# A car of mass 1000 kg traveling at 30 m/s has its speed reduced to 10 m/s by a constant braking...

## Question:

A car of mass 1000 kg traveling at 30 m/s has its speed reduced to 10 m/s by a constant braking force over 75 m. Find the initial and final kinetic energy?

## Kinetic Energy:

First we have to understand the kinetic energy of an object for solving this problem:

Suppose v is the velocity of the object and m is the mass of the object, then we can write the relation between the kinetic energy and velocity that is defined as: {eq}\displaystyle \text{ K. E. } = \frac{1}{2} \ m \ v^2 {/eq}.

Given:

The initial velocity of the car is: {eq}u = 30 \ \rm m / s {/eq}, the final velocity of the car is: {eq}v = 10 \ \rm m / s {/eq} and the mass of the car is: {eq}m = 1000 \ \rm kg {/eq}.

First we will compute the initial kinetic energy when the velocity of the car is {eq}30 \ \rm m / s {/eq}.

As we know the formula for the kinetic energy is:

{eq}\begin{align*} \displaystyle k_1 &= \frac{1}{2} \ m \ v^2 &\text{(Where } m = 1000 \ \rm kg \text{ and } v = u = 30 \ \rm m / s \text{)}\\ &= \frac{1}{2} \times 1000 \times 30^2 &\text{(Plugging in the values of mass and velocity)}\\ &= 500 \times 900 \\ k_1 \ &\boxed{ = 450,000 \ \rm J} \end{align*} {/eq}

Now we will compute the final kinetic energy when the velocity of the car is {eq}10 \ \rm m / s {/eq}.

As we know the formula for the kinetic energy is:

{eq}\begin{align*} \displaystyle k_2 &= \frac{1}{2} \ m \ v^2 &\text{(Where } m = 1000 \ \rm kg \text{ and } v = 10 \ \rm m / s \text{)}\\ &= \frac{1}{2} \times 1000 \times 10^2 &\text{(Plugging in the values of mass and velocity)}\\ &= 500 \times 100 \\ k_2 \ &\boxed{ = 50,000 \ \rm J} \end{align*} {/eq}