Question #3ef6b

1 Answer
Nov 27, 2016

Let

#m_1kg->"1st mass having weight " 9.9N#

#m_2kg->"2nd mass having weight " 22.2N#

So #m_1=9.9/g=9.9/10=0.99kg#

and #m_2=20.2/g=20.2/10=2.02kg#

Since #m_2>m_1#
The combined system will have downward acceleration on #m_2# and same upward acceleration on #m_1#. Let the acceleration of the combined system be a and tension on connecting rope be #T#

Now considering the force on #m_1# we can write

#T-m_1g=m_1a........(1)#

Agaun considering the force on #m_2# we can write

#m_2g-T=m_2a........(2)#

Adding (1) and (2) we get

#(m_2+m_1)a=(m_2g-m_1g)=(20.2-9.9)#

#=>a=10.3/(2.02+0.99)ms^-2#

#=>a=10.3/3.01ms^-2~~3.42ms^-2##