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A 4kg mass sits on a frictionless table....
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[QUOTE="jbriggs444, post: 6810785, member: 422467"] Let me do this quickly in my head. Indeed, I get something different. So let us check your work. As I understand it, we have two forces on the 4 kg mass. One of 20 N is applied directly with no mechanism specified. The other from the tension in the string. So here, F is the force of gravity on the dangling 2 kg mass. But here, F is the net force on... something. It is wise to avoid using the same variable name in the same problem to denote two different quantities. Apparently that something has a mass of 4+2. So it is the sum of the two masses. [B]But you only counted one of the two forces.[/B] Note that you wrote 20/4+2 where you presumably meant 20/(4+2) = ##\frac{20}{4+2}##. Order of operations is important. [/QUOTE]
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A 4kg mass sits on a frictionless table....
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