In the absence of mass m1 in the Atwood machine, and with m2 greater than m1, the acceleration of block m2 simplifies to the gravitational constant g, acting upward due to the net gravitational force.
In the given Atwood machine scenario, where two blocks of masses m1 and m2 are connected by a massless rope over a frictionless pulley, we explore a special case where m1 is not present, and m2 is greater than m1. Taking upward as the positive direction, and considering the gravitational constant g to be positive, we can analyze the acceleration of the block with mass m2 in this simplified system.
In this special case, with only m2 present, the net force acting on m2 is solely due to its weight. According to Newton's second law (F = ma), the force acting on m2 is given by the gravitational force:
F = m2 * g
Here, m2 is the mass of block m2 and g is the gravitational constant. As m2 is greater than m1, the net force is directed downward. The resulting acceleration (a) is determined by the ratio of the net force to m2:
a = F / m2 = (m2 * g) / m2 = g
In this special scenario, the acceleration of the block with mass m2 is equal to the gravitational constant g, and it acts in the positive direction, which is upward.
The question probable may be:
(Figure 1) shows an Atwood machine that consists of two blocks (of masses 11 and m₂) tied together with a massless rope that passes over a fixed, perfect (massless and frictionless) pulley. In this problem you'll investigate some special cases where physical variables describing the Atwood machine take on miting values. Often, examining special cases will simplify a problem, so that the solution may be found from inspection or from the results of a problem you've already seen. In the figure mass 2 is greater than mass 1
For problem, take upward to be the positive direction and take the gravitational constant, g, to be positive. for the same special case (the block of mass m1 not present), what is the acceleration of the block of mass m2?