A $650\;KW$ power engine of a vehicle of mass $1.5 \; * \; 10^5\; kg$ is rising on an inclined plane of inclination $1$ in $100$ with a constant speed of $60\;km/hr$. Find the frictional force between the wheels of the vehicle and the plane.

Given,

Power of engine $(P) = 650\;kw = 650000\;W$

Mass of vehicle $m = 1.5 \; * \; 10^5\; kg$

Inclination $Sin\;\theta = \frac{1}{100}$

Speed of vehicle $(v) = 60\;km/hr = $$\frac{60\;*\;1000}{60\;*\;60} = \frac{50}{3}m/s$

Frictional force $(f_r) = \;?$

We have, total upwarding force $F = f_r + mg\;sin\;\theta$    ⇒    $\frac{P}{v}$$ = f_r + mg\;sin\;\theta$

or, $f_r = \frac{P}{v} - mg\;sin\;\theta$  =  $\frac{650000\;*\;3}{50}$$ - 1.5\;*\;10^5\;*\;10\;*\;$$\frac{1}{100}$ = $24000\;N$

Hence, fractional force between the wheels of the vehicles and the plane is $24000\;N$.

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