The free body diagram for the block of mass <em>M</em> consists of four forces:
• the block's weight, <em>Mg</em>, pointing downward
• the normal force of the table pushing upward on the block, also with magnitude <em>Mg</em>
• kinetic friction with magnitude <em>µMg</em> = 0.2 <em>Mg</em>, pointing to the left
• tension of magnitude <em>T</em> pulling the block to the right
For the block of mass <em>m</em>, there are only two forces:
• its weight, <em>mg</em>, pulling downward
• tension <em>T</em> pulling upward
The <em>m</em>-block will pull the <em>M</em>-block toward the edge of the table, so we take the right direction to be positive for the <em>M</em>-block, and downward to be positive for the <em>m</em>-block.
Newton's second law gives us
<em>T</em> - 0.2<em>Mg</em> = <em>Ma</em>
<em>mg</em> - <em>T</em> = <em>ma</em>
where <em>a</em> is the acceleration of either block/the system. Adding these equations together eliminates <em>T</em> and we can solve for <em>a</em> :
<em>mg</em> - 0.2 <em>Mg</em> = (<em>m</em> + <em>M</em>) <em>a</em>
<em>a</em> = (<em>m</em> - 0.2<em>M</em>) / (<em>m</em> + <em>M</em>) <em>g</em>
<em>a</em> = 1.96 m/s²
Then the tension in the string is
<em>T</em> = <em>m</em> (<em>g</em> - <em>a</em>)
<em>T</em> = 78.4 N