Two weights with masses m1 = 0.4 and m2 = 0.6 kg are connected by a thread thrown over a fixed and weightless block.

Two weights with masses m1 = 0.4 and m2 = 0.6 kg are connected by a thread thrown over a fixed and weightless block. What distance L will the load m2 travel after the start of movement in the first second? (t = 1.0 sec)

In this case, each load can be considered separately, using the elastic force of the thread in the equations, or it is possible – as a system of bodies, on which the gravity force applied to the first load (mg = 4H) acts in one direction, and the gravity force applied to the other side. to the second load (mg = 6N). The difference of these forces (2N) will force the entire system of weights with a mass (0.4 kg + 0.6 kg = 1 kg) to move with acceleration. According to Newton’s 2nd Law, a = F / (m1 + m2) = 2 m / s ^ 2 (second squared). Initial speed of both weights = 0. Using the formula S = at ^ 2/2 (the acceleration is multiplied by the time squared, and all this is divided by 2), we find the path of the load m2 (as well as the associated m1) in 1 s: S = 1m.