The block in the figure is initially at rest on an inclined plane at the equilib
ID: 2139565 • Letter: T
Question
The block in the figure is initially at rest on an inclined plane at the equilibrium position that it would have if there were no friction between the block and the plane. How much work is required to move the block 25 cm down the plane if the frictional coefficient is ? = 0.?
How much work is required to move the block 25 cm down the plane if the frictional coefficient is ? = 0.16?
The block in the figure is initially at rest on an inclined plane at the equilibrium position that it would have if there were no friction between the block and the plane. How much work is required to move the block 25 cm down the plane if the frictional coefficient is ? = 0.? How much work is required to move the block 25 cm down the plane if the frictional coefficient is ? = 0.16?Explanation / Answer
a)eq. pos be xo
mgsin30=kxo
xo=.245m
Change in PE=.5k((.25+.245)^2-.245^2)=3.7
Work done by gravity=mgsin30x=2.45J
Work done=3.7-2.45=1.25J
b)Work done against friction=Umgcos30x=0.679 J
Work done=1.25+0.679=1.929 J
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