Posted by **Benji** on Thursday, November 16, 2006 at 5:33pm.

A light rope is attached to a block with a mass of 6 kg that rests on a horizontal, frictionless surface. THe horizontal rope passes over a frictionless, massless pulley, and a block of mass m is suspended from the other end. When the blocks are released, the tension in the rope is 18 N. a) what is the acceleration of the 6 kg block ? b) the mass m of the hanging block?

a) is 3 m/s

b) is 2.65 kg

I don't get the equations I should have used

Use Newton's second law:

F = m a

The tension of 18 N pulls on the 6 kg block, so it's acceleration is:

a = 18 N/(6 kg) = 3 m/s^2

Now the length of the rope can't change and it is pulled tight by the falling weight. This means that if the mass on the surface is accelerating at 3 m/s^2, the mass m is dropping downward with this acceleration.

The force on the hanging rope is (let's take the downward directon as positive):

F = m g - 18 N

But this has to equal m*a by Newton's second law and we know that a = 3 m/s^2.

m g - 18 N = m*(3 meters/s^2) --->

m = 18 N/[g - 3 m/s^2] = 2.64 kg

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