On the table lies a load with a mass of m1 = 70 g. A thread is tied to this load, thrown over a fixed block

On the table lies a load with a mass of m1 = 70 g. A thread is tied to this load, thrown over a fixed block, which is on the same level as the load. A load weighing m2 = 28 g was suspended at the other end of the thread. Under the influence of the load m2, the system began to move. Find the path S covered by a load of mass m1 in t = 1.5 s. The coefficient of friction when the load moves on the table is k = 0.35.

There is a load on the table with a mass of m₁ = 70 g = 0.07 kg, on which the action of vertical forces is balanced: the force of gravity m₁ ∙ g and the reaction of the support N, where g = 9.8 N / kg. We get that N = m₁ ∙ g. A thread is tied to this load, thrown over a fixed block, which is at the same level with the load, and causing its movement, which means that, according to Newton’s 2 law, the resultant force m₁ ∙ a = T – Ftr acts on the load, where a is the acceleration of the load, T – rope tension force, Ftr = k ∙ N – friction force, k – friction coefficient when the load moves on the table. Combining the formulas, we get the equation m₁ ∙ a = Т – k ∙ m₁ ∙ g. A load weighing m₂ = 28 g = 0.028 kg was suspended to the other end of the thread, on which, according to Newton’s 2 law, the resultant force m₂ ∙ a = m₂ ∙ g – T acts on the load. Adding the equations term by term, we obtain: (m₁ + m₂) ∙ a = T – k ∙ m₁ ∙ g + m₂ ∙ g – T. From here we find the acceleration of the motion of bodies: a = (m₂ – k ∙ m₁) ∙ g: (m₁ + m₂). Under the action of the load m₂, the system began to move. To find the path S, traversed by a load of mass m₁, in time t = 1.5 seconds, we use the formula for the path with uniformly accelerated rectilinear motion: S = vо ∙ t + a ∙ t² / 2, where the initial velocity of the body is vо = 0 m / s … Then S = (m₂ – k ∙ m₁) ∙ g ∙ t²: (2 ∙ (m₁ + m₂)). It is known from the problem statement that the coefficient of friction when the load moves on the table is k = 0.35. Substitute the values ​​of physical quantities in the formula and make calculations:
S = (0.028 – 0.35 ∙ 0.07) ∙ 9.8 ∙ (1.5) ²: (2 ∙ (0.028 + 0.07));
S ≈ 0.39 (m).
Answer: the distance traveled by a load of mass m₁ in time t = 1.5 seconds is equal to ≈ 0.39 meters.