An 80 N crate slides with constant speed a distance of 5 m downward along a rough slope
that makes an angle of 30° with the horizontal. What is the work (in J) done by the force
of gravity?
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The net result is a negative acceleration of 1.5 so...
let F be the man's force
mg sin 30 is the component of gravity parallel to the slope
80/9.8is the crate's mass
F−gravity=ma
F−mgsin30=80/9.8
F−80sin30=80/9.8
F=40+80/9.8
W=(40+80/9.8)
W=200J
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