A properly completed concept map is shown in the figure.

b) $F_{net}ne 0 rightarrow F_{net}=mcdot a$ (2nd Newton’s law)

c) $F_{friction}=F_{push} rightarrow F_{net}= 0 rightarrow v=const$ (1st Newton’s law)

We know that when riding a bike we have to press the pedals to maintain a constant speed which means we act with force while the acceleration is $a = 0$. Of course, we know that the force of air resistance and friction acts in the opposite direction, and that we actually overcome these two forces, that is, we maintain the sum of forces equal to 0.

$$
4<1<3<2
$$

A possible solution is also:

$$
4 <1 = 3 <2
$$

if the rope tension on block no. 2 is $F_T = 0$. And another possible solution would be:

$$
4 <1 <3 = 2
$$

If the rope tension on block no. 3 were equal to weight of blue block: $F_T = F_g$.

$$
begin{align*}
& F_{g(b)} tag{weight of bird} \
& F_{sb} tag{normal force exerted by statue on bird} \
& F_{bs} tag{ force exerted by bird on statue} \
& F_{g(s)} tag{ weight of statue} \
& F_{gs} tag{ force exerted by ground on statue}
end{align*}
$$

Interaction pair is: $F_{sb}$ and $F_{bs}$.

Note that the weight of the bird and the weight of the statue are not in a realistic ratio but this is for the purpose of clarity.

$$
F=ma
$$

We can therefore calculate the force of the man

$F_{Man}=75(10)=750$ N

And the force fo the Lion

$F_{Lion}=170(10)=1700$

To calculate the difference in forces we simply subtract the force of the man from the force of the lion

$$
F=F_{Lion}-F_{Man}=1700-750=boxed{950text{ N}}
$$

$$
begin{align*}
& 2 mathrm{OZ} \
& 340 mathrm{g}
end{align*}
$$

During a free fall in a vacuum, the only force acting on the body is the gravitational force $F_g = m cdot g$ if we include this force in Newton’s second law, we can calculate the acceleration.

$$
begin{align*}
F&=mcdot g \
mcdot a&= mcdot g \
a&=g
end{align*}
$$

$textbf{Air force is also included}$ in this task. The key thing is that objects are of equal surface area, volume and have equal acceleration. It is also crucial that the force of air resistance does not depend on the mass of the object but on its volume, area and speed. As a result, the balls will fall with an acceleration of less than $9.8 mathrm {m / s^2}$ until they reach the terminal velocity. $textbf{The difference occurs after reaching the terminal speed.}$ The terminal velocity is the one at which the force of gravity $F_g$ equals the force of air resistance $F_ {ar}$. A $textbf{heavier ball}$ is subjected to a $textbf{higher gravitational force}$, which means that a $textbf{higher force of air resistance}$ will be required, which ultimately means that the heavier ball will have a $textbf{higher terminal velocity.}$

So while it will $textbf{accelerate equally}$, the $textbf{lighter ball will reach its terminal speed earlier}$ and will continue to move constantly while the heavier one will continue to accelerate to its terminal speed and continue to continue constantly but faster, which tells us that the $textbf{heavier ball will fall sooner.}$

b) $v_{ball}=0 mathrm{m/s}$

c) $a=g=9.8 mathrm{m/s^2}$

$$
begin{align*}
m&=1 mathrm{kg} \
g&=9.8 mathrm{m/s^2} \
F_{net}&=? \
\
F_{net}&=F_g \
&=mcdot g \
&=1 mathrm{kg} cdot 9.8 mathrm{m/s^2} \
&=9.8 mathrm{N}
end{align*}
$$

Known:

$$
begin{align*}
F&=5 mathrm{N} \
m&=40 mathrm{kg}
end{align*}
$$

Unknown:

$$
begin{align*}
a&=? \
\
F&=mcdot a \
a&=dfrac{F}{m} \
&=dfrac{5 mathrm{N}}{40 mathrm{kg}} \
&=boxed{0.125 mathrm{m/s^2}}
end{align*}
$$

Known:

$$
begin{align*}
a&=3 mathrm{m/s^2} \
m&=2300 mathrm{kg}
end{align*}
$$

Unknown:

$$
begin{align*}
F&=? \
\
F&=mcdot a \
&=2300 mathrm{kg}cdot 3 mathrm{m/s^2} \
&=boxed{6900 mathrm{N}}
end{align*}
$$

Known:

$$
begin{align*}
F{net}&=0.17 mathrm{N} \
m&=13 mathrm{g}=1.3*10^{-2} mathrm{kg}
end{align*}
$$

