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?

Answer 1

net work= friction work+ changein PE

net work= .350*distance+mg*distance

Answer 2

0.1

Answer 3

when they fall the ball

Answer 4

(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

Answer 5

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.

Answer 6

the website finnytown states the answer is 0.00J but i cannot figure out why.

Answer 7

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

Answer 8

Or do you add the forces because you are overcoming friction??
Which makes it 0.972 + 0.462 = 1.434J ???

Answer 9

Its zero work

Answer 10

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