The boy acts on a box moving on a horizontal floor with a mass of m = 20 kg by a force directed towards the movement of the box and equal in absolute value to 50 N. The coefficient of friction of the box on the floor is μ = 0.2. In this case, for some time the box has moved at a distance of L = 2 m. What work will be done during this time: a) a boy; b) gravity; c) friction force; d) the sum of all the forces acting on the box? Will the speed of the box increase or decrease during this time?
m = 20 kg.
g = 10 m / s2.
μ = 0.2.
F = 50 N.
L = 2 m.
Mechanical work A, performed by force F, is determined by the formula: A = F * L * cosα, where L is the displacement of the body under the action of the force, ∠α is the angle between the direction of the force F and displacement L
Since the boy acts by force on the box in the direction of movement, then ∠α = 0 °.
Am = F * L * cos0 ° = F * L.
Am = 50 N * 2 m = 100 J.
The force of gravity m * g is always directed vertically downward, and the box is moved horizontally, therefore ∠α = 90 °.
At = m * g * L * cos90 ° = 0 J.
Friction force Ftr directed in the opposite direction of the movement of the box, therefore, ATr = Ftr * L * cos180 ° = – Ftr * L.
Ftr = μ * m * g = 0.2 * 20 kg * 10 m / s2 = 40 N.
Atr = – 40 N * 2 m = – 80 J.
The resultant of all forces Fр will be a vector sum: Fр = F + Fтр + m * g + N. Since m * g = – N, and F and Fтр are oppositely directed, then Fр = F – Fтр.
Ap = (50N – 40 N) * 2 m * cos0 ° = 20 J.
The boy’s speed will increase.
Answer: Am = 100 J, Am = 0 J, Atr = – 80 J, Ap = 20 J.