The aim of this article is to find the **Kinetic Energy** of an object in motion in $BTU$.

The basic concept behind this article is the understanding of **Kinetic Energy K.E.** and its **unit conversion**.

**Kinetic Energy** is defined as the energy that an object carries while in motion. All moving objects possess **kinetic energy**. When a **net force** $F$ is applied to an object, this **force** transfers **energy**, and resultantly **work** $W$ is done. This energy called **Kinetic Energy K.E.** changes the state of the object and causes it to **move** at a certain **speed**. This **Kinetic Energy K.E.** is calculated as follows:

\[Work\ Done\ W\ =\ F\ \times\ d\]

Where:

$F\ =$ **Net Force Applied to the Object**

$d\ =$ **Distance traveled by the Object**

Since:

\[F\ =\ m\ \times\ a\]

So:

\[W\ =\ (m\ \times\ a)\ \times\ d\]

As per the **Equation of Motion**:

\[2\ a\ d\ =\ {v_f}^2\ -\ {v_i}^2\]

And:

\[a\ =\ \frac{{v_f}^2\ -\ {v_i}^2}{2d}\]

Substituting in the equation for **work done**, we get:

\[W\ =\ m\ \times\ d\ \times\ \left(\frac{{v_f}^2\ -\ {v_i}^2}{2d}\right)\]

\[W=\frac{1}{2}\ m\times({v_f}^2\ -\ {v_i}^2)\]

If the object is initially at rest, then $v_i=0$. So, simplifying the equation, we get:

\[K.E.\ \ =\ \frac{1}{2}\ m\ {\ v}^2\]

Where:

$m$ is the **mass of the object**, and $v$ is the **velocity of the object**.

The **SI Unit** for **Kinetic Energy K.E.** is **Joules** $J$ or $BTU$ **(British Thermal Unit)**.

## Expert Answer

Given that:

**Mass of the Object** $m\ =\ 10\ lbm$

**Velocity of the Object** $v\ =\ 50\ \dfrac{ft}{s}$

We need to find the **Kinetic Energy K.E.** which is calculated as follows:

\[K.E.\ \ =\ \frac{1}{2}\ m{\ v}^2\]

Substituting the given values in the above equation, we get:

\[K.E.\ \ =\ \frac{1}{2}\ (10\ lbm){\ (50\ \frac{ft}{s})}^2\]

\[K.E.\ \ =\ 12500\ lbm \frac{{\rm ft}^2}{s^2}\]

We need to calculate the **Kinetic Energy K.E.** in $BTU$ – **British Thermal Unit.**

As we know:

\[1\ BTU\ =\ 25037\ lbm \frac{{\rm ft}^2}{s^2}\]

\[1\ lbm \frac{{\rm ft}^2}{s^2}\ =\ \frac{1}{25037}\ BTU\]

Hence:

\[K.E.\ \ =\ 12500\ \times\ \frac{1}{25037}\ BTU\]

\[K.E.\ \ =\ 0.499\ BTU\]

## Numerical Result

The **Kinetic Energy** of the Object in **BTU** is as follows:

\[K.E.\ \ =\ 0.499\ BTU\]

## Example

If an object having a **mass** of $200kg$ is moving at the **speed** of $15\dfrac{m}{s}$, calculate its **Kinetic Energy** in **Joules**.

**Solution**

Given that:

**Mass of the Object** $ m\ =\ 200\ kg $

**Velocity of the Object** $ v\ =\ 15\ \dfrac{m}{s} $

We need to find the **Kinetic Energy K.E.** which is calculated as follows:

\[ K.E.\ \ =\ \frac{1}{2}\ m{\ v}^2 \]

Substituting the given values in the above equation, we get:

\[ K.E.\ \ =\ \frac{1}{2}\ (200\ kg){\ (15\ \frac{m}{s})}^2 \]

\[ K.E.\ \ =\ 22500\ kg\ \frac{m^2}{s^2} \]

As we know:

The **SI unit** of **Kinetic Energy** is **Joule** $J$ which is expressed as follows:

\[ 1\ Joule\ J\ =\ 1\ kg\ \frac{m^2}{s^2} \]

Hence:

\[ K.E.\ \ =\ 22500\ J \]

\[ K.E.\ \ =\ 22500\ \frac{J}{1000} \]

\[ K.E.\ \ =\ 22.5\ KJ \]