From $textbf{Newton’s second law}$ for multiple forces, the net force acting on an object is equal to the mass of the object times its acceleration:

$$
begin{gather*}
sum vec{textbf{F}} = mvec{textbf{a}}tag{1}
end{gather*}
$$

The mass of the firefighter is $m = 97.0$ kg and the force exerted on him by the pole is $F_p = 650$ N.

Same as in Guided Example 5.7, there are only two forces acting on the firefighter, one exerted by the pole directed upward and the other is its weight directed downward.

We take upward as the positive direction, so we apply $textbf{Newton’s second law}$ to the firefighter in the $y$-direction, so we get:

$$
begin{align*}
sum F_y = F_p – w &= ma\\
F_p – mg &= ma\\
F_p & = m(a + g)\\
a & = dfrac{F_p}{m} – g
end{align*}
$$

Now, we plug our values for $F_p$ and $m$, so we get the acceleration of the firefighter:

$$
begin{gather*}
a = dfrac{650text{ N}}{97.0text{ kg}} – 9.81text{ m/s}^2 = -3.10text{ m/s}^2\\
therefore quad large boxed{a = -3.10text{ m/s}^2}
end{gather*}
$$

This means that the acceleration is directed upward and it continues to slow the firefighter’s motion until it stops.

From $textbf{Newton’s second law}$ for multiple forces, the net force acting on an object is equal to the mass of the object times its acceleration:

$$
begin{gather*}
sum vec{textbf{F}} = mvec{textbf{a}}tag{1}
end{gather*}
$$

The weight of an object is the mass of the object time the acceleration due to gravity:

$$
begin{gather*}
w = mgtag{2}
end{gather*}
$$

The force you exert on the suitcase is $F_l = 105$ N and its acceleration is $a = 0.705text{ m/s}^2$.

(a) Since the suitcase is set to move when you lift it, this means that the acceleration is in the same direction as the force you exert.

There are two forces acting on the suitcase, one is the force you exert which is directed upward and the other is the weight of the suitcase which is directed downward.

Taking upward as the positive direction, we apply $textbf{Newton’s second law}$ to the suitcase in the $y$-direction, so we get:

$$
begin{align*}
sum F_y = F_l – w & = m a\\
F_l – mg & = ma
end{align*}
$$

Solving for $m$, we get:

$$
begin{align*}
F_l &= ma + mg\\
& = m(a + g)\\
m & = dfrac{F_l}{a + g}
end{align*}
$$

Now, we plug our values for $a$ and $F_l$, so we get the mass of the suitcase:

$$
begin{gather*}
m = dfrac{105text{ N}}{( 0.705 + 9.81) text{ m/s}^2} = 9.99text{ kg}\\
therefore quad large boxed{m = 10.0text{ kg}}
end{gather*}
$$

$$
begin{gather*}
w = (10.0text{ kg})cdot (9.81text{ m/s}^2) = 98.1text{ N}\\
therefore quad large boxed{w = 98.1text{ N}}
end{gather*}
$$

(b) $w = 98.1$ N