Asked by Brian

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?
Answered by bobpursley
net work= friction work+ changein PE

net work= .350*distance+mg*distance

Answered by Anonymous
0.1
Answered by tt
when they fall the ball
Answered by 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
Answered by 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.
Answered by Anonymous
the website finnytown states the answer is 0.00J but i cannot figure out why.
Answered by 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
Answered by Cindy
Or do you add the forces because you are overcoming friction??
Which makes it 0.972 + 0.462 = 1.434J ???
Answered by Uknown
Its zero work
Answered by anonymous
how are there forces in constant speed, and also how is there a friction force on a gravity force (along the y axis). if the ball rises, that means the work done by the force of gravity would be negative, meaning that negative (force of gravity times distance) plus negative (force of friction times distance). so you add those two negatives resulting in -1.4322J
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