HW3_Sol_F10 - HW#3: Solutions to Problems Chapter 8 1. (a)...

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HW#3: Solutions to Problems Chapter 8 1. (a) The only force that does work on the ball is the force of gravity; the force of the rod is perpendicular to the path of the ball and so does no work. In going from its initial position to the lowest point on its path, the ball moves vertically through a distance equal to the length L of the rod, so the work done by the force of gravity is 2 (0.341 kg)(9.80 m/s )(0.452 m) 1.51 J Wm g L == = . (b) In going from its initial position to the highest point on its path, the ball moves vertically through a distance equal to L , but this time the displacement is upward, opposite the direction of the force of gravity. The work done by the force of gravity is 2 (0.341 kg)(9.80 m/s )(0.452 m) 1.51 J. g L =− (c) The final position of the ball is at the same height as its initial position. The displacement is horizontal, perpendicular to the force of gravity. The force of gravity does no work during this displacement. (d) The force of gravity is conservative. The change in the gravitational potential energy of the ball-Earth system is the negative of the work done by gravity: 2 (0.341 kg)(9.80 m/s )(0.452 m) 1.51 J Um g L ∆= = = as the ball goes to the lowest point. (e) Continuing this line of reasoning, we find 2 (0.341 kg)(9.80 m/s )(0.452 m) 1.51 J g L + = = as it goes to the highest point. (f) Continuing this line of reasoning, we have U = 0 as it goes to the point at the same height. (g) The change in the gravitational potential energy depends only on the initial and final positions of the ball, not on its speed anywhere. The change in the potential energy is the same since the initial and final positions are the same. 2. We use Eq. 8-17, representing the conservation of mechanical energy (which neglects friction and other dissipative effects).
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(a) In Problem 8-2, we found U A = mgh (with the reference position at C ). Referring again to Fig. 8-33, we see that this is the same as U 0 which implies that K A = K 0 and thus that v A = v 0 = 17.0 m/s. (b) In the solution to Problem 8-2, we also found Um g h B = 2. In this case, we have KU KU mv mgh mv mg h BB B 00 0 22 1 2 1 + = + += + F H G I K J which leads to 2 0 (17.0 m/s) (9.80 m/s )(42.0 m) 26.5 m/s. B vv g h =+ = + = (c) Similarly, 2 0 2 (17.0 m/s) 2(9.80 m/s )(42.0 m) 33.4 m/s. C g h =+= + = (d) To find the “final” height, we set K f = 0. In this case, we have mv mgh mgh ff f 0 2 1 2 0 + = + + which yields 2 2 0 2 (1 7 . 0 m/ s) 42.0 m 56.7 m. ( 9 . 8 0 m / s ) f v hh g =+ = + = (e) It is evident that the above results do not depend on mass. Thus, a different mass for the coaster must lead to the same results. 3. We convert to SI units and choose upward as the + y direction. Also, the relaxed position of the top end of the spring is the origin, so the initial compression of the spring (defining an equilibrium situation between the spring force and the force of gravity) is y 0 = –0.100 m and the additional compression brings it to the position y 1 = –0.400 m. (a) When the stone is in the equilibrium ( a = 0) position, Newton's second law becomes
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net spring G Fm a g k = −= −− = 0 0100 8 00 9 8 0 (. )( .) ( . ) where Hooke's law (Eq. 7-21) has been used. This leads to a spring constant equal to 784 N/m k = .
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This note was uploaded on 01/26/2011 for the course PHYSICS 2101 taught by Professor Dowling during the Spring '10 term at LSU.

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HW3_Sol_F10 - HW#3: Solutions to Problems Chapter 8 1. (a)...

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