Determine the work that the engine performs when a tractor with a mass of 2 tons is uniformly moving on a path

Determine the work that the engine performs when a tractor with a mass of 2 tons is uniformly moving on a path of 1 km, if the friction force is 0.08 of the tractor’s gravity.

m = 2 t = 2000 kg.

g = 10 N / kg.

S = 1 km = 1000 m.

Fcopr = 0.08 * Ft.

A -?

We express the work of the traction force of the tractor engine F by the formula: A = F * S, where S is the path traveled by the tractor.

Since the tractor moves uniformly, then according to Newton’s 1 law, the action of all forces on it is compensated.

The traction force of the tractor F is compensated by the resistance force Fcopr, the gravity Ft is compensated by the reaction of the road N: F = Fcopr, Ft = N.

Fт = m * g.

Since according to the condition of the problem Fcopr = 0.08 * Ft, then F = 0.08 * m * g.

The operation of the tractor engine A will be determined by the formula: A = 0.08 * m * g * S.

A = 0.08 * 2000 kg * 10 N / kg * 1000 m = 1600000 J.

Answer: the tractor engine performs A = 1600000 J of mechanical work.