hw4_sol_s11

# hw4_sol_s11 - Phys 2101 Homework4 Solution Spring 11 These...

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Phys 2101 Homework4 Solution Spring 11 These solutions use the parameter values from the problems printed in the book, not those that appear in your personal homework assignment. The logic used to get to the answer is the same, however. 1. P.8-2. We use Eq. 7-12 for W g and Eq. 8-9 for U . (a) The displacement between the initial point and A is horizontal, so = 90.0° and 0 g W (since cos 90.0° = 0). (b) The displacement between the initial point and B has a vertical component of h /2 downward (same direction as F g ), so we obtain 25 11 (825 kg)(9.80 m/s )(42.0 m) 1.70 10 J 22 gg W F d mgh . (c) The displacement between the initial point and C has a vertical component of h downward (same direction as F g ), so we obtain 3.40 10 J W F d . (d) With the reference position at C , we obtain B U (e) Similarly, we find (825 kg)(9.80 m/s )(42.0 m) A U (f) All the answers are proportional to the mass of the object. If the mass is doubled, all answers are doubled. 2. P. 8-6. We use Eq. 7-12 for W g and Eq. 8-9 for U . (a) The displacement between the initial point and Q has a vertical component of h R downward (same direction as F g ), so (with h = 5 R ) we obtain 4 4(3.20 10 kg)(9.80 m/s )(0.12 m) 0.15 J W F d mgR . (b) The displacement between the initial point and the top of the loop has a vertical component of h 2 R downward (same direction as F g ), so (with h = 5 R ) we obtain

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22 3 3(3.20 10 kg)(9.80 m/s )(0.12 m) 0.11 J gg W F d mgR . (c) With y = h = 5 R , at P we find 5 5(3.20 10 kg)(9.80 m/s )(0.12 m) 0.19 J U mgR . (d) With y = R , at Q we have (3.20 10 kg)(9.80 m/s )(0.12 m) 0.038 J U . (e) With y = 2 R , at the top of the loop, we find 2 2(3.20 10 kg)(9.80 m/s )(0.12 m) 0.075 J U . (f) The new information ( ) v i 0 is not involved in any of the preceding computations; the above results are unchanged. 3. (a) We take the reference point for gravitational energy to be at the lowest point of the swing. Let be the angle measured from vertical. Then the height y of the pendulum ―bob‖ (the object at the end of the pendulum, which in this problem is the stone) is given by L (1 cos ) = y . Hence, the gravitational potential energy is mg y = mgL (1 cos ). When = 0º (the string at its lowest point) we are told that its speed is 8.0 m/s; its kinetic energy there is therefore 64 J (using Eq. 7-1). At = 60º its mechanical energy is E mech = 1 2 mv 2 + mgL (1 cos ) . Energy conservation (since there is no friction) requires that this be equal to 64 J. Solving for the speed, we find v = 5.0 m/s. (b) We now set the above expression again equal to 64 J (with being the unknown) but with zero speed (which gives the condition for the maximum point, or ―turning point‖ t hat it reaches). This leads to max = 79 . (c) As observed in our solution to part (a), the total mechanical energy is 64 J.
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## This note was uploaded on 05/09/2011 for the course PHYS 2101 taught by Professor Grouptest during the Spring '07 term at LSU.

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hw4_sol_s11 - Phys 2101 Homework4 Solution Spring 11 These...

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