qz4sol_7193f10

# qz4sol_7193f10 - Since m 2> m 1 we take our positive...

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TA: Tomoyuki Nakayama Monday, September 27, 2010 PHY 2048: Physic 1, Discussion Section7193 Quiz 4 (Homework Set #5) Name: UFID: Formula sheets are not allowed. Calculators are allowed. Do not store equations in your calculator. You need to show all of your work for full credit. ________________________________________________________________________________ The figure below right shows Atwood’s machine , in which two containers are connected by a cord of negligible mass passing over a frictionless pulley. At time t = 0, container 1 has mass 1.50kg and container 2 has mass 3.00 kg, but container 1 is losing mass through a leak at a constant rate of 0.250 kg/s. a) At what rate is the acceleration magnitude of the containers changing at t = 2.00 s?
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Unformatted text preview: Since m 2 > m 1 , we take our positive directions for m 1 and m 2 upward and downward respectively. In these coordinate systems, m 1 and m 2 have the same acceleration. Newton’s 2 nd law yields m 1 a = T – m 1 g, m2a = m 2 g – T Solving these equations for a, we get (m 1 + m 2 )a = (m 2 – m 1 )g ⇒ a = (m 2 – m 1 )g/(m 1 + m 2 ) = (1.5 + 0.25t)g/(4.5 – 0.25t) The rate of change in the acceleration is da/dt = 1.5g/(4.5 – 0.25t) 2 At t = 2 s, we have da/dt(t = 2 s) = 1.5g/(4.5-0.25×2) 2 = 0.919 m/s 3 b) When does the acceleration reach its maximum value? The acceleration of the blocks increases with time. Therefore, the maximum acceleration is achieved when the mass of block 1 becomes zero. 1.5 – 0.25t = 0 t = 6.00 s...
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## This note was uploaded on 01/25/2011 for the course PHY 2048 taught by Professor Field during the Summer '08 term at University of Florida.

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