When the vehicle with a mass of 5t runs on a straight road, the resistance is 0.1 times of the vehicle weight, and the rated power of the engine is 60kW If the car starts to do uniform acceleration linear motion from standstill, it will reach rated power after 6 seconds. G = 10m / s 1) The maximum speed a car can reach 2) When the speed of the car is 10m / s, the power has reached the rated power. What is the acceleration at this time? 3) The acceleration of the car in the process of uniform acceleration Little girl, thank you first!

When the vehicle with a mass of 5t runs on a straight road, the resistance is 0.1 times of the vehicle weight, and the rated power of the engine is 60kW If the car starts to do uniform acceleration linear motion from standstill, it will reach rated power after 6 seconds. G = 10m / s 1) The maximum speed a car can reach 2) When the speed of the car is 10m / s, the power has reached the rated power. What is the acceleration at this time? 3) The acceleration of the car in the process of uniform acceleration Little girl, thank you first!

1. According to the law of conservation of energy: 60000 = FV, where f = 50000 * 0.1 = 5000n, i.e. resistance, so v = 60000 / F = 12m / S2, according to the law of constant dynamic energy: 60000 = f * 10 + ma * 10, where m is the vehicle mass, a is the vehicle acceleration, a = (60000-50000) / 5000 * 10 = 0.2m/s & sup2; 3, v = ATA = V / T = 2m / S & sup2

The rated power of automobile engine is 30kW and the mass is 2000kg. When the automobile runs on the horizontal road, the resistance is 0.1 times of the vehicle weight. (1) the maximum speed that the automobile can reach on the road? (2) What is the acceleration when the vehicle speed is 10m / S?

(1) When the vehicle has the maximum speed, the traction and resistance are balanced. From P = FV = FVM, the maximum speed of the vehicle is: VM = pf = 30 × 1030.1 × 2000 × 10m / S = 15m / s, (2) when the speed v = 10m / s, the traction f = PV = 30 × 10310n = 3000n, so the acceleration is: a = f − FM = 3000

The rated power of automobile engine is 60kW, the mass of automobile is 5T, and the resistance in motion is always 0.1 times of the vehicle weight
(1) If the car starts at constant power, what is the maximum speed that the car can reach? When the car's speed reaches 5m / s,
(2) If the vehicle starts at a constant acceleration of 0.5m/s, how long can this process last? What is the work done by the engine traction during this process? What is the maximum speed that the vehicle can reach? (G is taken as 10m / S & # 178;)

(1) The maximum velocity is: F = f = mg * 0.1 = 5000n, v = P / F = 60000 / 5000 = 12m / s
F=P/V=60000/5=12000 N F-f=ma a=1.4 m/s^2
（2）F-f=ma F=5000+5000*0.5=7500 N
v=P/F=60000/7500=8 m/s t=v/a=8/0.5=16 s s=vt/2=64 m
W-fs=1/2mv^2 W=480000 J
Vmax = v = 8 m / s

When an automobile engine output rated power is 40kW, the resistance is 2
The rated output power of an automobile engine is 40kW. When driving on the horizontal long straight highway, the resistance is 2 × 103n, and the mass of the automobile is 103kg. When its uniform speed is 10m / s, the engine power is 103kg__________ At a certain moment, the driver increases the accelerator to make the engine reach the rated power, and the traction of the car reaches the maximum__________ N. The ground acceleration of the car is__________ M / S2. Error correction

20；4000；2
① P = VF
F drag = f drag
The solution is p = 20000w = 20kW
② The solution of P = VF leads to f = 4000N
③ F drag '- f drag = ma, the solution is a = 2m / S & # 178;