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qz3sol_3705s11

# qz3sol_3705s11 - Newton’s 2 nd law yields M-g/6 = T –...

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TA: Tomoyuki Nakayama Monday, February 14, 2011 PHY 2048: Physic 1, Discussion Section 3705 Quiz 3 (Homework Set #5) Name: UFID: Formula sheets are not allowed. Do not store equations in your calculator. You have to solve problems on your own; memorizing final algebraic expressions from homework assignments and just plugging numbers into them will not give you full credit. Leave all your work. ________________________________________________________________________________ A cord is used to vertically lower an initially stationary block of mass M = 15.0 kg at a constant downward acceleration of g /6. a) When the block has fallen a distance d = 2.00 m, what is the work done by the cord's force on the block? We take our positive direction upward. Then the downward acceleration is expressed as –g/6.
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Unformatted text preview: Newton’s 2 nd law yields M(-g/6) = T – Mg ⇒ T = 5Mg/6 = 122.5 N In our coordinate system, the displacement of the block is –d. Thus the work done by the tension is W T = T(-d) = -245 J Note that the sign of work does not depend on your coordinate system. It is always negative if force and displacement are directed in opposite directions. b) When the block has fallen a distance d = 2.00 m, what is the speed of the block? The net force exerted on the block is F net = M(-g/6) The initial kinetic energy of the block is zero. The work-energy theorem yields ∆ K = W net (1/2)Mv 2 – 0 = M(-g/6)(-d) v = √ (2(g/6)d) = 2.56 m/s...
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