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

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Unformatted text preview: Physics 325, Fall 2010 Homework Solutions #4 1. (15 points) 1) a) vector F ( x, y, z ) = ( 9 x 2 y 2 z 2 xyz 3 ) x + ( 6 x 3 yz x 2 z 3 ) y + ( 3 x 3 y 2 3 x 2 z 2 y ) z vector vector F = ( 6 x 3 y 3 x 2 z 2 6 x 3 y + 3 x 2 z 2 ) x + ( 9 x 2 y 2 6 xyz 2 9 x 2 y 2 + 6 xyz 2 ) y + ( 18 x 2 yz 2 xz 3 18 x 2 yz + 2 xz 3 ) z = vector b) Using the path described in class, U ( x, y, z ) = I x + I y + I z where I x = integraldisplay x vector F ( x , , 0) xdx = 0 Its zero because z = 0 along this path. I y = integraldisplay y vector F ( x, y , 0) ydy = 0 Its zero because z = 0 along this path. I z = integraldisplay z vector F ( x, y, z ) zdz = ( 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. U ( x, y, z ) = ( 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 and U ( d, d, d ) = 2 d 6 . So T (0 , , 0) = 1 2 mv 2 and T ( d, d, d ) = T (0 , , 0) U ( d, d, d ) + U (0 , , 0) = 1 2 mv 2 + 2 d 6 . Solving for v ( d, d, d ) = radicalbig 2 T ( d, d, d ) /m = radicalbig v 2 + 4 d 6 /m . 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 . The force of gravity is directed downwards along....
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Physics 325 HW 4 Solutions - Physics 325, Fall 2010...

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