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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Thermodynamics

A vehicle (mass=1000 kg) , moving at a speed of $\;36 \;kmh^{-1}\;$ is stopped in 2 s . Assuming that $\;50 \%\;$ of the mechanical energy lost appears as thermal energy , the average rate of production of thermal energy in $\;cal s^{-1}\;$ is -

$(a)\;2 \times 10^{3} \;cals^{-1}\qquad(b)\;3 \times 10^{3} \;cals^{-1}\qquad(c)\;4 \times 10^{3} \;cals^{-1}\qquad(d)\;5 \times 10^{3} \;cals^{-1}$

1 Answer

Answer : (b) $\;3 \times 10^{3} \;cals^{-1}$
Explanation :
K.E lost = $\;\large\frac{1}{2} \times 1000 \times (36 \times \large\frac{5}{18})^2 J$
$=5 \times 10^{4}J$
$50 \%\;$ of KE lost in 2 secs
rate of production in $\;cal\;s^{-1}$
$=\large\frac{50}{100}\times \large\frac{5 \times 10^{4}}{2} \times \large\frac{1}{4.18}$
$=2.99 \times 10^{3}\;cal s^{-1}\;.$
answered Mar 12, 2014 by yamini.v
 

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