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 13:30:16 · 2 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 05:08:53 · answer #1 · answered by odu83 7 · 0⤊ 0⤋
the galaxies rotate at speeds inconsistent with their obvious mass is in view that we do see it all. I am regarding it as being the theoretical Dark Matter. There are very powerful proofs that shows that darkish topic exist. One the is the inconsistent pace of and obvious mass. darkish topic makes up approximately seventy five% to eighty% of the problem within the Universe...
2016-09-05 11:27:32 · answer #2 · answered by weigel 4 · 0⤊ 0⤋