Question

The average velocity of the wehicle is 180km/hr. It
requires 2000w for operation of 30 min. Find
magnitude of constant tractive efforts.
​

Answer 1

Tractive efforts. = 72 KN

Given:

The average velocity of the vehicle is 180km/hr.

Requires 2000w.

Time = 30 min.

To find:

Magnitude of constant tractive efforts.

Formula used:

Work done= power × Time

Work done=  Force × Distance traveled

Explanation:

Average velocity of the vehicle is 180km/hr. = $\frac{180}{3.6}$  m/s = 50 m/s

Power =  2000w.

Time = 30 min.= 30×60 = 1800 sec.

Work done= power × Time

Work done= 2000 ×​ 1800

Work done= 3.6× $10^{6}$ Nm.

Work done=  Force × Distance traveled

Distance traveled  =  Because per second

Force = $\frac{Work\ done}{Average\ velocity}$ =  $\frac{3.6\times 10^{6} }{50}$

Tractive efforts. = 72 KN

To learn more....

1)A train weighing 1000ton is moving on a horizontal track with a uniform velocity of 10m/s. the total resistance to the motion of the train is 5kgf per ton. if g= 10m/s2, then what is the power of the engine?

https://brainly.in/question/2950084

2)Calculate the power of an engine which can pull a train of mass 25000 quintal up to incline of 1 in 100 at the rate of 10.8 km/h resistance due to friction is 2 N/quintal

https://brainly.in/question/5663895