Question

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?

Answer 1

a) 1.5696 m/s^2

b) 2.389716 N

c) 1.12880 sec

Step-by-step 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

`a= (m_2-m_1)/(m_1+m_2)g`

`a= (290-210)/(210+290)*9.81`

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

⇒
`t= \sqrt{(2)/(1.5696) }`

t = 1.12880 sec