Two masses 5 and 7 kg are originally at rest on a frictionless surface. the masses are connected by a light cord. a second cord is attached to the 7 kg mass and pulled with a horizontal force of 30 n such that the masses accelerate in a single direction. what is the tension in the cord that connects the two masses?

12.5 N

Explanation:

We can find out about the tension of the cord pulling the 5kg mass by first calculate the horizontal acceleration of said mass, which is the same as the horizontal acceleration of the whole system.

Total system mass M = 5 + 7 = 12 kg

Total system acceleration = F/M = 30/12 = 2.5 (m/s2)

The tension force acting on the 5kg mass in order to exert a 2.5 m/s2 acceleration is

f = ma = 5*2.5 = 12.5 N

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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