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Response Details:
M = 4 kg, m = 1.5 kg, d = 150 mm = 0.150 m, k = 300 N/m
when B moves x down, A will move x/2, so the speed vA = vB/2
a) initial energy = mgd
final energy = MvA2/2 + mvB2/2 + k(d/2)2/2
mgd = MvA2/2 + mvB2/2 + kd2/8
2mgd = MvA2 + mvB2 + kd2/4
2mgd = M(vB/2)2 + mvB2 + kd2/4
2mgd = (M/4 + m)vB2 + kd2/4
vB = √[(2mgd - kd2/4)/(M/4 + m)] (1)
vB = 1.04 m/s
b) when v is max, net force = 0, kd = 2T = 2mg
sub it into (1)
vB = √[3mgd/(2(M/4 + m))] = 1.15 m/s
c) when d = max, v = 0
from (1)
2mgd - kd2/4 = 0
d = 8mg/k = 0.392 m
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