Kyle pushed a 50 kg sack of rice across a level floor with a horizontal force of 35.0 N against a frictional force of 12.0 N. He succeeded in moving the sack to a distance of 5.0m. How much work was done by (a) Kyle, (b) friction, (c) force of gravity on the sack of rice?
Best answer gets brainliest ty po
Best answer gets brainliest ty po
Answers & Comments
Let's calculate the work done in each scenario:
(a) To find the work done by Kyle, we use the formula: work = force x distance. So, the work done by Kyle is 35.0 N x 5.0 m = 175.0 Joules.
(b) The work done by friction can be calculated using the same formula: work = force x distance. The force of friction is 12.0 N and the distance is 5.0 m, so the work done by friction is 12.0 N x 5.0 m = 60.0 Joules.
(c) The force of gravity doesn't do any work in this case because the motion is horizontal. Therefore, the work done by the force of gravity on the sack of rice is 0 Joules.
I hope that helps!