Calculate the work done by the brakes of a car mass \[1000\,kg\] when its speed is reduced from \[20\,m/s\] to \[10\,m/s\] ?
Answer
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Hint: We can start by writing down the data given in the question. Then we move onto finding the work done by the brakes of the given car. We know that the work done can be defined as the difference between the initial and final kinetic energy. We can find the individual kinetic energies and find the difference between them to find the work done with the given data thus, getting us the required solution.
Complete step by step answer:
Let us start by writing down the given values in the question,
The value of initial velocity is given as, \[{v_i} = 20\,m/s\]
The value of final velocity is given as, \[{v_f} = 10\,m/s\]
The mass of the car is given as, \[m = 1000\,kg\]
Now we have these values, we can find the final and initial kinetic energies.The initial kinetic energy can be found out using the formula,
\[K{E_i} = \dfrac{1}{2}m{v_i}^2\]
When we substitute the value, we get
\[K{E_i} = \dfrac{1}{2}m{v_i}^2 \\
\Rightarrow K{E_i} = \dfrac{1}{2} \times 1000kg \times {\left( {20} \right)^2} \\
\Rightarrow K{E_i} = 2 \times {10^5}\]
Now that we have the value of the initial kinetic energy, we can find the value of final kinetic energy using the formula, \[K{E_f} = \dfrac{1}{2}m{v_f}^2\]
We substitute the values and get,
\[K{E_f} = \dfrac{1}{2}m{v_f}^2 \\
\Rightarrow K{E_f} = \dfrac{1}{2} \times 1000kg \times {\left( {10} \right)^2} \\
\Rightarrow K{E_f} = 5 \times {10^4}\]
Now that we have the value of initial and final kinetic energies, we can find the difference between them to find the work done
\[W = K{E_f} - K{E_i} \\
\Rightarrow W= 200,000 - 50,000 \\
\therefore W= 150,000J\]
Therefore, the work done by the brakes of a car mass \[1000\,kg\] when its speed is reduced from \[20\,m/s\] to \[10\,m/s\] is \[150,000\,J\].
Note: We can see the difference in the values of power in the initial and final kinetic energies, in order to avoid further confusion, we convert the value in power. Kinetic energy is defined as the energy by virtue of which the body remains to be in motion. The work done by the net force on a particle equals the change in the kinetic energy.
Complete step by step answer:
Let us start by writing down the given values in the question,
The value of initial velocity is given as, \[{v_i} = 20\,m/s\]
The value of final velocity is given as, \[{v_f} = 10\,m/s\]
The mass of the car is given as, \[m = 1000\,kg\]
Now we have these values, we can find the final and initial kinetic energies.The initial kinetic energy can be found out using the formula,
\[K{E_i} = \dfrac{1}{2}m{v_i}^2\]
When we substitute the value, we get
\[K{E_i} = \dfrac{1}{2}m{v_i}^2 \\
\Rightarrow K{E_i} = \dfrac{1}{2} \times 1000kg \times {\left( {20} \right)^2} \\
\Rightarrow K{E_i} = 2 \times {10^5}\]
Now that we have the value of the initial kinetic energy, we can find the value of final kinetic energy using the formula, \[K{E_f} = \dfrac{1}{2}m{v_f}^2\]
We substitute the values and get,
\[K{E_f} = \dfrac{1}{2}m{v_f}^2 \\
\Rightarrow K{E_f} = \dfrac{1}{2} \times 1000kg \times {\left( {10} \right)^2} \\
\Rightarrow K{E_f} = 5 \times {10^4}\]
Now that we have the value of initial and final kinetic energies, we can find the difference between them to find the work done
\[W = K{E_f} - K{E_i} \\
\Rightarrow W= 200,000 - 50,000 \\
\therefore W= 150,000J\]
Therefore, the work done by the brakes of a car mass \[1000\,kg\] when its speed is reduced from \[20\,m/s\] to \[10\,m/s\] is \[150,000\,J\].
Note: We can see the difference in the values of power in the initial and final kinetic energies, in order to avoid further confusion, we convert the value in power. Kinetic energy is defined as the energy by virtue of which the body remains to be in motion. The work done by the net force on a particle equals the change in the kinetic energy.
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