# A 8000-N car is traveling at 12 m/s along a horizontal road when the brakes are applied. The car...

## Question:

A 8000-N car is traveling at 12 m/s along a horizontal road when the brakes are applied. The car skids to a stop in 4.0 s. How much kinetic energy does the car lose in this time?

{eq}a. \ 4.8 \times 10^4 \ J \\ b. \ 1.2 \times 10^5 \ J \\ c. \ 4.8 \times 10^6 \ J \\ d. \ 5.8 \times 10^5 \ J \\ e. \ 5.9 \times 10^4 \ J {/eq}

## Kinetic Energy:

The energy of the body which is obtained by the motion or the velocity of the body is termed as kinetic energy. The measurable unit used to express the kinetic energy is Joule.

Given data

• The value of the weight force is {eq}W = 8000\;{\rm{N}} . {/eq}
• The value of the velocity of the car is {eq}{v_c} = 12\;{\rm{m/s}} {/eq}
• The value of the time of skid is {eq}t = 4\;\sec . {/eq}

The expression for the mass of the car is,

{eq}\begin{align*} W &= {m_{car}}g\\ {m_{car}} &= \dfrac{W}{g} \end{align*} {/eq}

Substitute the value in the above equation.

{eq}\begin{align*} {m_{car}} &= \dfrac{{8000}}{{9.81}}\\ &= 815.49\;{\rm{kg}} \end{align*} {/eq}

The expression for the lose of the kinetic energy is,

{eq}K.E = \dfrac{1}{2}{m_{car}}{v_{car}}^2 {/eq}

Substitute the value in the above equation.

{eq}\begin{align*} K.E &= \dfrac{1}{2}\left( {815.49} \right){\left( {12} \right)^2}\\ &= 58715.28\;{\rm{J}} \end{align*} {/eq}

Thus, the value of the loss of kinetic energy is {eq}58715.28\;{\rm{J}}. {/eq}