A 710 kg 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?
I may be wrong but P=F*v*cos(theta)
so 500(23.0 m/s)(cos(0))= 500(23)(1)
=11500W
1hp = 745.7W 11500W/745.7 = 15.42hp
To find the power required by the car's engine to drive the car, we can use the formula:
Power = Force × Velocity
First, let's calculate the power required on level ground.
(a) On level ground:
The car is subject to a drag force of 500 N. Since the car is moving at a constant speed, this drag force must be balanced by the engine's power.
Power (level ground) = Force (drag force) × Velocity
Power (level ground) = 500 N × 23 m/s
Power (level ground) = 11500 N·m/s or 11500 watts (W)
Therefore, the power required from the car's engine to drive the car on level ground is 11500 watts.
(b) Up a hill with a slope of 2.0:
When the car drives up a hill with a slope, an additional force is acting against the car's motion due to gravity. This force is proportional to the car's weight and the slope of the hill.
First, let's calculate the additional force due to gravity:
Force (gravity) = mass × gravity × sin(θ)
where
mass = 710 kg (mass of the car)
gravity = 9.8 m/s² (acceleration due to gravity)
θ = angle of the slope = 2.0 (slope)
Force (gravity) = 710 kg × 9.8 m/s² × sin(2.0)
Force (gravity) ≈ 2127 N
Since the car is still moving at a constant speed, this additional force due to gravity must also be balanced by the engine's power.
Power (uphill) = Force (drag force + gravity) × Velocity
Power (uphill) = (500 N + 2127 N) × 23 m/s
Power (uphill) = 2627 N × 23 m/s
Power (uphill) = 60421 N·m/s or 60421 W
Therefore, the power required from the car's engine to drive the car up a hill with a slope of 2.0 is 60421 watts.