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

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

The acceleration of the boat is $a = 0.530text{ m/s}^2$ and the three forces acting on it have magnitudes of $F = 85.0$ N.

Since the three forces are in the same direction (to the right) and have the same magnitude, the net force is to the right and has a magnitude equals the sum of the three forces,

$$
begin{align*}
sum F_x &= F + F + F = 3F\\
& = 3 (85.0text{ N}) = 255text{ N}
end{align*}
$$

Rearrainging equation (1), we get:

$$
begin{gather*}
m = dfrac{sum F_x}{a}
end{gather*}
$$

Now, we plug our values for $sum F_x$ and $a$, so we get the mass of the boat:

$$
begin{gather*}
m = dfrac{255text{ N}}{0.530text{ m/s}^2} = 481text{ kg}\\
therefore quad large boxed{m = 481text{ kg}}
end{gather*}
$$

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

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

The mass of the car is $m = 1500$ kg and its acceleration is $a = 3.80text{ m/s}^2$.

We simply plug our values for $m$ and $a$ into equation (1), so we get the force that must be exerted on the car:

$$
begin{gather*}
F = (1500text{ kg})cdot (3.80text{ m/s}^2) = 5700text{ N}\\
therefore quad large boxed{F = 5700text{ N}}
end{gather*}
$$

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

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

The magnitude of the exerted force on the sled by you is $F = 120$ N, the acceleration of the sled is $a = 2.50text{ m/s}^2$ and the mass of the sled is $m_{text{sled}} = 7.40$ kg.

Since the exerted force by you is the only force acting on the sled (the ice is frictionless), we can use $textbf{Newton’s second law}$ for a single force.

Rearranging equation (1), we get:

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

Now, we plug our values for $F$ and $a$, so we get $m$:

$$
begin{gather*}
m = dfrac{120text{ N}}{2.50text{ m/s}^2} = 48.0text{ kg}
end{gather*}
$$

This is the mass of the sled and your sister combined, $m = m_{text{sled}} + m_{text{sister}}$.

Thus, the mass of your sister is:

$$
begin{gather*}
m_{text{sister}} = m – m_{text{sled}} = 48.0text{ kg} – 7.40text{ kg}\\
therefore quad large boxed{m_{text{sister}} = 40.6text{ kg}}
end{gather*}
$$