A wooden block of mass `0*8` kg is dragged along a level frictionless surface by a hanging block of mass `0*2` kg as shown in the figure . Calculate the tension in the string and the acceleration of blocks .

Explanation

` m_(1) = 0*8 kg, m_(2) = 0*2 "kg"`
The forces on the two blocks and tension in the string are shown in the figure .
Let the two blocks move with common magnitude of acceleration a .
Along the horizontal , the block of mass`m_(1)` is acted upon by the force due to tension in the spring .
` :. T= m_(1)a " " ...(i) `
Since the block `m_(2)` moves with an acceleration a in downward direction .
` :. m_(2)g - T = m_(2)a " " ` ...(ii)
From (i) and (ii) , we get
` a = (m_(2)g)/(m_(1) +m_(2))= (0*2 xx 9*8)/(0*8 +0*2)= 1.96 m//s^(2)`
` a = 1. 96 m//s^(2)`
From equation (i),
` T = m_(1)a = 0*8 xx 1*96 `
` T = 1.568 N `