A workers pushes a 1.5 x 10^3 N crate with a horizontal force of 345 N a distance of 24.0 m The coefficient of kinetic force 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 done by the floor on the crate?
c. What is the net work done on the crate?
A 0.075 kg ball in a kinetic sculpture is raised 1.32m above the ground by a motorized vertical conveyor belt. A constant frictional force of 0.350 N acts in the direction opposite the conveyor belt's motion. What is the net work done on the ball?
2007-11-05 14:16:26 · 1 answers · asked by redalert 2 in Science & Mathematics ➔ Physics
a) Work=F*d
345*24
=8280 J
b) The frictional work is
1500*0.22*24
=7920 J
c) The net will be translated into kinetic energy
360 J
The ball gets raised 1.32 m so the work is equivalent to PE gain
0.075*9.81*1.32
0.971 J
j
2007-11-06 09:55:09 · answer #1 · answered by odu83 7 · 0⤊ 0⤋