First slide

Power

Question

An engine of one metric ton is going up an inclined plane 1 in 2 at the rate of 36 km/hr. If the coefficient of friction is (1/√3), the power of the engine in watt is

Moderate

A

98 watt
B

98 k watt
C

980 watt
D

72 k watt

Solution

The situation is shown in fig,
$Power P = F \times V$
$Given v = 36 km / hr = 36 \times \frac{5}{18} m / s$
$\begin{aligned} F & = mgsin θ + f = mgsin θ + μmgcos θ \\ = mg (\sin θ + μcos θ) \\ = 1000 \times 9 \cdot 8 (0 \cdot 5 + \frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{2}) \\ (∵ \sin θ = \frac{1}{2} and \cos θ = \frac{\sqrt{3}}{2}) \\ = 9800 N \end{aligned}$
$∴ P = 9800 \times 10 = 98 k watt.$

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