Unknown:

$$
begin{align*}
a&=? \
\
F&=mcdot a tag{Newton’s second law} \
\
a&=dfrac{F}{m} \
&=dfrac{0.17 mathrm{N}}{1.3*10^{-2} mathrm{kg}} \
&=boxed{13.08 mathrm{m/s^2}}
end{align*}
$$

$$
F = m g = (70) times (9.8) = 686 N
$$

Known:

$$
begin{align*}
F_g&=2450 mathrm{N} \
g&=9.8 mathrm{m/s^2}
end{align*}
$$

Unknown:

$$
begin{align*}
m&=?\
\
F_G&=mcdot g \
m&=dfrac{F_G}{g} \
&=dfrac{2450 mathrm{N}}{9.8 mathrm{m/s^2}} \
&=boxed{250 mathrm{kg}}
end{align*}
$$

1 newton = 0.224808943 pounds force

$$
==> 100.0 N = 22.5 pounds
$$

Known:

$$
begin{align*}
m&= 7.5 mathrm{kg} \
F&=78.4 mathrm{N} \
end{align*}
$$

Unknown:

$$
begin{align*}
g&=?\
\
F_G&=mcdot g \
g&=dfrac{F_G}{m} \
&=dfrac{F}{m} \
&=dfrac{78.4 mathrm{N}}{7.5 mathrm{kg}} \
&=boxed{10.45 mathrm{m/s^2}}
end{align*}
$$

Known:

$$
begin{align*}
m&=873 mathrm{kg} \
v_f&=26.3 mathrm{m/s} \
t_f&=0.59 mathrm{s} \
m_D&=68 mathrm{kg}
end{align*}
$$

Unknown:

$$
begin{align*}
a&=? \
F_C&=? tag{force on car}\
F_{sd}&=? tag{force of seat on driver}
end{align*}
$$

$$
begin{align*}
a&=dfrac{v_f-v_i}{t_f-t_i} \
&=dfrac{26.3 mathrm{m/s}}{0.59 mathrm{s}} \
&=boxed{44.58 mathrm{m/s^2}}
end{align*}
$$

$$
begin{align*}
F_{sd}&=mcdot a \
&=873 mathrm{kg}cdot 44.58 mathrm{m/s^2} \
&=boxed{3.892cdot 10^{4} mathrm{N}}
end{align*}
$$

$$
begin{align*}
F_C&=mcdot a \
&=68 mathrm{kg}cdot 44.58 mathrm{m/s^2} \
&=boxed{3031 mathrm{N}}
end{align*}
$$

b) $F_C=3.892*10^4 mathrm{N}$

c) $F_{sd}=3031 mathrm{N}$

$$
begin{align*}
m&=53 mathrm{kg} \
g&=9.8 mathrm{m/s^2} \
a_A&=0 mathrm{m/s^2} \
a_B&=-2 mathrm{m/s^2} \
a_C&=-2 mathrm{m/s^2} \
a_D&=0 mathrm{m/s^2} \
a_E&=a \
F_s&=?
\
F_s&=F_g+ma \
&=mcdot (g+a) \
\
F_{s(A)}&=mcdot (g+0) \
&=53 mathrm{kg}cdot 9.8 mathrm{m/s^2} \
&=519.4 mathrm{N} \
\
F_{s(B)}&=mcdot (g-2) \
&=53 mathrm{kg}cdot 7.8 mathrm{m/s^2} \
&=413.4 mathrm{N} \
\
F_{s(C)}&=mcdot (g-2) \
&=53 mathrm{kg}cdot 7.8 mathrm{m/s^2} \
&=413.4 mathrm{N} \
\
F_{s(D)}&=mcdot (g+0) \
&=53 mathrm{kg}cdot 9.8 mathrm{m/s^2} \
&=519.4 mathrm{N} \
\
F_{s(E)}&=mcdot (g+a)
end{align*}
$$

No, it can not hold.

$$
F = W = m g = (0.50)*(9.8) = 4.9 N
$$

$$
begin{align*}
g_M&=0.38cdot g_E \
g_E&=9.8 mathrm{m/s^2} \
m_A&=6 mathrm{kg} \
g_P&=0.08cdot g_M \
m_B&=7 mathrm{kg}
end{align*}
$$

$$
begin{align*}
F_{g(A)}&=? \
F_{g(B)}&=?
end{align*}
$$

$$
begin{align*}
F_g&=mcdot g
end{align*}
$$

$$
begin{align*}
F_{g(A)}&=m_Acdot g_M \
&=m_Acdot 0.38cdot g_E \
&=6 mathrm{kg}cdot 0.38cdot 9.8 mathrm{m/s^2} \
&=boxed{22.34 mathrm{N}}
end{align*}
$$

$$
begin{align*}
F_{g(B)}&=m_Bcdot g_P \
&=m_Bcdot 0.08cdot 0.38cdot g_E \
&=7 mathrm{kg}cdot 0.08cdot 0.38cdot 9.8 mathrm{m/s^2} \
&=boxed{2.085 mathrm{N}}
end{align*}
$$

Known:

$$
begin{align*}
m&=65 mathrm{kg} \
h&=10 mathrm{m} \
d&=2 mathrm{m} \
end{align*}
$$

