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Sagot :
a) The car's engine needs 11500 W of power to move the vehicle on level terrain.
b) A car's engine needs 17,084 W of power to move the vehicle up a hill with a 2.0° grade.
a) The energy needed to run the car's engine is provided by
P = Fv
where
The force which the engine needs to exert is F.
The car's velocity is v = 23 m/s.
Since the automobile is driving at a constant speed and hence experiencing no acceleration, the net force acting on it must also be zero. Since there is a 500 N drag force against the car's velocity, this implies that the engine's force applied forward must likewise be 500 N:
F = 500 N
Consequently, the power lost by the engine is
P = (500 N)(23 m/s)= 11500 W.
b) Both the drag force and the weight component that is parallel to the gradient are working against the motion of the automobile in this scenario.
This element is provided by
[tex]W_p[/tex] = mgsin∅
where
m = 710 kg, g = 9.8 m/s2, and ∅ = 2°, respectively, are the mass, acceleration of gravity, and slope of the hill, respectively.
[tex]W_p[/tex] = 710 × 9.8 m/s² × sin2° = 242.8 N
The drag force (500 N) and so this force is now added to determine the total rearward force acting against the velocity of the car:
F = 500 N + 242.8 N = 742.8 N
Because the force the engine applies must be equal, the power that is revoked will also be equal.
P = (742.8 N)(23 m/s) = 17,084 W
Learn more about the force at
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The question is -
A 710kg car drives at a constant speed of 23m/s. It is subject to a drag force of 500 N. What power is required from the car's engine to drive the car (a) on level ground? (b) up a hill with a slope of 2.0∘ ?
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