Question

a car weighing 1000kg is going up an incline with a slope of 2 in 25 at a steady speed of 18kmph .If g=10the power of its engine is

Answer 1

slope of inclined plane , tanθ = 2/25

so, sinθ = 2/√(25² + 2²) = 2/√629

now, weight of car along inclined plane, F = mgsinθ

here, m = 1000kg, g = 10m/s² and sinθ = 2/√(629)

so, F = 1000 × 10 × 2/√(629)

≈ 797.452 N

velocity of car along plane , v = 18km/h

= 18 × 5/18 = 5m/s

so, power of its engine = weight of car along plane × velocity of car along plane.

= F × v

= 797.452 × 5 W

= 3,987.26 W = 3.98726 kW ≈ 4 kW

hence, power of car engine = 4kW