**From the initial release until the wood reaches the state of rest, what amount of work is done by friction?**

This problem aims to familiarize with the concepts of **dynamic motion** which are part of classical dynamic **physics.** To better understand this topic you should be familiar with **kinetic** **energy, kinetic friction,** and **energy lost** due to **friction.**

The first term we should be familiar with is **kinetic energy,** which is the **energy** that the object maintains due to its **motion.** It is defined as the **work** needed to **accelerate** an object of some certain **mass** from **rest** to its given **velocity.** The object sustains this **kinetic energy** unless its **velocity** shifts after attaining it during its **acceleration.**

Another terminology to keep in touch with is **kinetic** **friction** which is described as a **force** acting between **rolling** surfaces. A **body rolling** on the surface undergoes a **force** in the **opposing direction** of its motion. The amount of **force** will rely on the coefficient of **kinetic friction** between the two surfaces.

## Expert Answer

The **Kinetic friction coefficient** is denoted by $\mu_k$ and its value is $0.20$.

The **M****ass** of the wood is $m$ and is given by $2.0 \space Kg$.

The **H****eight** above the rough bottom is $h$ and its value is $4.0 \space m$.

The **Gravitational** force is $g$ and is given as $9.8 m/s^2$.

**Part a:**

First, we will find the distance $d$, from the initial state, where the wood eventually comes to rest.

According to the Law of conservation of Energy,

**Initial** Energy = **Final** Energy,

**OR,**

**Gravitational Potential **Energy = **Friction** Energy.

\[ mgh = \mu_kgdm \]

**Inserting** the values:

\[ (2.0)(9.8)(4) = (0.2)(9.8)(2.0)d \]

Making $d$ the subject:

\[ d = \dfrac{78.4}{3.92} \]

\[ d = 20 \space m \]

**Part b:**

To find the total amount of **work done** by **friction,** we will find the total initial energy that will be the total **work** friction has done.

The initial energy is **Gravitational Potential Energy** given by:

\[ P.E. = mgh\]

**Inserting** the values:

\[= (2.0)(9.8)(4.0) \]

\[= 78.4 \space J \]

## Numerical Result

The **distance** where the **wood** eventually comes to **rest** is $20 \space m$.

The total amount of **work done** by friction is $78.4 \space J$.

## Example

A piece of the **log** having the mass $1.0 \space kg$ falls against a surface. The log has **entirely smooth curved** sides and a rough **horizontal** bottom that is $35 \space m$ long. The **kinetic friction** coefficient of the log is $0.15$. The starting point of the log is $3 \space m$ beyond the rough **bottom.** Find how much work **friction** has to do to stop the log.

To find the total amount of work done by **friction,** we will find the total **initial energy** that will be the total work friction has done.

The total work done by **friction** is the **initial** energy, that is **Gravitational Potential** Energy, and is given by:

\[P.E. = mgh\]

**Inserting** the values:

\[ = (1.0)(9.8)(3.0)\]

\[ P.E.= 29.4 \space J\]