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Physics 325 HW 4 Solutions - Physics 325 Fall 2010 Homework...

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Physics 325, Fall 2010 Homework Solutions #4 1. (15 points) 1) a) vector F ( x, y, z ) = α ( 9 x 2 y 2 z 2 xyz 3 ) ˆ x + α ( 6 x 3 yz x 2 z 3 ) ˆ y + α ( 3 x 3 y 2 3 x 2 z 2 y ) ˆ z vector ∇ × vector F = α ( 6 x 3 y 3 x 2 z 2 6 x 3 y + 3 x 2 z 2 ) ˆ x + α ( 9 x 2 y 2 6 xyz 2 9 x 2 y 2 + 6 xyz 2 ) ˆ y + α ( 18 x 2 yz 2 xz 3 18 x 2 yz + 2 xz 3 ) ˆ z = vector 0 b) Using the path described in class, U ( x, y, z ) = I x + I y + I z where I x = integraldisplay x 0 vector F ( x , 0 , 0) · ˆ xdx = 0 It’s zero because z = 0 along this path. I y = integraldisplay y 0 vector F ( x, y , 0) · ˆ ydy = 0 It’s zero because z = 0 along this path. I z = integraldisplay z 0 vector F ( x, y, z ) · ˆ zdz = α ( yx 2 z 3 3 x 3 y 2 z ) So U ( x, y, z ) is just I z plus a constant. But this constant needs to be zero in order for U (0 , 0 , 0) = 0. U ( x, y, z ) = α ( yx 2 z 3 3 x 3 y 2 z ) c) The bead feels the force F as well as any normal forces from the wire. The normal forces from the wire do no work, and so the change in potential energy of the bead is minus the change in kinetic energy. U (0 , 0 , 0) = 0 and U ( d, d, d ) = 2 αd 6 . So T (0 , 0 , 0) = 1 2 mv 2 0 and T ( d, d, d ) = T (0 , 0 , 0) U ( d, d, d ) + U (0 , 0 , 0) = 1 2 mv 2 0 + 2 αd 6 . Solving for v ( d, d, d ) = radicalbig 2 T ( d, d, d ) /m = radicalbig v 2 0 + 4 αd 6 /m .
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2. (15 points) A block of mass m slides down a tilted surface. The angle subtended by the surface and the horizontal (the positive x -axis) is θ . The force of gravity is directed downwards along the negative y -axis. The force of friction with the surface is assumed negligible. The force of air resistance acting on the sliding block is not negligible and it is equal to vector F
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