Unknown:

$$
begin{align*}
a&=?\
F_w&=? tag{stopping force of water}
end{align*}
$$

In order to calculate the speed at which a diver strikes the surface of the water we must consider its motion. When he jumps from the tower he accelerates with an initial speed of zero. If we ignore the resistance of the air, the only force acting on it and which accelerates it is the gravitational force. if we incorporate this force into Newton’s second law we get:

$$
begin{align*}
F&=mg \
ma&=mg\
a&=g=9.8 mathrm{m/s^2}
end{align*}
$$

The velocity at uniformly accelerated motion is calculated as follows:

$$
begin{align*}
v^2&=2cdot acdot s \
v^2&=2cdot gcdot h \
&=2cdot 9.8 mathrm{m/s^2}cdot 10 mathrm{m} \
v^2&=196 mathrm{m^2/s^2} \
v&=boxed{14 mathrm{m/s}}
end{align*}
$$

$$
begin{align*}
v^2 &=2cdot acdot s \
a&=dfrac{v^2}{2cdot s} \
a&=dfrac{v^2}{2cdot d} \
&=dfrac{196 mathrm{m^2/s^2}}{2cdot 2 mathrm{m}} \
&=-49 mathrm{m/s^2} tag{we must add negative signe because of direction}\
\
F_w&=mcdot a \
&=65 mathrm{kg}cdot -49 mathrm{m/s^2} \
&=boxed{-3185 mathrm{N}}
end{align*}
$$

Where x is the displacement, a is the acceleration, t is the time duration, $v_0$ is the initial velocity.

$x = (1/2) a t^2 + v_0 t$

$40.0 = (1/2) a (3.0)^2 + (0.0) (3.0)$

$==> a = 8.889 m/s^2$

The net force:

$$
F_{net} = m a = (710)*(8.8889) = 6300 N
$$

Known:

$$
begin{align*}
m_1&=6 mathrm{kg} \
m_2&=7 mathrm{kg} \
end{align*}
$$

Unknown:

$$
begin{align*}
F_{12}&=? tag{force exerted by upper block on the lower block} \
F_{21}&=? tag{force with which the lower block acts on the upper} \
\
F_{12}&=m_cdot g \
&=6 mathrm{kg}cdot -9.8 mathrm{m/s^2} \
&=boxed{-58.8 mathrm{N}} \
\
F_{21}=-F_{12} \
&=boxed{58.8 mathrm{N}}
end{align*}
$$

Positive direction: upward

b) $F_{12}=-58.8 mathrm{N}$

Known:

$$
begin{align*}
m&=2.45 mathrm{mg}=2.45cdot 10^{-6} mathrm{kg} \
g&=9.8 mathrm{m/s^2}
end{align*}
$$

Unknown:

$$
begin{align*}
F_{rE}&=? tag{force of rain drop exerted on Earth}
end{align*}
$$

$$
F_g=mcdot g
$$

We also know if the Earth acts by force on a droplet then the droplet also acts by force on the Earth due to Newton’s third law.

$$
begin{align*}
F_{21}&=-F_{12} tag{third Newton’s law} \
F_{rE}&=-F_{Er} tag{$F_{Er} $force of Earth exerted on rain drop}
end{align*}
$$

$$
begin{align*}
F_{Er}&=mcdot g \
&=2.45cdot 10^{-6} mathrm{kg}cdot 9.8 mathrm{m/s^2} \
&=2.401cdot 10^{-5} mathrm{N} \
\
F_{rE}&=-F_{Er} \
&=boxed{-2.401cdot 10^{-5} mathrm{N}}
end{align*}
$$

Notice that force of gravity $F_{Er}$ is positive meaning that positive direction is toward the Earth, ie. downward.

$F_{55} = m g = (55)*(0.025) = 1.4 N$

The force on the 90 Kg man is equal to the force on the 55 Kg man:

$$
F_{90} = F_{55} = 1.4 N
$$

$$
begin{align*}
m_H&=75 mathrm{kg} \
m_L&=170 mathrm{kg} \
a&=10 mathrm{m/s^2}
end{align*}
$$

Unknown:

$$
begin{align*}
F_L/F_H&=?
end{align*}
$$

$$
begin{align*}
F&=mcdot a
end{align*}
$$

Then to get the ratio we simply divide the two forces.

$$
begin{align*}
F_L&=m_Lcdot a \
&=170 mathrm{kg}cdot 9.8 mathrm{m/s^2} \
&=1666 mathrm{N}
end{align*}
$$

$$
begin{align*}
F_H&=m_Hcdot a \
&=75 mathrm{kg}cdot 9.8 mathrm{m/s^2} \
&=735 mathrm{N}
end{align*}
$$

$$
begin{align*}
F_L/F_H&=dfrac{F_L}{F_H} \
&=dfrac{1666 mathrm{N}}{735 mathrm{N}} \
&=boxed{2.267}
end{align*}
$$

To have the same acceleration as a human sprinter, a lion must use approximately twice as much force because of its approximately twice as much mass.

