# A 1300-kg automobile travels at 80 km/h. 1. What is its kinetic energy? 2. What net work would be...

## Question:

A 1300-kg automobile travels at 80 km/h.

1. What is its kinetic energy?

2. What net work would be required to bring it to a stop?

## Kinetic Energy:

The kinetic energy occurs when an object or body is moving. Mathematically this is calculated using the formula {eq}K.E = \dfrac{1}{2}m{v^2} {/eq}, where m is the mass of an object, and v is the velocity of the object.

## Answer and Explanation:

**Given data:**

- The mass of automobile is: {eq}M = 1300\;{\rm{kg}} {/eq}

- The speed of automobile is: {eq}V = 80\;{{{\rm{km}}} {\left/ {\vphantom {{{\rm{km}}} {\rm{h}}}} \right. } {\rm{h}}} {/eq}

**(1)**

Recall the expression for the kinetic energy.

$$\color{red}{K.E = \dfrac{1}{2}M{V^2}} $$

By substituting the known values in above equation, we will get the kinetic energy.

$$\begin{align*} K.E &= \dfrac{1}{2}\left( {1300\;{\rm{kg}}} \right){\left( {80\;{{{\rm{km}}} {\left/ {\vphantom {{{\rm{km}}} {\rm{h}}}} \right. } {\rm{h}}}\left( {\dfrac{{\dfrac{5}{{18}}\;{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}}}}{{1\;{{{\rm{km}}} {\left/ {\vphantom {{{\rm{km}}} {\rm{h}}}} \right. } {\rm{h}}}}}} \right)} \right)^2}\\ &= 320987.65\;{\rm{N}} \cdot {\rm{m}}\\ &\approx 321\;{\rm{kJ}} \end{align*} $$

**Thus, the kinetic energy is** {eq}\color{blue}{321\;{\rm{kJ}}}. {/eq}

**(2)**

As per the energy conservation, we know that the change in kinetic energy is equal to the work done.

Recall the expression for the work done.

$$\color{red}{W = K.{E_f} - K.{E_i}} $$

Here initial kinetic energy of automobile is zero, {eq}K.{E_i} = 0 {/eq}.

Substitute the known value in above equation.

$$\begin{align*} W &= 321\;{\rm{kJ}} - 0\\ &= 321\;{\rm{kJ}} \end{align*} $$

**Thus, the required net work done to stop it is ** {eq}\color{blue}{321\;{\rm{kJ}}} {/eq}.

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from Physics 101: Help and Review

Chapter 17 / Lesson 4