Two blocks are connected by a very light string passing over a massless and frictionless pulley. the 20.0 n block moves 75.0cm to the right and the 12.0 n block moves 75.0cm downward.
find the total work done on 20.0n block if there is no friction between the table and the 20.0n block.

The string is assumed to be massless so the tension is the sting above the 12.0 N block has the same magnitude to the horizontal tension pulling to the right of the 20.0 N block. Thus, 
1.22 a = 12.0 - T  (eqn 1)
and for the 20.0 N block: 
2.04 a = T - 20.0 x 0.325 (using µ(k) for the coefficient of friction) 
2.04 a = T - 6.5  (eqn 2) 

[eqn 1] + [eqn 2] → 3.26 a = 5.5 
a = 1.69 m/s² 

Subs a = 1.69 into [eqn 2] → 2.04 x 1.69 = T - 6.5 
T = 9.95 N 

Now want the resultant force acting on the 20.0 N block: 
Resultant force acting on the 20.0 N block = 9.95 - 20.0 x 0.325 = 3.45 N 
Units have to be consistent ... so have to convert 75.0 cm to m: 

75.0 cm = 75.0 cm x [1 m / 100 cm] = 0.750 m 
work done on the 20.0 N block = 3.45 x 0.750 = 2.59 J

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