$F_{grav}$ = (4500)(9.8)

$F_{grav}$ = 44100 N

Let’s look at the first part: What’s the gravitational force on that helicopter? We know it’s mass and the acceleration due to gravity (9.8 m/$s^{2}$).

$F_{motion}$ = (4500)(2.0)

$F_{motion}$ = 9000 N

$F_{lift}$ = (44100) + (9000)

$F_{lift}$ = 53100 N

Known:

$$
begin{align*}
m_U&=4.6 mathrm{kg} \
m_M&=1.2 mathrm{kg} \
m_D&=3.7 mathrm{kg} \
end{align*}
$$

Unknown:

$$
begin{align*}
F_n? tag{all normal forces in the task}
end{align*}
$$

In this task, 6 normal forces act, three in the downward direction and three in the upward direction. It is enough to calculate the forces in only one direction. We will calculate the forces in the second using Newton’s third law. Let’s choose the direction down. Here we actually calculate the weights of the blocks because the forces with which the upper blocks act on the lower ones are their weights. We have to make sure that the second and third blocks have a block above them so we have to add their weight as well.

Notice that the positive direction is downward.

Positive direction: downward.

Known:

$$
begin{align*}
s&=402.3 mathrm{m} \
t&=4.936 mathrm{s} \
end{align*}
$$

Unknown:

$$
begin{align*}
a&=? \
v_f&=? \
\
s&=dfrac{1}{2} acdot t^2 \
a&=dfrac{2s}{t^2} \
&=dfrac{2cdot 402.3 mathrm{m}}{4.936^2 mathrm{s^2}} \
&=boxed{33.02 mathrm{m/s^2}} \
\
v&=at \
\
v_f&=at \
&=33.02 mathrm{m/s^2}cdot 4.936 mathrm{s}\
&=boxed{163 mathrm{m/s^2}}
end{align*}
$$

Known:

$$
begin{align*}
F_g&=2.75cdot 10^6 mathrm{N} tag{weight of jet} \
F_t&=6.35cdot 10^6 mathrm{N} tag{thrust force}\
v_f&=285 mathrm{km/h}=79.17 mathrm{m/s} \
end{align*}
$$

Unknown:

$$
begin{align*}
s&=? tag{required runway length}\
\
F_g&=mg \
m&=dfrac{F_g}{g} tag{mass of jet} \
&=dfrac{2.75cdot 10^6 mathrm{N}}{9.8 mathrm{m/s^2}} \
&=2.806cdot 10^5 mathrm{kg} \
\
F_t&=mcdot a \
a&=dfrac{F_t}{m} \
&=dfrac{6.35cdot 10^6 mathrm{N}}{2.806cdot 10^5 mathrm{kg}} \
&=22.63 mathrm{m/s^2} \
\
v^2&=2cdot a cdot s \
s&=dfrac{v^2}{2cdot a} \
&=dfrac{v_F^2}{2cdot a} \
&=dfrac{79.17 mathrm{m/s}^2}{2cdot 22.63 mathrm{m/s^2}}\
&=boxed{138.5 mathrm{m}}
end{align*}
$$

$$
begin{align*}
m&=873 mathrm{kg} \
Delta v&=26.3 mathrm{m/s} \
Delta t&=0.59 mathrm{s} \
v_f&=126.6 mathrm{m/s} \
s&=402.3 mathrm{m} tag{assumed race lenght} \
\
a_1&=dfrac{Delta v}{Delta t } tag{acceleration from task 68} \
&=dfrac{26.3 mathrm{m/s}}{0.59 mathrm{s}} \
&=44.58 mathrm{m/s^2} \
\
v^2&=2cdot acdot s \
a&=dfrac{v^2}{2s} \
a_2&=dfrac{v_f^2}{2s} tag{average acceleration of whole race} \
&=dfrac{126.6^2 mathrm{m^2/s^2}}{2cdot 402.3 mathrm{m}} \
&=19.92 mathrm{m/s^2}
end{align*}
$$

We can see that the average acceleration $a_2$ is less than the acceleration at the beginning of $a_1$ which means that the acceleration is certainly not constant throughout the race. Another way to determine whether the acceleration is constant is to measure the time in several places, measure the distances from the start to the measurement site and display the data in a $s-t^2$ diagram, if the direction cannot be drawn through the data, ie if the dependence is not linear. acceleration is not constant.

