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posted by  davviieee on 11/3/2009 4:03:44 PM  |  status: Live  |  Earned Karma: 25

What is the power needed to operate the log splitter's pump?

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General Physics Physics (8th) by Cutnell, Johnson 11 16 11/4/2009 at 1:00:00 PM
Question Details:
A log splitter uses a pump with hydraulic oil to push a piston, attached to which is a chisel. The pump can generate a pressure of 2.6x107 Pa in the hydraulic oil, and the piston has a radius of 0.030 m. In a stroke lasting 11 s, the piston moves 0.55 m. What is the power needed to operate the log splitter's pump?
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posted by watt (MNK) on 11/3/2009 5:53:35 PM  |  status: Live
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Response Details:
Pressure  P = 2.6 * 10 ^ 7 Pa
Radius  r = 0.03 m
Force F = P  A
             = P * ( πr ^ 2 )
             = 73513.26 N
distance S = 0.55 m
timet = 11 s
SO, power  P = FS / t
                      = 3675.66 watt
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