Two bars weighing 1 kg and 4 kg, connected by a cord, lie on the table. A force of 40 N was applied to the first of them

Two bars weighing 1 kg and 4 kg, connected by a cord, lie on the table. A force of 40 N was applied to the first of them, directed horizontally. With what acceleration do the bodies move if the coefficient of sliding friction of the bars on the table is 0.2?

Given:

m1 = 1 kilogram is the mass of the first body;

m2 = 4 kilograms – the mass of the second body;

g = 10 m / s ^ 2 – free fall acceleration;

F = 40 Newton – force applied to the first body and directed horizontally;

k = 0.2 is the coefficient of friction of the bars on the table surface.

It is required to determine the acceleration of bodies a (m / s ^ 2).

For the first body:

F – T – k * m1 * g = m1 * a (1);

For the second body:

T – k * m2 * g = m2 * a (2).

Substituting the value of T from equation (2) into equation (1), we obtain:

F – m2 * a – k * m2 * g – k * m1 * g = m1 * a;

F – g * k * (m2 + m1) = a * (m1 + m2);

a = (F – g * k * (m2 + m1)) / (m1 + m2) = (40 – 10 * 0.2 * (1 + 4)) / (1 + 4) = (40 – 10) / 5 = 30/5 = 6 m / s ^ 2.

Answer: bodies move with an acceleration of 6 m / s ^ 2.