Respuesta :
Answer:
The input energy is 1687.5 kJ.
Explanation:
Given that,
Mass of car = 1500 kg
Speed of car = 15 m/s
Efficiency = 10 %
Now output is equal to work done
We need to calculate the work done
Using work energy theorem.
Work done = change in kinetic energy
[tex]W=\dfrac{1}{2}mv_{f}^2-\dfrac{1}{2}mv_{i}^2[/tex]
[tex]W=\dfrac{1}{2}m(v_{f}^2-v_{i}^2)[/tex]
Put the value into the formula
[tex]W=\dfrac{1}{2}\times1500\times(15)^2[/tex]
[tex]W=168750\ J[/tex]
Now we know that efficiency is given by
[tex]\eta=\dfrac{Out-put}{in-put}\times100[/tex]
[tex]In-put=\dfrac{out-put}{\eta}\times100[/tex]
[tex]In-put=\dfrac{168750}{10}\times100[/tex]
[tex]In-put=1687500\ J[/tex]
Hence, The input energy is 1687.5 kJ.
The energy is transferred to the engine by the burning gasoline is 1687500 J.
The work done can be calculated by the work-energy theorem,
[tex]\bold {W = \dfrac 1{2}m(v_f^2- v_i^2) }[/tex]
Where, m - mass
[tex]\bold {\v_f}[/tex][tex]\bold {v_f}[/tex] - final velocity
[tex]\bold {v_i}[/tex] - initial velocity
Put the values in the formula,
[tex]\bold {W = \dfrac 1{2}1500 (15^2- 0^2) }\\\\\bold {W = 168750\ J}[/tex]
The efficiency of car can be calculated by the formula,
[tex]\bold {\eta = \dfrac {E_o}{E_i}\times 100}\\\\\bold {E_i = \dfrac {E_o}{\eta} \times 100}\\\\\bold {E_i = \dfrac {168750}{10} \times 100}\\\\\bold {E_i = 1687500 J}[/tex]
Therefore, the energy is transferred to the engine by the burning gasoline is 1687500 J.
To know more engine efficiency,
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