T_{\mathrm{one}}=\text{736 North}$

$T_{\mathrm{two}}=\text{194 Northward}$

44.

(a) $\mathrm{vii.43\; m/southward}$

(b) 2.97 thousand

46.

(a) $\mathrm{iv.xx\; one\; thousand/s}$

(b) $\text{29.four}{\text{m/s}}^2$

(c) $4\text{.}\text{31}\times {\text{10}}^{\mathrm{three}}\text{N}$

48.

(a) 47.1 k/s

(b) $2\text{.}\text{47}\times {\text{10}}^3{\text{grand/s}}^2$

(c) $6.18\times {\text{10}}^{\mathrm{three}}\text{North}$ . The average force is 252 times the shell's weight.

52.

(a) $1\times {\text{10}}^{-\text{xiii}}$

(b) $\mathrm{i}\times {\text{10}}^{-\text{11}}$

54.

${\text{x}}^2$

ane.

Motorcar 10 is shown on the left, and Car Y is shown on the right.

i.

Car X takes longer to advance and does not spend whatsoever fourth dimension traveling at summit speed. Motorcar Y accelerates over a shorter time and spends time going at superlative speed. So Car Y must encompass the straightaways in a shorter time. Curves have the same fourth dimension, and so Car Y must overall take a shorter time.

ii.

The just difference in the calculations for the fourth dimension of 1 segment of linear dispatch is the difference in distances. That shows that Car X takes longer to accelerate. The equation $\frac{d}{\mathrm{four}{v}_c}=t_c$ corresponds to Auto Y traveling for a time at tiptop speed.

Substituting $a=\frac{v_c}{t_1}$ into the displacement equation in part (b) ii gives $D=\frac{\mathrm{iii}}{\mathrm{ii}}{\mathrm{five}}_c{t}_1$ . This shows that a motorcar takes less time to reach its maximum speed when information technology accelerates over a shorter distance. Therefore, Car Y reaches its maximum speed more quickly, and spends more than time at its maximum speed than Car X does, as argued in part (b) i.

iii.

A body cannot exert a force on itself. The militarist may advance as a outcome of several forces. The hawk may accelerate toward Globe as a result of the force due to gravity. The hawk may accelerate every bit a result of the boosted force exerted on it by wind. The hawk may accelerate as a outcome of orienting its body to create less air resistance, thus increasing the internet force forward.

5.

(a) A soccer player, gravity, air, and friction normally exert forces on a soccer ball being kicked.

(b) Gravity and the surrounding h2o commonly exert forces on a dolphin jumping. (The dolphin moves its muscles to exert a force on the water. The water exerts an equal force on the dolphin, resulting in the dolphin's motion.)

(c) Gravity and air exert forces on a parachutist drifting to Globe.

ix.

The diagram consists of a black dot in the centre and two small red arrows pointing up (Fb) and down (Fg) and two long ruby-red arrows pointing correct (Fc = 9.0 N) and left (Fw=13.0 N).

In the diagram, F _g represents the force due to gravity on the balloon, and F_b represents the buoyant force. These two forces are equal in magnitude and reverse in management. F _c represents the force of the electric current. F _w represents the force of the wind. The net force on the balloon volition be $F_{\mathrm{due\; west}}-F_c=\mathrm{iv.0}$ N and the balloon will accelerate in the management the wind is blowing.

11.

Since $m=F/a$ , the parachutist has a mass of $539\text{North/9}{\text{.viii km/southward}}^{\text{ii}}=55\text{kg}$ .

For the commencement ii s, the parachutist accelerates at ix.viii m/s².

$$\begin{array}{c}v=at\\ =9.8\frac{\text{m}}{{\text{s}}^{\text{2}}}•2\text{s}\\ =19.6\frac{\text{m}}{\text{south}}\end{array}$$

Her speed after two s is nineteen.half dozen thou/s.

