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A student tries to raise a chain consisting of three identical links. Each link has a mass of 300 g . The three-piece chain is connected to a string and then suspended vertically, with the student holding the upper end of the string and pulling upward. Because of the student's pull, an upward force of 15.0 N is applied to the chain by the string. Use Newton's laws to answer the following questions.
Find the acceleration of the chain.Take the free fall acceleration to be g.Find the force exerted by the top link on the middle link.

Respuesta :

Answer:

a=6.65 m/s²

F(top)=9.96 N

Explanation:

Given that

m= 300 g= 0.3 kg

F= 15 N

Lets acceleration of the chain is a

So from  Newton's law

F(net) = m(total) a

F(net) - 3 m g= 3 m a

Here  F(net) = F

Take g=9.81 m/s²

So now by putting the values

F - 3 m g= 3 m a

15 - 3 x 0.3 x 9.81 = 3 x 0.3 x a

a=6.85 m/s²

F(top) - 2 m g= 2 m a

F(top)= 2 ma + 2 mg

F(top)=2 x 0.3 (9.81 +6.85 )

F(top)=9.96 N

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