Question

The average velocity of the vehicle is 180 km/hr. It requires 2000 W for operation of 30 min. Find the magnitude of constant tractive effort.​

Answer 1

The magnitude of Tractive effort is 72 Kilo Newton

Explanation:

Given as :

The average velocity of vehicle = v = 180 km/h  = $\dfrac{180\times 1000}{3600}$ = 50 m/s

The power required for operation = p = 2000 watts  = 2000 Nm/s

The time period of operation = t = 30 min = 1800 sec

Let The magnitude of Tractive effort  = $\left | F \right |$ Newton

According to question

Work done = Power × time

i.e  work done = p × t

Or,work done = 2000 Nm/s × 1800 sec

∴ work done = 36 × $10^{5}$  = 3.6 × $10^{6}$ Nm

Again

magnitude of Tractive effort = $\dfrac{work done}{velocity}$

i.e   $\left | F \right |$  = $\dfrac{3.6\times 10^{6}}{50 }$

∴   $\left | F \right |$   = 72000

i.e   $\left | F \right |$  = 72 kN

So, The magnitude of Tractive effort  = $\left | F \right |$  = 72 Kilo Newton

Hence, The magnitude of Tractive effort is 72 Kilo Newton Answer

Answer 2

Thus the magnitude of constant tractive effort is 40 N.

Explanation:

We are given that:

• Average velocity of the vehicle "v" = 180 km/hr = 50 m/s
• Power of the vehicle "P" = 2000 W
• Time "t" = 30 min

Solution:

We need to find the magnitude of constant tractive effort.​

F is Tractive effort

P = F x v

F = P / v

F = 2000 / 50 = 40 N

Thus the magnitude of constant tractive effort is 40 N.