Tractor power 75kW. At what speed does the tractor move if it overcomes the resistance force of 15,000 N?

Given:
P = 75 kW = 75 * 10 ^ 3 W.
Fc = 15000 H.
v =?
1. The tractor does the work:
A = P * t, where P is the useful power, t is the operating time of the tractor.
2. Expression for defining work:
A = F * S, where F is the applied force, S is the displacement.
3. Expression for determining the path:
S = v * t, v – speed of movement, t – time.
4. Equate the work and substitute the expression for the path:
A = P * t = F * S.
P * t = F * v * t.
5. Let’s reduce the time and express the speed from the resulting expression:
P = F * v.
v = P / F.
6. Substitute the numerical values:
v = P / F = 75 * 10 ^ 3/15000 = 3 m / s.
Answer: tractor speed 3 m / s.