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An engine of weight 6.5 metric ton is going up an incline of $\large\frac{5}{13}$ at the rate of $9\; km/hr$ .The power of engine , if co-efficient of friction is $\large\frac{1}{12}$ is

\[(a)\;83.5\;KW \quad (b)\;97.5\;KW \quad (c)\;60.5\;KW \quad (d)\;73.5\;KW\]

1 Answer

$\sin \theta=\large\frac{5}{13}$
$\cos \theta=\sqrt {1-\sin ^ \theta}$
$\qquad= \sqrt {1-\bigg(\large\frac{5}{13}\bigg)^2}$
Force against which the engine has to work
$F=mg \sin \theta+\mu mg \cos \theta$
$\quad=6500 \times 9.8 \bigg[\large\frac{5}{13}+\frac{1}{12} \times \frac{12}{13}\bigg]$
$\quad=29400 \;N$
Engine power $=Fv$
$\quad=29400 \times 2.5 \qquad 9km/hr=2.5 m/s$
$\quad=73500 \;W$
$\quad=73.5 KW$
Hence d is the correct answer.


answered Jul 22, 2013 by meena.p
edited Feb 10, 2014 by meena.p

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