in the follpowing figure assume that there is no friction force on the 20.0N- bl

Question

in the follpowing figure assume that there is no friction force on the 20.0N- block that sits on the tabletop. The pulley is light and frictionless, a 12N block hangs off the other side of the pulley. A.) Calculate the tension T in the light string that connects the blocks. B.) For a displacement in which the 12.0N- block descends 1.20m, calculate the total work done on the 20.0N- block. C.) For a displacement in which the 12.0- block descends 1.20 , calculate the total work done on the 12.0- block. D.) For the displacement in part , calculate the total work done on the system of the two blocks. How does your answer compare to the work done on the 12.0-block by gravity? E.) If the system is released from rest, what is the speed of the 12.0N- block when it has descended 1.20m?

Explanation / Answer

(a)
You have to find the masses which are
m1 = 20N / 9.8m/s^2 = 2.04 kg and
?m2 = 12N / 9.8m/s^2 = 1.22 kg
T = ((m1*m2)/(m1 + m2))*g
T = ((2.4888) / (3.26))*9.8
T = (0.76319) * 9.8
T = 7.5 N

(b)
Work = Force * Displacement
W = 7.5N * 1.2m
W = 9 J (joules)

(c)
W = F_w*D - T*D
W =12 N * 1.2 m - 7.5 N * 1.2 m
W = 5.4 J

(d)
Workdone = F*d = 12 * 1.2 = 14.4 J

(e)
K = 1/2*totalmass*velocity^2
12*1.2 = 1/2(3.26)v^2
14.4 = 1/2(3.26)v^2
v = ((2*14.4)/3.26)^1/2
v = 2.97m/s

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