This problem aims to find the length of a rough patch using the concept of the work-energy theorem and the Principle of Energy Conservation. It also covers the study of the non-conservative force of friction between ice and skates.
The most important concept discussed here is the work-energy theorem, most commonly known as the principle of work and kinetic energy. It is defined as the net work done by the forces on an object equal to the change in the kinetic energy of that object.
It can be represented as:
\[ K_f – K_i = W \]
Where $K_f$ = Final kinetic energy of the object,
$K_i$ = Initial kinetic energy and,
$W$ = total work done by the forces acting on the object.
The force of friction is defined as the force induced by two rough surfaces that contact and slide creating heat and sound. Its formula is:
\[ F_{fric} = \mu F_{norm} \]
Expert Answer
To start with, when the ice-skater encounters a rough patch, he undergoes the effect of three forces that act on her, the first is the force of gravity, its own weight or the normal force, and lastly the force of friction. The gravity and the normal force cancel out each other because both are perpendicular to each other. So the only force acting on the skater is the force of friction, represented as $F_f$, and is given by:
\[F_f=\mu mg\]
According to the problem statement, the force of friction is $25\%$ to the weight of the skater:
\[F_f=\dfrac{1}{4}weight\]
\[F_f=\dfrac{1}{4}mg\]
So from the above equation, we can assume that the value of $\mu$ is $\dfrac{1}{4}$.
As the force of friction is always opposite to the displacement, a negative effect will be observed by the skater, which will result in work done as:
\[W_f = -\mu mgl\]
Where $l$ is the total length of the rough patch.
Also, we are given the initial and final speeds of the skater:
$v_i=3 m/s$
$v_f=1.65 m/s$
So according to the work-energy theorem,
\[ W_f = W_{\implies t}\]
\[ \mu mgl = K_{final} – K_{initial}\]
\[ \mu mgl = \dfrac{1}{2}mv_f^2 – \dfrac{1}{2}mv_i^2\]
\[ \mu mgl = \dfrac{1}{2}m(v_f^2 – v_i^2)\]
\[ l= \dfrac{1}{2\mu mg}m(v_f^2 – v_i^2)\]
\[ l = \dfrac{1}{2\mu g}(v_f^2 – v_i^2)\]
Substituting the values of $m$, $v_f$, $v_i$ and $g$ into the above equation:
\[ l = \dfrac{1}{2\times 0.25 \times 9.8}(3^2 – 1.65^2)\]
\[ l = \dfrac{1}{4.9}(9 – 2.72)\]
\[ l = 1.28m\]
Numerical Result
The total length of the rough patch comes out to be:
\[ l = 1.28m\]
Example
A worker carries a $30.0kg$ crate over a distance of $4.5m$ at a constant velocity. $\mu$ is $0.25$. Find the magnitude of force to be applied by the worker and calculate the work done by friction.
To find the friction force:
\[ F_{f} = \mu mg\]
\[ F_{f} = 0.25\times 30\times 9.8\]
\[ F_{f} = 73.5N \]
The work done by the friction force can be calculated as:
\[ W_f = -r F_f \]
\[ W_f = -4.5\times 73.5 \]
\[ W_f = -331 J\]