Asked by Bill
A dock worker slides a 65.0 Kg crate at a constant speed 35.5 m across a cement floor with a coefficient of kinetic friction of .633. How much work does friction do on the crate? How much work does the dock worker do on the crate?
Answers
Answered by
R_scott
friction and the worker do the same amount of work
work = m * g * μ * d = 65.0 * 9.81 *.633 * 35.5 = ? Joules
work = m * g * μ * d = 65.0 * 9.81 *.633 * 35.5 = ? Joules
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