presended by Sapling Leaming ness center, a man ies face down on a bench and ift
ID: 1780332 • Letter: P
Question
presended by Sapling Leaming ness center, a man ies face down on a bench and ifts a weight with one lower leg While exercising in a fit by contracting the muscles in the back of the upper leg. Find the magnitude of the angular acceleration produced given the mass lifted is 13.5 kg at a distance of 28.0 crm from the knee joint, the moment of lower leg is 0.900 kg m2·the muscle fore is 1410 N, and its effective peperdoar lever arm is 3.45 cm. Number rad/s How much work is done if the leg rotates through an angle of 20 0 with a constant force exerted by the muscle? Number Check Als GNaotExplanation / Answer
a) Torque = F r - m g x
= (1410 * 0.0345) - (13.5 * 9.8 * 0.28) = 11.6 N.m
Inet = I + m x2 = 0.900 + (13.5 * 0.282) = 1.96 kg.m2
torque = I * alpha
angular acceleration = 11.6 / 1.96
= 5.92 rad/s2
b) theta = 20 deg = 0.349 rad
Work done = 11.6 * 0.349
Work done = 4.05 J
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