Two weights with masses m1 = m2 = 100 g each are suspended at the ends of a thread thrown over the block

Two weights with masses m1 = m2 = 100 g each are suspended at the ends of a thread thrown over the block (the block mass is negligible). What additional weight m3 must be put on one of the weights for the system to start moving and move with acceleration a = 0.2 m / s2?

m1 = 100 g = 0.1 kg.

m2 = 100 g = 0.1 kg.

g = 9.8 m / s ^ 2.

a = 0.2 m / s ^ 2

m3 -?

A stationary block only changes the direction of the force.

The first weight is acted upon by the force of gravity m1 * g and the tension force of the rope N.

m1 * a = N – m 1 * g.

N = m1 * a + m 1 * g.

The second weight is acted upon by gravity m * g and rope tension N.

m = m2 + m3.

m * a = m * g – N.

m * a = m * g – m1 * a – m 1 * g.

m1 * a + m 1 * g = m * g – m * a.

m1 * (a + g) = m * (g – a).

m = m1 * (a + g) / (g – a).

m = 0.1 kg * (0.2 m / s ^ 2 + 9.8 m / s ^ 2) / (9.8 m / s ^ 2 – 0.2 m / s ^ 2) = 0.104 kg.

m3 = m – m2.

m3 = 0.104 kg – 0.1 kg = 0.04 kg.

Answer: for any body you need to add a body with a mass of m3 = 0.04 kg.




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