$$
begin{align*}
m_B&=65 mathrm{kg} \
m_G&=45 mathrm{kg} \
a_G&=-3 mathrm{m/s^2} \
a_B&=? \
\
F_{12}&=-F_{21} tag{Third Newton’s law} \
F_{BG}&=-F_{GB} tag{ the force with which the boy acts on the girl through the force of tension and vice versa} \
m_Bcdot a_B&=-(m_Gcdot a_G) \
a_B&=dfrac{-(m_Gcdot a_G)}{m_B} \
&=dfrac{-(45 mathrm{kg}cdot -3 mathrm{m/s^2})}{65 mathrm{kg}} \
&=2.077 mathrm{m/s^2}
end{align*}
$$

$$
begin{align*}
F_g&=588 mathrm{N} \
a&=3 mathrm{m/s^2} \
F_{wP}&=? tag{force with which wall acts on Pratish}
end{align*}
$$

We must first calculate Pratisha’s mass from the weight. When we calculate the mass we can calculate the force required for a given acceleration.

$$
begin{align*}
F_g&=mg \
m&=dfrac{F_g}{g} \
&=dfrac{588 mathrm{N}}{9.8 mathrm{m/s^2}} \
&=60 mathrm{kg} \
\
F&=mcdot a \
F_{wP}&=mcdot a \
&=60 mathrm{kg}cdot 3 mathrm{m/s^2} \
&=180 mathrm{N}
end{align*}
$$

Known:

$$
begin{align*}
v_i&=30 mathrm{m/s} \
v_f&=0 mathrm{m/s} \
Delta t&=0.005 mathrm{s} \
m&=0.145 mathrm{kg} \
end{align*}
$$

Unknown:

$$
begin{align*}
a&=? \
F_{PB}&=? tag{force with which player acts on ball} \
F_{BP}&=? tag{force with which ball acts on player}
end{align*}
$$

Suppose the positive direction is in the direction of motion of the ball before it is caught. Acceleration is calculated from the data on the change in speed.

$$
begin{align*}
a&=dfrac{v_f-v_i}{Delta t} \
&=dfrac{0 mathrm{m/s}-30 mathrm{m/s}}{0.005 mathrm{s}} \
&=boxed{-6000 mathrm{m/s^2}}
end{align*}
$$

The force exerted on the ball required to stop it is calculated according to Newton’s second law:

$$
begin{align*}
F&=mcdot a \
F_{PB}&=mcdot a\
&=0.145 mathrm{kg}cdot -6000 mathrm{m/s^2} \
&=boxed{-870 mathrm{N}}
end{align*}
$$

The force acting on the player is calculated using Newton’s third law:

$$
begin{align*}
F_{BP}&=-F_{PB} \
&=boxed{870 mathrm{N}}
end{align*}
$$

b) $F_{PB}=-870 mathrm{N}$

c) $F_{BP}=870 mathrm{N}$

Known:

$$
begin{align*}
m&=0.25 mathrm{kg} \
F&=12 mathrm{N} \
Delta t&=9 mathrm{s} \
end{align*}
$$

Unknown:

$$
begin{align*}
a&=? \
v_f&=?
end{align*}
$$

Friction can be ignored, gravity is reversed by the force with which air acts upwards. The only force we are interested in is the one with which he was pushed. Acceleration is then calculated from Newton’s second law. When we calculate the acceleration, we can easily calculate the final speed of the puck.

$$
begin{align*}
F&=mcdot a \
a&=dfrac{F}{m} \
&=dfrac{12 mathrm{N}}{0.25 mathrm{kg}} \
&=boxed{48 mathrm{m/s^2}} \
\
v&=acdot t \
v_f&=acdot t \
&=48 mathrm{m/s^2}cdot 9 mathrm{s} \
&=boxed{432 mathrm{m/s}}
end{align*}
$$

By calculation, the final speed turns out to be very high, even unrealistically high. Although the force is not large the time of action is relatively large. Anyone who has played air hockey knows that the force on the puck acts only for a fraction of a second because the puck is started by hitting.

b) $v_f=432 mathrm{m/s}$

Known:

$$
begin{align*}
F_{s(v=0)}&=836 mathrm{N} tag{the force shown by the scale} \
F_{s(A)}&=936 mathrm{N} \
F{s(B)}&=782 mathrm{N} \
end{align*}
$$

Unknown:

$$
begin{align*}
a_A&=? \
a_B&=?
end{align*}
$$

At rest the scale shows the amount of student weight. When the elevator accelerates, the scale shows the net force of the student’s weight and the inertial force that we will calculate using Newton’s second law.

b) $a_B=0.633 mathrm{m/s^2}$

c) More time is needed to stop

Known:

$$
begin{align*}
m&=5 mathrm{kg} \
F_{up}&=98 mathrm{N} \
t&=10 mathrm{s} \
end{align*}
$$

Unknown:

$$
begin{align*}
a&=? \
v_f&=?
end{align*}
$$

We must first calculate the net force on the instruments which is actually the difference between the upward force and the gravitational force. After that, we get the acceleration using Newton’s second law.

