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Two blocks are initially at rest on frictionless surfaces and are connected by a string that passes over a fri?

Two blocks are initially at rest on frictionless surfaces and are connected by a string that passes over a fri? Topic: On rest
June 21, 2019 / By Rick
Question: Two blocks are initially at rest on frictionless surfaces and are connected by a string that passes over a frictionless pulley in the figure below, where θ1 = 29.5°, θ2 = 44.5°, m = 1.21 kg. Find the tension in the string.
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Best Answers: Two blocks are initially at rest on frictionless surfaces and are connected by a string that passes over a fri?

Originally Answered: Two blocks are initially at rest on frictionless surfaces and are connected by a string that passes over a fri?
Logic would tell U that after being released the heavier mass would fall and the lighter one would rise giving the system "motion" which would give it kinetic energy (KE); so KE must be >0. There are only two choices with KE>0 can U now choose the correct one? Use logic again =>Conservation of ALL mechanical energy must hold. Comment: IF the pulley is FRICTIONLESS what's the point of having it be a "pulley" since it won't ROTATE under these conditions - a frictionless stick or rod would be just as good as a pulley in this case...or is the pulley to be FRICTIONLESS ONLY in its bearing? Ask UR teacher :>)
Originally Answered: Two blocks are initially at rest on frictionless surfaces and are connected by a string that passes over a fri?
2 gadgets are appropriate by a basic string passing over a basic frictionless pulley, putting from a ceiling. the article of mass 3.00kg is released from resting on the floor. the 2d merchandise of 5.00kg, putting on the different aspect of the pulley, begins to bypass from its unique acceptable of four.00 m above the floor. utilising the isolated device type locate the utmost acceptable which the three kg merchandise rises. because the 5 kg merchandise speeds up down 4.00 meters, the three kg merchandise speeds up up 4.00 meters. Then the three kg merchandise decelerates until eventually its very last speed = 0 m/s the internet stress on both gadgets = Weight of 5 kg merchandise – Weight of three kg merchandise = 5 * 9.8 – 3 * 9.8 = 19.6 N information superhighway stress = complete mass * acceleration 19.6 = 8 * a a = 19.6 ÷ 8 = 2.40 5 m/s^2 the three kg merchandise speeds up at 2.40 5 m/s^2 for a distance of four.00 meters. very last speed^2 – initial speed^2 = 2 * a * d initial speed = 0 a = 2.40 5 m/s^2 d = 4 m very last speed^2 = 2 * 2.40 5 * 4 very last speed = (2 * 2.40 5 * 4)^0.5 = 19.6^0.5 Then the three kg merchandise decelerates at 9.8 m/s until eventually its speed = 0 m/s. very last speed^2 – initial speed^2 = 2 * a * d very last speed^2 = 0 initial speed^2 = 19.6 a = -9.8 0 – 19.6 = 2 * -9.8 * d d = a million meter complete distance the three kg merchandise moved up = 4 + a million = 5 meters the answer is 5.00 m.

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