The boy pulls the sled along a horizontal surface at a constant speed, applying a force of 100N

The boy pulls the sled along a horizontal surface at a constant speed, applying a force of 100N to the rope. The rope forms an angle of 60 degrees to the horizon. What kind of work does the friction force do when the slide is moved at a distance of 10m?

V = const.

F = 100 N.

S = 10 m.

Atr -?

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

Atr = Ftr * S * cosα.

The friction force Ffr is always directed in the opposite direction of the sled movement, that is, the angle ∠α between the direction of the friction force Ffr and the movement of the sledges S is ∠α = 180 °.

cos180 ° = – 1.

Since, according to the condition of the problem, the sleds are pulled uniformly rectilinearly, then according to Newton’s 1 law, the sliding friction force Ffr is balanced by the boy’s force F: Ftr = F.

Atr = – F * S.

Atr = – 100 N * 10 m = -1000 J.

Answer: when the sled is moving, the friction force performs the mechanical work Atr = -1000 J.