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

Question

A car weighing 1000 kg is going up an incline with a slope of 2 in 25 at a steady speed of 18 kmph. If g=10 ms- 2 , the power of its engine is (A) 4 kW (B) 50 kW (C) 625 kW (D) 25 kW

Tardigrade Admin · Accepted Answer

Power of engine can also be expressed as,

P = F \to \cdot v \to

Component of force along the inclined plane,

F = m g s in θ = 1000 \times 10 \times \frac{2}{25} = 800 N

Component of velocity along the plane is,

v = 18 km p h = 18 km p h \times \frac{1000}{3600} \frac{m / s}{km p h} = 5 m / s

Power

P = F v cos ϕ,

ϕ

is the angle between

F \to an d v \to

P o w er = 800 \times 5 (∵ ϕ = 0^{\circ}) = 4 kW

Q. A car weighing 1000 kg is going up an incline with a slope of 2 in 25 at a steady speed of 18 kmph. If $g = 10 m s^{- 2}$ , the power of its engine is

$4 kW$

$50 kW$

$625 kW$

$25 kW$

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

4kW

50kW

625kW

25kW

Q. A car weighing 1000 kg is going up an incline with a slope of 2 in 25 at a steady speed of 18 kmph. If $g = 10 m s^{- 2}$ , the power of its engine is

$4 kW$

$50 kW$

$625 kW$

$25 kW$