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: 
<u>Do Keep Change Flip (KCF)</u>
Keep: 5/6
Change: ÷ into ×
Flip: 3/1 into 1/3
Your new problem should be: 5/6×1/3
<u>Multiply</u>
5/6×1/3=5/18
2x - 6 = 2 • (x - 3)
2 = 0
x-3 = 0
x=3
Answer:
Doesn't make sense.
Step-by-step explanation:
-1 ≤ cosθ ≤ 1
=> cosθ = 3√3 doesn't make sense.