$$
begin{align*}
F_{net}&=F_{up}-F_g \
&=98 mathrm{N}-5 mathrm{kg}cdot 9.8 mathrm{m/s^2} \
&=49 mathrm{N}
end{align*}
$$

$$
begin{align*}
F&=mcdot a \
a&=dfrac {F}{m} \
\
a&=dfrac{F_{up}}{m} \
&=dfrac{49 mathrm{N}}{5 mathrm{kg}} \
&=boxed{9.8 mathrm{m/s^2}} \
\
v&=acdot t \
v_f&=acdot t \
&=9.8 mathrm{m/s^2}cdot 10 mathrm{s} \
&=boxed{98 mathrm{m/s}}
end{align*}
$$

After releasing the instruments, $textbf{only gravity}$ acts on them and they start to slow down.

$$
begin{align*}
v&=acdot t \
t&=dfrac{v}{a} \
\
t&=dfrac{v}{g} \
&=dfrac{98 mathrm{m/s}}{9.8 mathrm{m/s^2}} \
&=boxed{10 mathrm{s}}
end{align*}
$$

We see that the deceleration time is equal to the acceleration time.

b) $v=98 mathrm{m/s}$

c) Force of gravity

d) The instruments will change direction after $10 mathrm {s}$

Known:

$$
begin{align*}
F&=4.5 mathrm{N} \
a_1&=2.5 mathrm{m/s^2} \
m_2&=4 mathrm{kg} \
end{align*}
$$

Unknown:

$$
begin{align*}
a_2&=? tag{acceleration of combination}
end{align*}
$$

To calculate the acceleration of a combination of blocks we need to know the total mass which means we need to calculate the mass of the first block. We will calculate this using Newton’s 2nd law, then only the total mass and the total force are included in the same law.

$$
begin{align*}
F&=mcdot a \
m&=dfrac{F}{a} \
\
m_1&=dfrac{F}{a_1} \
&=dfrac{4.5 mathrm{N}}{2.5 mathrm{m/s^2}} \
&=1.8 mathrm{kg} \
\
a&=dfrac{F}{m} \
a_2&=dfrac{F}{m_1+m_2} \
&=dfrac{4.5 mathrm{N}}{1.8 mathrm{kg}+4 mathrm{kg}} \
&=boxed{0.7759 mathrm{m/s^2}}
end{align*}
$$

Known:

$$
begin{align*}
m_1&=4.3 mathrm{kg} \
m_2&=5.4 mathrm{kg} \
F&=22.5 mathrm{N} \
end{align*}
$$

Unknown:

$$
begin{align*}
F_{12}&=? tag {force of $4.3 mathrm{kg}$ block on $5.4 mathrm{kg}$ block} \
F_{21}&=? tag {force of $5.4 mathrm{kg} $ block on $4.3 mathrm{kg}$ block }
end{align*}
$$

It is first necessary to calculate the acceleration of the system, which we will do using Newton’s second law.

$$
begin{align*}
F&=mcdot a \
a&=dfrac{F}{m} \
\
a&=dfrac{F}{m_1+m_2} \
&=dfrac{22.5 mathrm{N}}{4.3 mathrm{kg}+5.4 mathrm{kg}} \
&=boxed{2.32 mathrm{m/s^2}}
end{align*}
$$

$$
begin{align*}
F&=mcdot a \
\
F_{12}&=m_2cdot a \
&=5.4 mathrm{kg}cdot 2.32 mathrm{m/s^2} \
&=boxed{12.53 mathrm{N}} \
\
F_{21}&=-F_{12} \
&=boxed{-12.53 mathrm{N}}
end{align*}
$$

b) $F_{12}=12.53 mathrm{N}$

c) $F_{21}=-12.53 mathrm{N}$

Known:

$$
begin{align*}
m_1&=3 mathrm{kg} \
m_2&=5 mathrm{kg} \
end{align*}
$$

Unknown:

$$
begin{align*}
F_T&=? \
a&=?
end{align*}
$$

To solve this problem, consider which forces act on a particular block. The forces acting on the block are the gravitational force and the tension force of the rope. The force of gravity varies from block to block because they are not of equal mass but the tension have to be the same.

To calculate the rope tension force we will have to solve the equations for both blocks at the same time.

$$
begin{align*}
F_T&=F_{G1}+F \
&=F_{G1}+m_1cdot a \
F_T&=F_{G2}+F \
&=F_{G2}+m_2cdot -a\
\
a&=dfrac{F_T-F_{G1}}{m_1} \
a&=dfrac{F_{G2}-F_T}{m_2}
end{align*}
$$

Pay attention to the sign of acceleration in individual blocks.

