Posted by Brian on .
A 0.075 kg ball in a kinetic sculpture moves at a constant speed along a motorized conveyor belt. The ball rises 1.32 m above the ground. 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?

Physics 
bobpursley,
net work= friction work+ changein PE
net work= .350*distance+mg*distance 
Physics 
Anonymous,
0.1

Physics 
tt,
when they fall the ball

Physics 
Darby,
(0.075)(9.81)(1.32)= 0.971 J
then, (0.350)(1.32)= 0.462J
then, 0.971 0.462= 0.509J ~ work done on the ball 
Physics 
anonymous,
F=mg, so force = mass times gravity.
0.075 times 9.8 = 0.735
W=FD, so work= force times displacement.
0.735 times 1.32= 0.9702 J.
So, 0.9702 is the first amount of work.
Second, frictional force times displacement.
0.350 times 1.32 = 0.462 J.
Subtract them to get the net force, so:
0.9702  0.462 = 0.51 J
0.51 J is therefore the answer. 
Physics 
Anonymous,
the website finnytown states the answer is 0.00J but i cannot figure out why.

Physics 
ANON,
The work is 0.00J because the sculpture's speed is constant, and if speed is constant a= 0m/s^2 which makes the F= 0N and W=0J

Physics 
Trolly,
I don't know