Question

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.

Answer 1

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