The rated power of a certain type of automobile engine is 60kW, and the resistance is 2400N when driving on the horizontal road, then the speed of the automobile when driving at constant speed under the rated power is 2400N______ m/s．

The rated power of a certain type of automobile engine is 60kW, and the resistance is 2400N when driving on the horizontal road, then the speed of the automobile when driving at constant speed under the rated power is 2400N______ m/s．

When the car reaches the maximum speed, the traction and resistance are equal. From P = FV = FV, we can get: v = pf = 60 × 1032400m / S = 25m / S; so the answer is: 25

A vehicle with a mass of 4.0 × 103kg and a rated engine power of 60kW moves uniformly in a straight line with an acceleration of a = 0.5m/s2 from standstill. When it moves on the horizontal plane, its resistance is 0.1 times of the vehicle weight. Take 10m / S2 as G to calculate (1) the output power of the engine at the end of 2S; (2) the driving time of the vehicle moving uniformly in a straight line with an acceleration of 0.5m/s2; (3) the driving time of the vehicle at the end of 2S The maximum speed that can be driven on this road

(1) The resistance of the car is: F = 0.1mg = 4000N. According to Newton's second law, the solution of f-f = ma is: F = F + Ma = 6000n2s, the speed at the end of N2S is: v = AT1 = 0.5 × 2m / S = 1m / s, the output power of the engine at the end of 2S after starting is: P = FV = 6kW (2)

The rated power of automobile engine is 60kW, the maximum speed of driving on the straight road is 12mgs, and the mass of automobile is 5000kg. If the automobile starts to do the uniform acceleration linear motion with acceleration of 0.5m/s2 from the static state, and runs at the rated power after reaching the rated power, the resistance will not change during the motion Traction force; (3) the maximum speed of the car's uniform acceleration linear motion

(1) According to P = FV, the maximum speed of a car is f = f = pvmax = 60 × 10312n = 5000n (2) according to Newton's second law, the traction force of a car is f '- F = ma, so the traction force of a car at a uniform acceleration of a = 0.5m/s2 is f' = F + Ma = 5000 + 5000 × 0.5N = 7500n (3) according to P = FV, the uniform acceleration speed of a car is f '= F + Ma = 5000 + 5000 × 0.5N = 7500n (3) The maximum speed of uniform acceleration is Vmax ′ = PF ′ = 60 × 1037500m / S = 8m / s

The rated power of automobile engine is 60kW, and the mass of automobile is 5T. When the automobile runs on the horizontal road, the resistance is 0.1 times of the vehicle weight, g = 10m / S2,
① What is the maximum speed that a car can reach after starting from standstill with rated power?
② When the speed is 10m / s, what is the acceleration of the car?
③ How long can the process last if the car starts to do uniform acceleration motion from standstill with an acceleration of 0.5m/s2?

1. Let P = FV and f be the resistance. When the power is constant and F = f, the speed is the maximum
Then Vmax = P / F
F = 0.1mg = 0.1 × 5 × 1000 × 10 = 5000n, v = 12m / S (unit)
2. When v = 10m / S
F=P/V=6000N
Acceleration a = (f-f) / M = 0.2m/s2
3. When a = 0.5m/s2
Because the maximum speed Vmax = 12m / S
From t = V / a = 12 / 0.5 = 24s