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# Please solve

Asked by Amuwadunsin 26th February 2020, 2:31 PM
Answered by Expert
Answer:
For m2 , we have,  m2g - T = m2 a .......................(1)
For m1 , we have ,    T  = m1 a  .....................(2)

Where T is tension in the string and a is acceleration of system.

By adding (1) and (2),  we get  a =  [ m2 / ( m1 + m2 ) ] g  ......................(3)

If m1 starts from rest and travel a distance d in t seconds with acceleration a,

then d = (1/2) a t2 ,  or   t = ( 2d / a )1/2 = [ 2d (m1 + m2 ) / ( m2 g ) ]1/2
---------------------------------------------------------------
for block of mass m1 , we have ,  T - ( m1 g )= m1 a ......................(1)

for block of mass m2 , we have ,  ( m2 g ) - T = m2 a ...................(2)

Where T is tension in the string and a is acceleration of system.

By adding both eqn. above , we get ,  a  = [ ( m2 - m1 ) / ( m1 + m2 ) ] g

If mass m2 hitting the ground after travelling a distance x starting from rest,

we have v2 = 2 a x  ,   where v is final speed when m2 is hitting the ground

hence v = { 2 [ ( m2 - m1 ) / ( m1 + m2 ) ] g x }1/2
Answered by Expert 27th February 2020, 9:30 PM
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