Physics
tcmusni
2020-04-29 09:18:00
A worker pushes a 1.50 x 10^3 N crate with a horizontal force of 345 N a distance of 24.0 m. Assume the coefficient of kinetic friction between the crate and the floor is 0.220. a) How much work is done by the worker on the crate? b) How much work is fone by the floor on the crate? c) What is the net work done on the crate?
ANSWERS
nannie
2020-04-29 09:52:31

(a) work=Fd  345x24=8280J  (b) work=Force of friction*d  Force of friction =coefficient*normal force=.22x1.5x10^3=330  330*d=7920J  (c) net work =8280-7920=360J

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