Q 11-05-038NEETJEE MainMedium
A car of mass $1000$ kg moves up a slope rising $1$ m for every $20$ m along the road at a steady $36$ km/h. Friction and air resistance together are $500$ N. The power delivered by the engine is ($g = 10\ \text{m/s}^2$)
Answer: (A) $10$ kW
$v = 10$ m/s, $\sin\theta = \dfrac{1}{20}$. At steady speed the engine force balances gravity along the slope plus resistance:
$$F = mg\sin\theta + 500 = 500 + 500 = 1000\ \text{N}$$
$$P = Fv = 1000 \times 10 = 10\ \text{kW}$$
Solution by Sreeraj P, M.Sc Physics