Two weights with masses of 300 g and 400 g are connected by a weightless inextensible thread thrown over the block

Two weights with masses of 300 g and 400 g are connected by a weightless inextensible thread thrown over the block. find the thread tension and the acceleration with which the weights move.

Given:

m1 = 300 grams = 0.3 kilograms – the mass of the first cargo;

m2 = 400 grams = 0.4 kilograms – the mass of the second cargo;

g = 10 Newton / kilogram – acceleration due to gravity (approximate value).

It is required to determine a (m / s2) – the acceleration with which the system of weights moves, and also T (Newton) – the thread tension.

By the condition of the problem, we will not take into account the mass of the thread and the block itself, as well as the friction force. Then, proceeding from Newton’s second law, we have a system of two equations:

m2 * g – T = m2 * a (1) – for the second body;

T – m1 * g = m1 * a (2) – for the first body.

From equation (2) we find the thread tension force:

T = m1 * a + m1 * g.

Substituting the found value into equation (1), we get:

m2 * g – m1 * a – m1 * g = m2 * a;

m2 * g – m1 * g = m2 * a + m1 * a;

g * (m2 – m1) = a * (m2 + m1);

a = g * (m2 – m1) / (m2 + m1) = 10 * (0.4 – 0.3) / (0.4 + 0.3) = 10 * 0.1 / 0.7 =

= 1 / 0.7 = 1.4 m / s2 (result has been rounded to one decimal place).

Then the tension force of the thread will be equal to:

T = m1 * a + m1 * g = m1 * (a + g) = 0.3 * (1.4 + 10) = 0.3 * 11.4 = 3.42 Newton.

Answer: the system of bodies moves with an acceleration equal to 1.4 m / s2, the tension force of the thread is 3.42 Newtons.