**– A railway station has a loading yard for goods transport, a small 0.2kg documents package is released from rest to a point A on a booking place which is one-quarter of a circle having the radius of 1.6 m. The package size is much smaller compared to a radius of 1.6 m. Therefore, the package is treated as a particle. It slides down to the booking station and reaches point B with a final speed of 4.8 m/s. After point B, the package slides on a level surface and covers a final distance of 3.0 m to reach point C, where it comes to rest.**

**– What is the coefficient of kinetic friction on the horizontal surface?**

**– How much work is done on the package by friction as it slides down the circular arc from A to B?**

The aim of this question is to become familiar with the basic concepts of physics that include the **work done, friction, and kinetic energy**. A practical example of these concepts is given at the truck loading station. The relation of **work done** and **kinetic friction** with the **mass, radius, position**, and **speed** of a body should be known.

## Expert Answer

To calculate the required answer, we have the following data.

\[ Mass,\ m = 2\ kg \]

\[ Radius,\ r = 1.6\ m \]

\[ Package’s\ Size,\ p = 1.6\ m \]

\[ Speed,\ s = 4.80\ m/s \]

\[ Distance,\ d = 3\ m \]

a ) On the **horizontal** surface, the **kinetic energy** becomes equal to the **friction’s work** done.

Since:

\[ \text{Kinetic Energy,}\ K_e = \dfrac{1}{2}\ mv^2 \]

\[ \text{Friction,}\ F_w = u_f \times m \times g \times d \]

Where $u_f$ is the **friction work,**

Hence:

\[\dfrac{1}{2} mv^2 = u_f \times m \times g \times d\]

\[u_k = \dfrac{v^2}{2g \times d}\]

\[\dfrac{4.8^2}{2 \times 9.81 \times 3}\]

\[u_k = 0.39\]

b ) **Work done** on the package by **friction** as it slides down the circular arc from $A$ to $B$ is equal to the **potential energy** at a point $A$. The **potential energy** in a circular arc is $mgh$.

\[ \text{Potential Energy} = \text{Work done by Friction} + \text{Kinetic Energy} \]

\[mgh = W.F_{A-B} + \dfrac{1}{2} mv^2\]

\[W.F_{A-B} = mgh – \dfrac{1}{2} mv^2\]

\[W.F_{A-B} = (0.2) (9.81 \times 1.6 – \dfrac{1}{2} (4.8)^2)\]

\[W.F_{A-B} = 0.835J\]

## Numerical Results

(a) The **coefficient of kinetic friction** on the horizontal surface is calculated as:

\[u_k = 0.39\]

(b) Work done on the package by **friction** as it slides down the **circular arc** from $A$ to $B$.

\[W.F_{A-B} = 0.835J\]

## Example

A **ball** of $1kg$ **swings** in a **circle vertically** at a string that is $1.5m$ long. When the ball reaches the circle’s bottom, the **string** has a **tension** of $15N$. Calculate the **ball speed.**

As we have the following data given:

\[ Mass = 1kg \]

\[ Radius = 1.5m \]

\[ Tension = 15N \]

\[ g = 9.8 m/s^2 \]

We have the formula of **Tension**, so we can calculate $v$ as:

\[ T = \dfrac{mv^2}{r} – mg \]

\[ v = 3.56 m/s \]