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 
<span>Units have to be consistent ... so have to convert 75.0 cm to m: </span>
75.0 cm = 75.0 cm x [1 m / 100 cm] = 0.750 m 
<span>work done on the 20.0 N block = 3.45 x 0.750 = 2.59 J</span>
        
             
        
        
        
Answer:
They will leave at 14:25 (2:25) at the same time
Step-by-step explanation:
 
        
                    
             
        
        
        
Answer:$181.81
Step-by-step explanation:300/198
=1.5151515….multiple by 120 
=181.818181
 
        
             
        
        
        
Answer:
2
Step-by-step explanation:
i think