# A box with a mass of m = 100kg is pulled along a horizontal plane with a rope tied at an angle of a = 80 °

**A box with a mass of m = 100kg is pulled along a horizontal plane with a rope tied at an angle of a = 80 ° to the horizon. Coefficient of friction between the box and the floor (mu) = 0.5. What is the smallest step A needs to be done to move the box a distance S = 100m in a straight line?**

m = 100 kg.

g = 9.8 m / s ^ 2.

μ = 0.5.

∠α = 80 °.

S = 100 m.

A -?

The work of force is determined by the formula: A = F * S * cosα, where F is the force with which the load is pulled, S is the movement of the load, ∠α is the angle between force F and displacement S.

m * a = F + Ftr + N + m * g.

ОХ: 0 = F * cosα – Ftr.

OU: 0 = F * sinα + N – m * g.

F * cosα = Ftr.

N = m * g – F * sinα.

Ftr = μ * N = μ * (m * g – F * sinα) = μ * m * g – μ * F * sinα.

F * cosα = μ * m * g – μ * F * sinα.

F * cosα + μ * F * sinα = μ * m * g.

F * (cosα + μ * sinα) = μ * m * g.

F = μ * m * g / (cosα + μ * sinα).

A = μ * m * g * S * cosα / (cosα + μ * sinα).

A = 0.5 * 100 kg * 9.8 m / s ^ 2 * 100 m * cos80 ° / (cos80 ° + 0.5 * sin80 °) = 12621 J.

Answer: it is necessary to do the work A = 12621 J.