A car of mass $1000$ kg is traveling up a hill of 1 in $49$ , with the engine working at a constant power of $40 k W$ . If the resistance offered to the motion is of $800 N$ , then the maximum speed the car attain up the hill is

34 m s^{- 1}

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40 m s^{- 1}

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45 m s^{- 1}

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50 m s^{- 1}

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Solution

The correct option is A $40 m s^{- 1}$
For maximum speed, the power supplied by motor must be equal to the power consumed by friction and gravity forces
$P = P_{f r i c t i o n} + P_{g r a v i t y}$ ...(i)
For simplicity assuming height of slope is $1 m$ and base of slope is $49 m$ in length.
$P_{f r i c t i o n} = F_{f r i c t i o n} \times v P_{g r a v i t y} = \frac{m g h}{Δ t}$
from given data $F_{f r i c t i o n} = 800 N; m = 1000 k g;$
$Δ t = \frac{\sqrt{49^{2} + 1^{2}}}{v} ≃ \frac{49}{v}$
substituting values in equation (i)
$40000 = 800 \times v + \frac{1000 \times 9.8 \times 1}{\frac{49}{v}}$
solving equation for v $= 40 m / s$

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