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 workenergy 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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 Spring '08
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 Energy, Force, Mass, Potential Energy, initially stationary block

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