What is the kinetic energy of a car that travels at a speed of $20m{{s}^{-1}}$ and has a mass of 1200kg ?
Answer
562.5k+ views
Hint: The kinetic energy of a body is the energy associated with the motion of the body. The kinetic energy of a body moving with a constant velocity is given by the formula, half of mass times velocity squared. We will use this formula to calculate the kinetic energy of the car as the car is travelling with a constant speed of twenty meters per second.
Complete step-by-step answer:
Let us first define some terms that we are going to use later in our solution.
Let the kinetic energy of the car be denoted by ‘K’, then we need to solve for ‘K’.
Let the speed by which the car is travelling be ‘v’, then it is given to us as $20m{{s}^{-1}}$. And,
Let the mass of the car be ‘m’, then it is given to us as 1200kg.
Now, the expression of kinetic energy ‘K’ can be written as:
$\Rightarrow K=\dfrac{1}{2}m{{v}^{2}}$
Putting the values of all the known terms and calculating for ‘K’, we get:
$\begin{align}
& \Rightarrow K=\dfrac{1}{2}\times 1200\times {{\left( 20 \right)}^{2}}J \\
& \Rightarrow K=\dfrac{1}{2}\times 1200\times 400J \\
& \Rightarrow K=600\times 400J \\
& \Rightarrow K=240000J \\
& \therefore K=240kJ \\
\end{align}$
Hence, the kinetic energy of a car that travels at a speed of $20m{{s}^{-1}}$ and having a mass of 1200kg comes out to be 240kJ .
Note: The kinetic energy that we calculated for the car is the energy associated with the motion of the car. This means that, if the driver decides to stop the car at some point, then this amount of work should be done by the brakes and the friction combined to put the car to a stop. And, also since the car is assumed to be moving on a levelled road, the potential energy of the car is constant.
Complete step-by-step answer:
Let us first define some terms that we are going to use later in our solution.
Let the kinetic energy of the car be denoted by ‘K’, then we need to solve for ‘K’.
Let the speed by which the car is travelling be ‘v’, then it is given to us as $20m{{s}^{-1}}$. And,
Let the mass of the car be ‘m’, then it is given to us as 1200kg.
Now, the expression of kinetic energy ‘K’ can be written as:
$\Rightarrow K=\dfrac{1}{2}m{{v}^{2}}$
Putting the values of all the known terms and calculating for ‘K’, we get:
$\begin{align}
& \Rightarrow K=\dfrac{1}{2}\times 1200\times {{\left( 20 \right)}^{2}}J \\
& \Rightarrow K=\dfrac{1}{2}\times 1200\times 400J \\
& \Rightarrow K=600\times 400J \\
& \Rightarrow K=240000J \\
& \therefore K=240kJ \\
\end{align}$
Hence, the kinetic energy of a car that travels at a speed of $20m{{s}^{-1}}$ and having a mass of 1200kg comes out to be 240kJ .
Note: The kinetic energy that we calculated for the car is the energy associated with the motion of the car. This means that, if the driver decides to stop the car at some point, then this amount of work should be done by the brakes and the friction combined to put the car to a stop. And, also since the car is assumed to be moving on a levelled road, the potential energy of the car is constant.
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