A truck with a mass of 5t drives for 6km on a straight road at a constant speed for 10min. At this time, the engine power is 66kw?

A truck with a mass of 5t drives for 6km on a straight road at a constant speed for 10min. At this time, the engine power is 66kw?

1. Speed v = 6km / 10min = 10m / S
2.W=Pt=66KW×600s=39600J
3. The displacement of the car in the vertical direction is zero, so the work done by gravity is 0
4.P=Fv
F=P/v=66KW/10ms=6600N

A 1.5T minivan
A 1.5T pickup truck carrying 0.8t cargo is known to slide down a straight road for 500m under the action of 4000N traction, and the time taken is 400s (g = 10N / kg)
1. Work done by traction
2. Traction power

Work equals force times distance, w = 4000 * 500 = 2000000j
Power equals work divided by time, P = w / T = 2000000 / 400 = 5000W

The power of automobile engine is 60kW, and the mass of automobile is 4T. When it runs on a long straight road with a slope of 0.02 (that is, the sine value of the slope angle), the resistance is 0.1 times of the vehicle weight
(1) The maximum speed a car can reach
(2) How long can the process last if the car starts to make a uniform acceleration linear motion with an acceleration of 0.6m/s ^ 2 from a standstill? 3) when the vehicle's uniform acceleration speed reaches the maximum, how much work does the car do?
(4) What is the instantaneous power of the vehicle at the end of 10s?

1）F=4800N
P=F*V
V=12.5m/S
2）
F = 4800n + 0.6 * 4000kg = 7200n
V1 = P / F and T = V1 / A
3）Q=F*S=7200n*1/2mv1^2=
4) F joint * v10

The engine power of the car is 140W, and the mass of the car is 4T. When the car runs on a long straight road with a slope of 0.25, the resistance is 0.1 times of the vehicle weight
G = 10m / S2)
(1) What is the maximum speed the car can reach?
(2) How long can the process last if the car starts to move in a straight line with a constant acceleration of 0.6m/s2 from a standstill?
(3) How much work does the car do when its speed reaches the maximum value?

1. P = FV 140 = 4000 * 10 * 0.1 * V V = 0.035m/s. With common sense, the engine power = 140W in your question is wrong. It should be 140kW
1. The highest speed is 35m / s at 140kW
2. If the acceleration is 0.6, then the acceleration time is v = at t = 35 / 0.6 = 58.33 seconds
3. W = FS, cos α force is friction force, f = mg * friction coefficient
Because of the uniform acceleration, the distance can be calculated
Angle a is the slope
It's a long time since I graduated. I forgot how to put it in later