problem07_74

# University Physics with Modern Physics with Mastering Physics (11th Edition)

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7.74: a) From either energy or force considerations, the speed before the block hits the spring is . s m 30 . 7 ) 1 . 53 cos ) 20 . 0 ( 1 . 53 )(sin m 00 . 4 )( s m 80 . 9 ( 2 ) cos (sin 2 2 k = ° - ° = - = θ μ gL v b) This does require energy considerations; the combined work done by gravity and friction is ) cos )(sin ( k θ θ d L mg - + , and the potential energy of the spring is 2 2 1 kd , where d is the maximum compression of the spring. This is a quadratic in d , which can be written as . 0 ) cos (sin 2 k 2 = - - - L d mg k d The factor multiplying 2 d is 1 m 504 . 4 - , and use of the quadratic formula gives m 06 . 1 = d . c) The easy thing to do here is to recognize that the presence of the spring determines d , but at the end of the motion the spring has no potential energy, and the distance below the starting point is determined solely by how much energy has been lost
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Unformatted text preview: to friction. If the block ends up a distance y below the starting point, then the block has moved a distance d L + down the incline and y d L-+ up the incline. The magnitude of the friction force is the same in both directions, θ mg cos k , and so the work done by friction is θ mg y d L cos ) 2 2 ( k-+-. This must be equal to the change in gravitational potential energy, which is θ mgy sin-. Equating these and solving for y gives . tan 2 ) ( cos sin cos 2 ) ( k k k k + + = + + = d L d L y Using the value of d found in part (b) and the given values for k and gives m 32 . 1 = y ....
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