In the arrangement shown in the figure work dione by the string on the block of mass 0.36 kg during the first second after the blocks are released from state of rest is (Ignotre friction and mass of the string )
(acceleration due ot gravity g=`10 ms^(-2)`)

Options

A. 8J
B. 4J
C. 12J
D. 2J

Answer

A

Explanation

According to the question

Given mass of block `1,m_(1) = 0.36 kg`
mass of block `2, m_(2) = 0.72 kg`
and acceleration due to gravity g= 10m `// x^(2)`
Now Dynamic equation of block `m_(1) (m_(2) gt m_(2))`
`:. T - m_(1) g = m_(1) a`
Dynamic equation of block `m_(2) (m_(2) gt m_(1))`
`:. m_(2) g - T= m_(2) g`
thus from adding Eqs . (i ) and (ii) we get
`:. T = m_(1) g + m_(2) g - T = m_(1) a + m_(2) a`
or `a = ((m_(2) - m_(1))g )/( (m_(1) + m_(2)))`
`:. ` Acceleration a `= ((0.72 - 0.36))/( 1.08) xx 10`
`a= (3.6)/(1.08) m//s^(2)`
Now the lension force in the string is for block 1 st `m_(1) =0.36`
`T - m_(1)g =m_(1) a`
`(T- 3.6)/(0.36) = (36)/(1.08)`
`T = 3.6 + 0.36 xx (36)/(1.08)`
Distance travelled by mass `m_(1)` block By displacement - time relation
Displacement `s= ul + (1)/(2) at^(2)`
[by kinetic equation for uniformly motion ]
`s= (1)/(2) xx (3.6)/(1.08) xx (1)^(2) `
`s= (1)/(2) xx (3.6)/(1.08)`