The railcar moves uniformly on horizontal rails under the action of a traction force of 140 N

The railcar moves uniformly on horizontal rails under the action of a traction force of 140 N. Determine the work of each force applied to it on a 1 km track section.

F = 140 N.

g = 10 m / s2.

S = 1 km = 1000 m.

A -?

The mechanical work A of the force F is determined by the formula: A = F * S * cosα, S is the displacement of the body, ∠α is the angle between the direction of action of the force and displacement S.

Since the body moves in a straight line, its movement will coincide with the path traveled.

When moving, 4 forces act on the body.

The traction force F acts in the direction of the body’s motion, therefore ∠α = 0 °, cos0 ° = 1.

Traction work: A1 = F * S. A1 = 140 N * 1000 m = 140,000 J.

The friction force Ffr, directed in the opposite direction of the body’s motion, and numerically Ffr = F. Therefore, ∠α = 180 °, cos180 ° = -1.

Work of friction force A2: A2 = -F * S. A2 = – 140 N * 1000 m = -140000 J.

The force of gravity is m * g vertically downward, the force is the reaction of N rails directed vertically upward. They are directed perpendicular to the displacement S, ∠α = 90 °, cos90 ° = 0.

A3 = A4 = 0.

Answer: when the railcar is moving, the traction force does the work A1 = 140,000 J, the friction force does the work A2 = -140000 J, the gravity force and the reactions of the rails do not perform the work.