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a) The effort needed to move the load up the inclined plane at a constant velocity.
V.R=1/sin 30 =2
Therefore, (M.A/2)*20=75
M.A= (2*75)/20 =3/2
3/2 = 82N/effort
Effort= (82*2)/3 =540N
b) The workdone against friction in raising the mass through the height of 1.0m.
Work input=effort*effort distance = (540*4)/sin 30 =4320J
Work output=load*load distance= 81*2*4= 3240J
Therefore, workdone against friction= 4320-3240= 1180J
Moraa orina answered the question on April 9, 2018 at 16:54
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