At a speed of 15 m / s, the force of resistance to movement of a car weighing 900 kg is 600 N. What speed should the car’s engine develop if the force of resistance is 900 N? (The drag force is the sum of the asphalt friction and air drag.)
V1 = 15 m / s.
m = 900 kg.
μ = 0.04.
g = 10 m / s2.
Fcopr1 = 600 N.
Fcopr2 = 900 N.
According to the condition of the problem, the resistance force Ffr is the sum of the friction force Ffr and the air resistance force Fw: Fcopr = Ffr + Fw.
The friction force of the wheels on the asphalt Ftr does not depend on the speed of movement and is determined by the formula: Ftr = μ * m * g, where μ is the coefficient of friction, m * g is the force of gravity.
The force of air resistance Fв is directly proportional to the speed of movement V: Fв = k * V, where k is the coefficient of proportionality.
Fcopr1 = Ftr + Fv1 = μ * m * g + k * V1.
k * V1 = Fcopr1 – μ * m * g.
k = (Fcopr1 – μ * m * g) / V1.
k = (600 N – 0.04 * 900 kg * 10 m / s2) / 15 m / s = 16.
Fcopr2 = Ftr + Fv2 = μ * m * g + k * V2.
V2 = (Fcopr2 – μ * m * g) / k.
V2 = (900 N – 0.04 * 900 kg * 10 m / s2) / 16 = 33.75 m / s.
Answer: the car must reach a speed of V2 = 33.75 m / s.
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