Two bodies with masses m1 = 50 g and m2 = 100 g, connected by a weightless thread

Two bodies with masses m1 = 50 g and m2 = 100 g, connected by a weightless thread, lie on a smooth table. With what force can the first body be pulled parallel to the surface so that a thread that can withstand a load of 5 N does not break?

Given:

m1 = 50 grams = 0.05 kilograms – the mass of the first body;

m2 = 100 grams = 0.1 kilogram – the mass of the second body;

g = 10 m / s ^ 2 – acceleration of gravity;

T = 5 Newton – the maximum load that the thread can withstand.

It is required to determine the force F (Newton) with which the first body can be pulled so that the thread does not break.

Since, according to the condition of the problem, the bodies lie on a smooth table, that is, the friction force can be ignored, and the thread itself is weightless, then:

Newton’s second law for the first body:

F – T = m1 * a;

Newton’s second law for the second body:

T = m2 * a, hence:

a = T / m2. Substitute in the first equation:

F – T = m1 * T / m2;

F = m1 * T / m2 + T = T * (m1 / m2 + 1) = 5 * (0.05 / 0.1 + 1) = 5 * (0.5 + 1) = 5 * 1.5 = 7.5 Newton.

Answer: You can pull with a force equal to 7.5 Newtons.

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