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

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:

$T = \dfrac{mv^2}{r} – mg$

$v = 3.56 m/s$