${large textbf{Knowns}}$

From $textbf{Newton’s second law}$ for a single force, the force $F$ acting on an object of mass $m$ and moving with an acceleration $a$ is given by:

$$
begin{gather*}
F = matag{1}
end{gather*}
$$

${large textbf{Given}}$

The mass of the canoe, in this case, is $m = 25000$ kg, and, from Guided Example 5.5, the force acting on the canoe 2 is $F_2 = 46.0$ N.

${large textbf{Calculations}}$

(a) Rearranging equation (1), we get:

$$
begin{gather*}
a = dfrac{F}{m}tag{2}
end{gather*}
$$

So, according to $textbf{Newton’s second law}$, the acceleration of an object is inversely proportional to its mass, $boldsymbol{a propto 1/m}$, when the force acting on it is constant.

Thus, if the force acting on the canoe does not change and its mass increases, its acceleration $textbf{decreases}$.

(b) We simply plug our values for $F$ and $m$ into equation (2), so we get the acceleration of canoe 2 with its new mass:

$$
begin{gather*}
a = dfrac{46.0text{ N}}{25000text{ kg}} = 0.0018text{ m/s}^2\\
therefore quad large boxed{a = 0.0018text{ m/s}^2}
end{gather*}
$$

Comparing with its acceleration when its mass was only $250$ kg, we find that the acceleration decreases to $0.01$ of its original accelration.

(a) According to $textbf{Newton’s second law}$, if the mass of canoe 2 increases, its acceleration will $textbf{decrease}$.

(b) $a = 0.0018text{ m/s}^2$

(a) The two forces acting on the book are:

1- $textbf{Gravitational force}$: exerted by the earth on the book and it is directed downward and its magnitude is called $textit{weight}$.

2- $textbf{Action force}$: exerted by your hand on the book and it is directed upward, this force is normally called $textit{normal force}$.

(b) According to $textbf{Newton’s second law}$ for multiple forces, since the book is at rest, the net force acting on it must equal $zero$;

Thus, the two forces acting on the book must be equal in magnitude and opposite in direction so that they cancel each other.

(c) $textbf{No}$, these force are not action-reaction pair.

This is because

– These two forces are different (gravitational and normal forces).

– Both are acting on the same object, not on $textit{different objects}$.

– They cancel each other.

And we know that an action-reaction pair are two forces of the same nature acting on different objects and they do not cancel each other.

(a) Gravitational and action (by your hand) forces.

(b) $textbf{Yes}$, they are equal in magnitude and opposite in direction.

(c) $textbf{No}$. They are different forces acting on the same object and cancel each other.

${large textbf{Knowns}}$

From $textbf{Newton’s second law}$, the force $F$ acting on an object of mass $m$ moving with an acceleration $a$ is given by:

$$
begin{gather*}
F = matag{1}
end{gather*}
$$

From $textbf{Newton’s third law}$:

– For every action force acting on an object, there is a reaction force acting on a different object.

– The action and reaction forces are equal in magnitude and opposite in direction.

${large textbf{Givens}}$

The mass of the parent is $m_p = 71.0$ kg, the mass of the child is $m_c = 19.0$ kg and the aceleration of the child is $a_c = 2.60text{ m/s}^2$.

${large textbf{Calculations}}$

First, we plug our values for $m_c$ and $a_c$ into equation (1), so we get the magnitude of the force acting on the child:

$$
begin{gather*}
F_c = (19.0text{ kg}) cdot (2.60text{ m/s}^2) = 49.4text{ N}
end{gather*}
$$

According to $textbf{Newton’s third law}$, the force exerted by the parent on the child equals in magnitude and opposite in direction to the force exerted by the child on the parent.

$$
begin{gather*}
F_p = -F_c = – 49.4text{ N}
end{gather*}
$$

Now, we plug this value and our value for $m_p$ into equation (1) and evaluate for $a_p$, so we get the acceleration of the parent:

$$
begin{gather*}
a_p = dfrac{F_p}{m_p} = dfrac{-49.4text{ N}}{71.0text{ kg}} = -0.696text{ m/s}^2\\
therefore quad large boxed{a_p = 0.696text{ m/s$^2$, opposite to that of the child}}
end{gather*}
$$

$a_p = 0.696text{ m/s}^2$ opposite to that of the child.