${large textbf{Knowns}}$

The kinetic friction force on an object is proportional to the normal force of that object:

$$
begin{gather*}
f_k = mu_k Ntag{1}
end{gather*}
$$

Where the proportionality constant is the coefficient of kinetic friction $mu_k$.

${large textbf{Givens}}$

The mass of the salt shaker is $m = 50.0$ g and the coefficient of kinetic friction between the shaker and the table is $mu_k = 0.120$.

${large textbf{Calculations}}$

Since there is no acceleration in the vertical direction, we apply $textbf{Newton’s first law}$ to the shaker in the $y$ direction, so we get:

$$
begin{align*}
sum F_y = N – w &= 0\\
N &= w = mg
end{align*}
$$

Now, we substitute for $N$ into equation (1), so we get the kientic friction on the shaker:

$$
begin{gather*}
f_k = mu_k m g
end{gather*}
$$

In the horizontal direction, the kinetic friction is the only force acting on the shaker.

Since the shaker is moving with a nonzero acceleration in the horizontal direction, we apply $textbf{Newton’s second law}$ in the $x$ direction, so we get:

$$
begin{align*}
sum F_x = f_k = ma
end{align*}
$$

Now, we substitute for $f_k$ and solving for $a$, so we get:

$$
begin{align*}
mu_k m g &= ma\\
a & = mu_k g
end{align*}
$$

Finally, we plug our value for $mu_k$, so we get:

$$
begin{gather*}
a = (0.120)cdot (9.81text{ m/s}^2) = 1.18text{ m/s}^2\\
therefore quad large boxed{a = 1.18text{ m/s}^2}
end{gather*}
$$

$$
a = 1.18text{ m/s}^2
$$

${large textbf{Knowns}}$

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

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

The kinetic friction force on an object is proportional to the normal force of that object:

$$
begin{gather*}
f_k = mu_k Ntag{1}
end{gather*}
$$

Where the proportionality constant is the coefficient of kinetic friction $mu_k$.

From the velocity-position equation, the equation that relates the initial and final velocities with the acceleration and the distance traveld is:

$$
begin{gather*}
v_f^2 = v_i^2 + 2a Delta xtag{2}
end{gather*}
$$

${large textbf{Givens}}$

The initial speed of the player is $v_i = 4.00$ m/s and the coefficient of kinetic friction between the player and the ground is $mu_k = 0.460$.

${large textbf{Calculations}}$

Since there is no acceleration in the vertical direction, we apply $textbf{Newton’s first law}$ to the player in the $y$ direction, so we get:

$$
begin{align*}
sum F_y = N – w &= 0\\
N &= w = mg
end{align*}
$$

Now, we substitute for $N$ into equation (1), so we get the kientic friction on the player:

$$
begin{gather*}
f_k = mu_k m g
end{gather*}
$$

In the horizontal direction, the kinetic friction is the only force acting on the player.

Since the player is moving with a nonzero acceleration in the horizontal direction, we apply $textbf{Newton’s second law}$ in the $x$ direction, so we get:

$$
begin{align*}
sum F_x = f_k = ma
end{align*}
$$

Then, we substitute for $f_k$ and solving for $a$, so we get:

$$
begin{align*}
mu_k m g &= ma\\
a & = mu_k g
end{align*}
$$

Now, we plug our value for $mu_k$, so we get the magnitude of the acceleration of the player:

$$
begin{gather*}
a = (0.460)cdot (9.81text{ m/s}^2) = 4.51text{ m/s}^2
end{gather*}
$$

We know that the friction force always acts in the opposite direction of the motion, so the acceleration of the player is also in the opposite direction of the motion, $vec{textbf{a}} = -4.51text{ m/s}$.

Since the player eventually comes to rest, $v_f = 0$, we plug our values for $v_f$, $v_i$ and $a$ into equation (2) and solve for $x$, so we get the distance the player slides before he stops:

$$
begin{align*}
0 &= (4.00text{ m/s})^2 + 2 (-4.51text{ m/s}^2) x\\
x &= dfrac{(4.00text{ m/s})^2}{2 (4.51text{ m/s}^2)} = 1.77text{ m}
end{align*}
$$

$$
therefore quad large boxed{x = 1.77text{ m}}
$$