$$
begin{align*}
dfrac{F_T-F_{G1}}{m_1}&=dfrac{F_{G2}-F_T}{m_2} \
dfrac{F_T-m_1g}{m_1}&=dfrac{m_2g-F_T}{m_2} \
m_2F_T-m_1m_2g&=m_1m_2g-m_1F_T tag{crosswise multiplication} \
(m_1+m_2)F_T&=2m_1m_2g \
F_T&=dfrac{2m_1m_2g}{(m_1+m_2)}
end{align*}
$$

$$
begin{align*}
a&=dfrac{F_T-F_{G1}}{m_1} \
&=dfrac{F_T-m_1g}{m_1} \
&=dfrac{36.75 mathrm{N}-3 mathrm{kg}cdot 9.8 mathrm{m/s^2}}{3 mathrm{kg}} \
&=boxed{2.45 mathrm{m/s^2}}
end{align*}
$$

b) $a=2.45 mathrm{m/s^2}$

$$
begin{align*}
m_A&=2 mathrm{kg} \
m_B&=3 mathrm{kg} \
end{align*}
$$

Unknown:

$$
begin{align*}
a&=? \
\
F_{net}&=mcdot a \
\
F_{net(A)}&=F_T-F_{gA} \
F_{net(B)}&=F_{GB}-F_T \
\
F_T&=m_Acdot a+F_{gA} \
F_T&=F_{GB}-m_Bcdot a\
\
m_A cdot a+F_{gA}&=F_{GB}-m_B cdot a\
a(m_A+m_B)&=F_{GB}-F_{GA} \
a&=dfrac{F_{GB}-F_{GA}}{(m_A+m_B)}\
&=dfrac{m_Bg-m_Ag}{m_A+m_B} \
&=dfrac{3 mathrm{kg}cdot 9.8 mathrm{m/s^2}-2 mathrm{kg}cdot 9.8 mathrm{m/s^2}}{3 mathrm{kg}+2 mathrm{kg}} \
&=boxed{1.96 mathrm{m/s^2}}
end{align*}
$$

$$
a=1.96 mathrm{m/s^2}
$$

$$
begin{align*}
m_A&=1 mathrm{kg} \
m_B&=4 mathrm{kg}
end{align*}
$$

Unknown:

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

$$
begin{align*}
F_{net(A)}&=F_{T(A)}-F_{GA} \
F_{net(B)}&=F_{GB}-F_{T(B)}
end{align*}
$$

We know that this difference of forces causes motion so we can replace the net force by another Newton’s law, and express the force of tension for a particular block.

$$
begin{align*}
F_{net}&=mcdot a \
\
F_{T(A)}&=m_Acdot a+F_{GA} \
F_{T(B)}&=F_{GB}-m_Bcdot a\
end{align*}
$$

$$
begin{align*}
F_{T(A)}&=F_{T(B)} \
m_A cdot a+F_{gA}&=F_{GB}-m_B cdot a\
a(m_A+m_B)&=F_{GB}-F_{GA} \
a&=dfrac{F_{GB}-F_{GA}}{(m_A+m_B)}
end{align*}
$$

$$
begin{align*}
a&=dfrac{m_Bg-m_Ag}{m_A+m_B} \
&=dfrac{4 mathrm{kg}cdot 9.8 mathrm{m/s^2}-1 mathrm{kg}cdot 9.8 mathrm{m/s^2}}{4 mathrm{kg}+1 mathrm{kg}} \
&=boxed{5.88 mathrm{m/s^2} }
end{align*}
$$

$$
begin{align*}
F_{t=0}&=0 mathrm{N} \
F_{t=0.001}&=10^4 mathrm{N} \
F_{t=0.002}&=0 mathrm{N}
end{align*}
$$

$$
begin{align*}
F_{avg}&=dfrac{F_{t=0}+F_{t=0.001}}{2} \
&=dfrac{0 mathrm{N}+10^4 mathrm{N}}{2} \
&=boxed{5000 mathrm{N}}
end{align*}
$$

Known values:

$$
begin{align*}
F_{avg}&=5000 mathrm{N}\
m&=0.145 mathrm{kg}
end{align*}
$$

Unknown:

$$
begin{align*}
a_{avg}&=? \
\
F&=mcdot a tag{Newton’s second law} \
a&=dfrac{F}{m} \
\
a_{avg}&=dfrac{F_{avg}}{m} \
&=dfrac{5000 mathrm{N}}{0.145 mathrm{kg}} \
&=boxed{3.448cdot 10^4 mathrm{m/s^2}}
end{align*}
$$

$$
begin{align*}
a_{avg}&=3.448cdot 10^4 mathrm{m/s^2} \
v_i&=-25 mathrm{m/s} \
t&=0.002 mathrm{s} \
v_f&=? \
\
v_f&=v_i+acdot t \
&=-25 mathrm{m/s}+3.448cdot10^4 mathrm{m/s^2}cdot 0.002 mathrm{s} \
&=boxed{43.97 mathrm{m/s}}
end{align*}
$$

b) $a_{avg}=3.448cdot 10^4 mathrm{m/s^2}$

c) $v_f=43.97 mathrm{m/s}$

Known:

$$
begin{align*}
m_1&=2 mathrm{kg} \
m_2&=4 mathrm{kg} \
m_3&=6 mathrm{kg} \
F&=36 mathrm{N} \
end{align*}
$$

