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MMLU-Pro / 9596 / A chain is wound on a spool fixed at the top of a frictionless inclined plane.…

Problem

Answer published by the source. Consult the official source to check your work against its answer.

question

A chain is wound on a spool fixed at the top of a frictionless inclined plane. The plane has a length l and is at an angle \texttheta with the horizontal. One end of the chain is initially at the lower edge of the plane, but is free to fall straight down. Find the velocity of this end after it has fallen y feet, assuming a link of the chain has no velocity until it leaves the spool.

category

physics

options

  1. v^2 = [{g(3y^3 + 4ly^2 (sin \theta + 1) + (9l^2y sin \theta))} / {4(l + y)^2}]
  2. v^2 = [{g(2y^3 + 3ly (sin \theta + 2) + (5l^2y sin \theta))} / {3(l + y)}]
  3. v^2 = [{g(2y^3 + 3ly^2 (sin \texttheta + 1) + (6l^2y sin \texttheta))} / {3(l + y)^2}]
  4. v^2 = [{g(2y^3 + 3ly^2 (sin \theta + 1) + (6l^2y cos \theta))} / {3(l + y)^2}]
  5. v^2 = [{g(y^3 + 2ly^2 (sin \theta + 1) + (4l^2y sin \theta))} / {2(l + y)^2}]
  6. v^2 = [{g(2y^3 + 2ly^2 (cos \theta + 1) + (7l^2y sin \theta))} / {3(l + y)^2}]
  7. v^2 = [{g(2y^3 + 3ly (sin \theta + 1) + (6l^2 sin \theta))} / {3(l + 2y)^2}]
  8. v^2 = [{g(2y^3 + 3ly^2 (tan \theta + 1) + (6l^2y sin \theta))} / {3(l + y)^2}]
  9. v^2 = [{g(2y^3 + 3ly^2 (sin \theta + 1) + (6l^3y sin \theta))} / {3(l + y)^2}]
  10. v^2 = [{g(2y^3 + 3ly (sin \theta + 1) + (6l^2y sin \theta))} / {3(l + y)^2}]

src

stemez-Mechanics

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Source and history

upstream_id / Original problem ID
9596
task_id
d39f3275-e791-51aa-b241-721d020a7110
task_key
default--test--9596
task_revision_id
1
config / split
default / test

Official source

initial import