MHT CET · Physics · Work Power Energy
A car of mass ' \(m\) ' moving with velocity ' \(u\) ' on a straight road in a straight line, doubles its velocity in time \(t\). the power delivered by the engine of a car for doubling the velocity is
- A \(\frac{3 \mathrm{mu}^2}{2 \mathrm{t}}\)
- B \(\frac{\mathrm{mu}^2}{2 \mathrm{t}}\)
- C \(\frac{2 \mathrm{mu}^2}{\mathrm{t}}\)
- D \(\frac{3 \mathrm{mu}^2}{\mathrm{t}}\)
Answer & Solution
Correct Answer
(A) \(\frac{3 \mathrm{mu}^2}{2 \mathrm{t}}\)
Step-by-step Solution
Detailed explanation
Initial kinetic energy \(\mathrm{k}_1=\frac{1}{2} \mathrm{mu}^2\)
Final kinetic energy \(\mathrm{k}_2=\frac{1}{2} \mathrm{mu}_2^2=\frac{1}{2} \mathrm{~m}(2 \mathrm{u})^2=\frac{1}{2}\left(4 \mathrm{mu}^2\right)\)
\(\therefore \mathrm{k}_2-\mathrm{k}_1=\frac{3}{2} \mathrm{mu}^2\)
Change in K.E. is work done
Power, \(\mathrm{P}=\frac{\text { workdone }}{\mathrm{t}}=\frac{3 \mathrm{mu}^2}{2 \mathrm{t}}\)
Final kinetic energy \(\mathrm{k}_2=\frac{1}{2} \mathrm{mu}_2^2=\frac{1}{2} \mathrm{~m}(2 \mathrm{u})^2=\frac{1}{2}\left(4 \mathrm{mu}^2\right)\)
\(\therefore \mathrm{k}_2-\mathrm{k}_1=\frac{3}{2} \mathrm{mu}^2\)
Change in K.E. is work done
Power, \(\mathrm{P}=\frac{\text { workdone }}{\mathrm{t}}=\frac{3 \mathrm{mu}^2}{2 \mathrm{t}}\)
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