A force (thrust) of 400 N is required to overcome road friction and air resistance in propelling an automobile at 80 km h⁻¹. What power (kW) must the engine develop?

(Ans: 8.9 Kw)



Given:

Force = F = 400 N

Velocity of Automobile =  V = 80 km h⁻¹ = 22.22 m s⁻¹ 


To Find:

Power of the engine =  P = ?   


Solution:

Relation of power in term of force `\vec F` and Velocity `\vec v` is given by

P = `\vec F` . `\vec v`  

P = F v cos ፁ 

but ፁ = 0° and cos 0° = 1, So

 P = F v 

putting values

 P = 400 N ｘ 22.22 m s⁻¹  

 P = 8,888 W

or

P = 8.888 ｘ10³ W

Or

P = 8.89 kW --------------Ans.

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