Question

a standard car develops 40hp. find the maximum speed the car can attain against a resistance of 20 kgwt due to air and friction. given efficiency of the engine is 25℅

Answer 1

Efficiency of engine = 25%

Power = 25 % of 40hp
= 25/100 × 40 = 10hp
= 10 × 746 watt [ as 1 hp = 746 watt]
= 7460 watt

Use formula, P = F.v
Where P is power, F is force and v is speed.

Here,
P = 7460 watt,
F = 20kgwt = 200N

So, 7460 = 200 × v
v = 7460/200 = 37.3 m/s

Hence, speed of the car is 37.3 m/s

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Answer 2

Speed of the car is 37.3 m/s