4)Given,
m = 2 kg ; M = 8 kg ; uk = 0.3 ; F = 10 N ; r = 3 m
Ff = 0.3 x 10 = 3 N
Fnet = 10 - 3 = 7 N
a = F/m = 7/2 = 3.5 m/s^2
a' = 3/8 = 0.38 m/s^2
t = sqrt (2r/(a - a'))
t = sqrt (2 x 3/(3.5 - 0.38)) = 1.4 s
Hence, t = 1.4 s
b)D = 1/2 at^2
D = 0.5 x 0.38 x 1.4^2 = 0.37 m
Hence, D = 0.37 m
Problem 3 (20 points) In the following figure, a horizontal force F is applied to a...
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4M
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