Unknown:

$$
begin{align*}
a&=? \
F_{T1}&=? \
F_{T2}&=?
end{align*}
$$

$$
begin{align*}
F=&mcdot a\
a&=dfrac{F}{m} \
\
a&=dfrac{F}{(m_1+m_2+m_3)} \
&=dfrac{36 mathrm{N}}{2 mathrm{kg}+4 mathrm{kg}+6 mathrm{kg}} \
&=boxed{3 mathrm{m/s^2}}
end{align*}
$$

$$
begin{align*}
F_{T1}&=F_1 \
&=m_1cdot a \
&=2 mathrm{kg}cdot 3 mathrm{m/s^2} \
&=boxed{6 mathrm{N}} \
\
F_{T2}&=F_2+F_{T1} \
&=m_2cdot a+F{T1} \
&=4 mathrm{kg}cdot 3 mathrm{m/s^2} +6 mathrm{N} \
&=boxed{18 mathrm{N}}
end{align*}
$$

b) $F_{T1}=6 mathrm{N}, F_{T2}=18 mathrm{N}$

Known:

$$
begin{align*}
m&=3.46 mathrm{kg}\
g&=9.8 mathrm{m/s^2} \
end{align*}
$$

Unknown:

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

$$
begin{align*}
2cdot F_T&=F_G \
F_T&=dfrac{F_G}{2} \
&=dfrac{mcdot g}{2} \
&=dfrac{3.46 mathrm{kg}cdot 9.8 mathrm{m/s^2}}{2} \
&=boxed{16.95 mathrm{N}}
end{align*}
$$

The experiment would be carried out very quickly and very easily. The device would simply be placed on the ground and the spring extension would be read. On the planet Pluto where $g_P=0.3 mathrm {m/ s^2}$ our spring would extend by $l=3 mathrm {mm}$ On a scale we would read this as $0.3 mathrm{m/s^2}$. The same procedure would apply to Mercury but due to $g_M =3.7 mathrm{m/s^2}$ the spring would be extended by $l =3.7 mathrm{cm}$.

$textbf {A body at rest or moving at a constant speed will remain in that state until a non-zero force begins to act on it}$

From this law we can read the following:

Force is needed to start moving or to change direction but not to move the body.

Care should be taken when claiming that no force is required to maintain speed. It is valid only when no force acts on the body. In the real world we know very well from experience that it takes some force to move at a certain speed. The reason for this is that in the real world a force of friction acts, which means that a force of equal amount and opposite direction should be applied.

Known:

$$
begin{align*}
t_i&=8:25 mathrm{A.M.} \
v_{avg}&=8 mathrm{km/h} \
s&=5.2 mathrm{km} \
end{align*}
$$

Unknown:

$$
begin{align*}
t_f&=? \
\
s&=vcdot t \
t&=dfrac{s}{v}\
\
t&=dfrac{s}{v_{avg}} \
&=dfrac{5.2 mathrm{km}}{8 mathrm{km/h}} \
&=0.65mathrm{h} \
&=39 mathrm{min}
\
t_f&=t_i+t \
&=8:25 mathrm{A.M.}+39 mathrm{min} \
&=boxed{9:04 mathrm{A.M.}}
end{align*}
$$

A faster car is one with a higher speed. If we know that in the $s-t$ graph the speed is actually the slope of the curve then we can easily compare the speeds of the car. At $t =7 mathrm {s}$ the car A has a higher speed.

As we have already said, velocity is the slope of the graph. The velocities A and B are equal to $t = 5 mathrm {s}$.

If we look at the curve showing the motion of the B car, we can see that the slope never increases but on the contrary falls all the time.

b) Car A

c) $t=5 mathrm{s}$

d) Never

e) Through out the movement, From $t=3 mathrm{s}$ to $t=8 mathrm{s}$.

Since we have no data for car B at the moment $t = 2 mathrm {s}$ we cannot determine the speed, we can say that it did not move because it did not cross any road and time passed so we can say that its speed was $v = 0 mathrm {m / s}$.

At the moment $t = 9 mathrm {s}$ the curve describing the motion of car B is almost parallel to the x axis which means that the distance traveled is constant over time which means that its velocity $v=0 mathrm{m / s}$.

At the moment $t= 2 mathrm{s}$ on the curve describing the motion of car A we can draw a tangent passing through the points $(2,0)$ and $4,4)$ which means that its slope ie speed is $v= 2 mathrm{m/s}$.

b) $v_{B,t=9 mathrm{s}}=0 mathrm{m/s}$

c) $v_{A,t=2 mathrm{s}}=2 mathrm{m/s}$