For a system released from rest at θ = 0° when a constant couple moment M = 100 N.m is applied, the speed of the 10-kg block is mathematically given as
V=4.33m/s
<h3>What is the speed of the 10-kg block?</h3>
Generally, the equation for the workdone is mathematically given as
W=T tehta<dtheta
W=100(90*\pi/180)
W=1.5707*100
W=1.57Nm
The change in potential energy across the pulley
dP=mgh
dp=10*9.81*111.8
dp=10.97J
For the thrid position, potential energy is
dP=mg(0.3)
dP=17.658J
dP'=17.658J-13.125
dP'=-4.532J
For 2nd position dP=0
The change in Kinectic energy across the pulley\
dK.E=0.5mv^2
For 1st
dk.E=0.5m(10)^2
2nd
dK.E=0.5Iw^2
dK.E=0.5(7.5*10^-3)(v^2/0.05)^2
3rd
2nd=3rd
In conclusion,
157.07=dKE+dP.E
157.07=5v^2+ (7.5*10^-3)(v^2/0.05)^2+10.97-4.53
V=4.33m/s
Read more about Kinetic energy
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V= 0.02 x 4000 = 80v
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