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:

\[ W_f = -r F_f \]

\[ W_f = -4.5\times 73.5 \]

\[ W_f = -331 J\]