A figure skater is gliding gracefully across the ice at a speed of \(8 \,\text{m}/\text{s}\) when she crosses an uneven surface that the Zamboni missed during its last resurfacing. The force due to friction on the rough surface, which is half the weight of the figure skater, causes the figure skater to slow down to \(5 \,\text{m}/\text{s}\). How long is the rough patch of ice?
Use the Work-Energy Theorem where the work is done by the force due to friction, and then solve for the displacement \(d\).
The Work-Energy Theorem states:
\begin{equation*}
W = \Delta K.
\end{equation*}
The work done by the force due to friction is \(-f_r d\), but the force due to friction is half the weight as stated in the problem. The last equation can then be written as:
\begin{equation*}
-\frac{mgd}{2}=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2.
\end{equation*}
Solving for \(d\), we get:
\begin{equation*}
d=\frac{v_f^2-v_i^2}{-g},
\end{equation*}
which, with numerical values, is:
\begin{equation*}
d \approx 4\, \text{m}.
\end{equation*}
The problem asks us to find the length of the uneven patch of ice in the way of the professional ice skater. To do that, we must relate the length of the path to other known variables, such as the change of speed by the ice skater. We know that the work performed by a force is related to the displacement during which the force acts, and so to find the length in question we must find the work of the relevant forces. We also know that the total work over a body is related to the change of kinetic energy, and so to the change in speed, by means of the work-energy theorem. Hence, the key to solving this problem will be to first find an expression for the net work over the skater, and then use that expression in the work-energy theorem, to solve for the length of the patch of ice.
The work-energy theorem states that
\begin{equation}
\label{workenergy}
W_{\text{net}}=K_f-K_i,
\end{equation}
where \(W_{\text{net}}\) is the sum of the works made by all the forces exerted over the ice skater, or, equivalently, the work done by the net force. The terms \(K_i\) and \(K_f\) are the kinetic energies before and after the work performed by all the forces. In this case, they correspond to the kinetic energies just when reaching the uneven ice patch (\(K_i\) ), and just after leaving it (\(K_f\) ).
To use the previous theorem, let us calculate the net work done on the ice skater. If an object moves along a trajectory described by a displacement vector \(\vec{d}\), then the work exerted by a force \(\vec{F}\) over such an object is defined to be
\begin{equation}
\label{work}
W=\vec{F}\cdot\vec{d},
\end{equation}
where the dot product is defined as \(\vec{F}\cdot\vec{d}=Fd \cos \theta \). Clearly, if the force in question is perpendicular to the displacement, then the work by that force is zero (because \(\cos 90^{\circ}=0\)). So we will only need to find the work of the forces that are parallel or anti-parallel to the direction of motion of the skater.
Over the skater, there are three forces in total; the weight pointing downwards, the normal force with the floor, pointing upwards, and the kinetic friction pointing against the direction of motion. If we use a coordinate system where X points along the direction of motion, then the force diagram is shown in figure 1.
Figure 1: Free-body diagram for the ice-skater where three forces are shown: the normal force with the floor, which is \(\vec{N}\), the weight \(\vec{W}=-mg\,\hat{\textbf{j}}\) and the friction \(\vec{f}_r\). The displacement \(\vec{d}\) is also shown in blue. The coordinate system is chosen such that the X axis point in the direction of motion, and the Y axis points upwards.
We see that the only force that goes along the same axis as the displacement vector \(\vec{d}\) is the friction \(\vec{f}_r\), which is anti-parallel to such displacement. The other forces (the weight and the normal force) are perpendicular to the displacement vector and so they do not contribute to the net work. From the prompt, we also know that the magnitude of the friction is half the weight \(mg\) of the ice skater. Here \(m\) is the mass of the ice skater and \(g\) the gravitational acceleration of Earth. We can then write the friction force as follows:
\begin{equation}
\vec{f}_r=-\frac{mg}{2}\,\hat{\textbf{i}},
\end{equation}
where we indicate that its direction is negative in X. Then, the work done by friction is
\begin{equation}
\label{workdone}
W_{\text{fr}}=\vec{f}_r\cdot \vec{d},
\end{equation}
where \(\vec{d}\) is the displacement vector, which in this case, has magnitude \(d\), the length of the uneven patch of ice, and its direction is the same as the motion of the ice skater. Hence we can write the displacement explicitly as
\begin{equation}
\vec{d}=d\,\hat{\textbf{i}}.
\end{equation}
Using the explicit expressions for the friction force and the displacement back in equation \eqref{workdone}, we get
\begin{equation}
W_{\text{fr}}=\left(-\frac{mg}{2}\,\hat{\textbf{i}}\right)\cdot \left(d\,\hat{\textbf{i}}\right),
\end{equation}
which, after performing the dot product, is
\begin{equation}
W_{\text{fr}}=-\frac{mgd}{2}\,\hat{\textbf{i}}\cdot\hat{\textbf{i}}.
\end{equation}
The expression above can be further simplified to
\begin{equation}
\label{network}
W_{\text{fr}}=W_{\text{net}}=-\frac{mgd}{2},
\end{equation}
where we used the fact that \(\hat{\textbf{i}}\cdot\hat{\textbf{i}}=1\) (this comes from the fact that \(\textbf{i}\) is of magnitude 1 and it is parallel to itself, and so the angle between \(\textbf{i}\) and itself is zero, which then leads to \( \cos 0^{\circ}=1 \)). Notice that we also used that this is the same as the net work )the weight and the normal do not perform work, as explained above).
Before using this net work in the work-energy theorem, let’s first write the kinetic energy in terms of known variables (the mass and the speed of the skater). The kinetic energy is
\begin{equation}
K=\frac{1}{2}mv^2,
\end{equation}
where \(m\) is the mass of the ice skater and \(v\) the speed. Using this expression and the explicit expression for the net work found in equation \eqref{network} into equation \eqref{workenergy}, we get
\begin{equation}
-\frac{mgd}{2}=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2,
\end{equation}
where \(v_i\) and \(v_f\) are the speeds before and after the ice skater passes through the uneven patch. We can simplify this expression by cancelling out the mass \(m\) and the \(1/2\) term:
\begin{equation}
-gd=v_f^2-v_i^2.
\end{equation}
Solving for \(d\), we obtain
\begin{equation}
d=\frac{v_f^2-v_i^2}{-g},
\end{equation}
which numerically is
\begin{equation}
d=\frac{(5\,\text{m/s})^2-(8\,\text{m/s})^2}{-(9.8\,\text{m/s}^2)},
\end{equation}
\begin{equation}
d\approx 4\,\text{m}.
\end{equation}
Thank you so much! Very well explained!