Celebrating a Soccer Goal

a) To begin, we have to identify all of forces exerted on the soccer player while he’s on his knees. We’ll have his weight \(\vec{W}\) pressing downwards, the normal force exerted upwards by the ground \(\vec{N}\), and the friction force \(\vec{f}_r\) along the opposite direction of the player’s initial velocity \(\vec{v}_i\).

*Show free body diagram of the player with the weight, normal force, and friction. Also indicate the direction of the initial velocity. Place coordinate system with positive X to the right*

Because the player does not move along the Y axis, the magnitude of the weight must be equal to the magnitude of the normal force \(N\). So, we can write in this particular case as

\begin{equation}\label{normal}
N=mg,
\end{equation}

where \(m\) is the player’s mass and \(g=9.8\,\text{m/s}^2\) is the gravitational acceleration on Earth.

The friction force that the ground exerts on the player can be written as the product of the kinetic friction coefficient \(\mu_k\) and the magnitude of the normal force \(N\). Its direction is contrary to the motion of the player, thus

\begin{equation}\label{frcition}
\vec{f}_r=-\mu_k N\,\hat{\textbf{i}}.
\end{equation}

We can then use the work-energy theorem to find a relation between the kinetic energy \(K\) of the player and the net work \(W_{\text{net}}\) completed by all forces exerted on the player. The relation reads,

\begin{equation}\label{wk}
W_{\text{net}}=K_f-K_i,
\end{equation}

where \(K_f\) is the kinetic energy when the player has traveled a distance of \(d\) and \(K_i\) is the player’s initial kinetic energy as soon as he falls to his knees. The net work can be calculated as the dot product between the sum of all forces (which are constant) and the distance vector \(\vec{d}\). So, we write,

\begin{equation}
W_{\text{net}}=\left(\vec{W}+\vec{N}+\vec{f}_r\right)\cdot\left(\vec{d}\right).
\end{equation}

Because the dot product of two perpendicular vector is zero, we find that \(\vec{W}\cdot \vec{d}=\vec{N}\cdot\vec{d}=0\), which implies that the neither the weight nor the normal force do any work when the player slides across the ground. The distance vector can be written as

\begin{equation}\label{distance}
\vec{d}=d\,\hat{\textbf{i}}.
\end{equation}

Hence, the net work will just be the dot product between the friction \eqref{frcition} and the distance \eqref{distance}, explicitly

\begin{equation}
W_{\text{net}}=\left(-\mu_k N\,\hat{\textbf{i}}\right)\cdot\left(d\,\hat{\textbf{i}}\right),
\end{equation}

which is

\begin{equation}\label{nw}
W_{\text{net}}=-\mu_k N d \,\hat{\textbf{i}}\cdot\hat{\textbf{i}}.
\end{equation}

With \(\hat{\textbf{i}}\cdot\hat{\textbf{i}}=1\), the equation for the net work reads \eqref{nw}

\begin{equation}
W_{\text{net}}=-\mu_k N d,
\end{equation}

which, after replacing the expression for the normal force given in \eqref{normal}, becomes

\begin{equation}\label{netw}
W_{\text{net}}=-\mu_k mg d.
\end{equation}

Because the net work is negative, this suggests that the kinetic energy is diminishing. The expression for the kinetic energy for an object of mass \(m\) and speed \(v\) is

\begin{equation}\label{kinetic}
K=\frac{1}{2}mv^2.
\end{equation}

Therefore, using expression \eqref{kinetic}, equation \eqref{wk} can be written as

\begin{equation}
W_{\text{net}}=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2.
\end{equation}

Using the result for the net work given in expression \eqref{netw} in the equation above, we obtain

\begin{equation}\label{we}
-\mu_k mg d=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2.
\end{equation}

Because we want the distance at which the player stops, then \(v_{f}=0\) and equation \eqref{we} is simplified to

\begin{equation}
-\mu_k mg d=-\frac{1}{2}mv_i^2,
\end{equation}

where we can cancel the minus sign and the mass \(m\), namely

\begin{equation}\label{kgd}
\mu_kgd=\frac{1}{2}v_i^2.
\end{equation}

Solving for \(d\) in equation \eqref{kgd} (by dividing both sides by \(\mu_k g\)), we have that

\begin{equation}
d=\frac{v_i^2}{2\mu_kg}.
\end{equation}

Using the numerical values

\begin{equation}
d=\frac{(4\,\text{m/s})^2}{2(0.2)(9.8\,\text{m/s})},
\end{equation}

\begin{equation}
d\approx 4.08\,\text{m}.
\end{equation}

b) To find the distance traveled by the player when his velocity has dropped to half of his initial velocity, we can use equation \eqref{we} and \(v_{f}=\frac{v_i}{2}\) to obtain

\begin{equation}
-\mu_k mgd=\frac{1}{2}m\left(\frac{v_i}{2}\right)^2-\frac{1}{2}mv_i^2,
\end{equation}

which simplifies to

\begin{equation}
-\mu_k mg d=-\frac{3}{8}m v_i^2,
\end{equation}

where we can cancel the minus sign and the mass \(m\), namely

\begin{equation}\label{kgd2}
\mu_kgd=\frac{3}{8}v_i^2.
\end{equation}

Solving for \(d\) in equation \eqref{kgd2} (by dividing both sides by \(\mu_k g\)), we have that

\begin{equation}
d=\frac{3v_i^2}{8\mu_kg}.
\end{equation}

Using the numerical values

\begin{equation}
d=\frac{3(4\,\text{m/s})^2}{8(0.2)(9.8\,\text{m/s})},
\end{equation}

\begin{equation}
d\approx3.06\,\text{m}.
\end{equation}

Thus the player has traveled more than half of the total distance, even when his velocity is just half the initial velocity. In fact, he had traveled \(3/4\) of the total path before stopping.