hw12 - TAM212 Introductory Dynamics Spring 2005 S...

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Unformatted text preview: TAM212 Introductory Dynamics Spring 2005 S. Balachandar Homework 12 (2.108, 2.115) 2.108 1.5m t2 4m When the block is moved to the left, the spring is stretched by a length t2 . √ δ (t2 ) = 42 + 1 . 52 − 1 . 5 = 2.77 m Change in kinetic energy = Work done by linear spring 1 1 m v2 − v2 = − k δ (t2 )2 − δ (t1 )2 2 1 2 2 1 200 2 1 v − 72 = − 80 2.772 − 02 2 9.81 2 2 80(2.77)2 2 v2 = 49 − 9.81 = 18.8 200 v2 = 4.34 m/s 2.115 Work done by gravity = Change in kinetic energy 1 2 5(mg) = mv 2 v = 9.90 m/s 1 Also, we can ﬁnd the equations of motion as 1 2 x −5 20 1 = x 10 1 = 10 y = y y ∴ y (0) = 0 & y (0) = 1 10 Now, remember that the radius of curvature is given by, At B, y 1 = ρ (1 + y 2 )3/2 1 1 = ρ 10 mg N F = ma v2 N − mg = m ρ 9.902 N − 52(9.81) = 52 ρ N = 1020 Newtons Force on the ground = 1020 N v2 Acceleration = = 9.80 m/s2 Up ρ 2 ...
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hw12 - TAM212 Introductory Dynamics Spring 2005 S...

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