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A person holds a 210-g block that is connected to a 290-g block by a string going over a light pulley with no friction in the bearing (an Atwood machine). After the person releases the 210-g block, it starts moving upward and the heavier block descends.

Part A: What is the acceleration of each block?
Part B: What is the force that the string exerts on each block?
Part C: How long will it take each block to traverse 1.0 m?

Respuesta :

Manetho

Answer:

a) 1.5696 m/s^2

b) 2.389716 N

c)  1.12880 sec

Explanation:

Given data:

A two mass system connected by string

m1=210 g

m2= 290 g

a) For the given situation acceleration of the system a is given by

[tex]a= \frac{m_2-m_1}{m_1+m_2}g[/tex]

[tex]a= \frac{290-210}{210+290}\times9.81[/tex]

a=1.5696 m/s^2

force exerted by string on each block

b) T= m1(g+a)

= 0.210(9.81+1.5696)

=2.389716 N

c) Initial velocity u= 0

distance s= 1 m

acceleration a= 1.5696 m/s^2

therefore, S=ut + 0.5at^2

⇒[tex]t= \sqrt{\frac{2}{1.5696} }[/tex]

  t = 1.12880 sec

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