From 2 s to 10 s, the net force on the parachutist is 539 N – 615 Due north, or 76 Northward upward.

$\begin{array}{ccc}a& =& \frac{F}{m}\\ & =& \frac{-76\text{N}}{55\text{}\text{}\text{}\text{}\text{kg}}\\ & =& -\mathrm{1.iv}\frac{\text{m}}{{\text{s}}^2}\end{array}$

Since $5=5_0+at$ , $v=\mathrm{nineteen.6}{\text{m/s}}^{\text{ii}}+(-\mathrm{1.four}{\text{yard/southward}}^{\text{2}})(\mathrm{viii}\text{s)}=8.4{\text{m/south}}^{\text{2}}$ .

At 10 s, the parachutist is falling to World at 8.four m/s.

thirteen.

The system includes the gardener and the wheelbarrow with its contents. The following forces are important to include: the weight of the wheelbarrow, the weight of the gardener, the normal force for the wheelbarrow and the gardener, the force of the gardener pushing against the ground and the equal forcefulness of the ground pushing dorsum against the gardener, and any friction in the wheelbarrow'southward wheels.

xv.

The system undergoing acceleration is the ii figure skaters together.

Net force = $120\text{N}\text{–}\text{5}\text{.0}\text{N}\text{=}\text{115}\text{Due north}$ .

Full mass = $\mathrm{xl}\text{kg + 50 kg = xc kg}$ .

Using Newton's 2nd police force, nosotros have that

$\begin{array}{ccc}a& =& \frac{F}{m}\\ & =& \frac{115\text{Northward}}{90\text{kg}}\\ & =& 1.28\frac{\text{thou}}{{\text{s}}^2}\end{array}$

The pair accelerates forward at one.28 m/s².

17.

The strength of tension must equal the forcefulness of gravity plus the force necessary to accelerate the mass. $F=m\mathrm{thousand}$ can be used to calculate the first, and $F=ma$ can be used to summate the second.

For gravity:

$\begin{array}{l}F=mg\\ =(120.0\text{kg)(ix}\text{.viii}{\text{m/s}}^{\text{ii}})\\ =\mathrm{1205.iv}\text{N}\end{array}$

For acceleration:

$$\begin{array}{l}F=\mathrm{thousand}a\\ =(120.0\text{kg)(i}\text{.iii}{\text{chiliad/s}}^{\text{2}})\\ =\mathrm{159.nine}\text{N}\end{array}$$

The total forcefulness of tension in the cable is 1176 N + 156 Northward = 1332 Due north.

21.

The diagram has a black dot and three solid red arrows pointing abroad from the dot. Arrow Ft is long and pointing to the left and slightly down. Pointer Fw is also long and is a chip below a diagonal line halfway between pointing up and pointing to the correct. A brusk arrow Fg is pointing downwards.

F _g is the force on the kite due to gravity.

F _w is the forcefulness exerted on the kite by the wind.

F _t is the force of tension in the string belongings the kite. It must remainder the vector sum of the other two forces for the kite to bladder stationary in the air.

27.

A gratuitous-torso diagram would show a northward force of 64 N and a westward strength of 38 N. The net force is equal to the sum of the two applied forces. It can be constitute using the Pythagorean theorem:

$\begin{array}{cc}{F}_{\text{net}}=& \sqrt{F_x{}^2+F_y{}^w}\\ & =\sqrt{{(38\text{N})}^{\mathrm{ii}}+{(64\text{}\text{Northward})}^2}\\ & =\mathrm{74.iv}\text{N}\end{array}$

Since $a=\frac{F}{\mathrm{yard}}$ ,

$\begin{array}{cc}a& =\frac{74.4\text{N}}{825\text{kg}}\\ & =0.09{\text{m/s}}^{\text{two}}\end{array}$

The bedrock volition accelerate at 0.09 m/southward².

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Source: https://openstax.org/books/college-physics-ap-courses/pages/chapter-4

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