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HW_6_sols_1

# HW_6_sols_1 - Physics 112 Homework#6 Solutions Spring 2008...

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Physics 112 Homework #6 Solutions, Spring 2008 March 3, 2008 Problem 1 At any point in the path of the remote control car’s motion around the cylinder, it must be accelerating toward the center at a radial = v 2 r in order to stay in a circular path. (a) At the bottom of the circle, the vertical forces are the weight downward and the normal force upward. X F y = ma y = N cyl on car - W earth on car N = N cyl on car m v 2 r = N - mg m v 2 r + mg = N N = 61 . 8 Newtons For part (b) the normal force from the cylinder must be pointing down (still against the car) while gravity also points down. X F y = ma y = - N cyl on car - W earth on car - m v 2 r = - N - mg m v 2 r = N + mg m v 2 r - mg = N N = 30 . 4 Newtons where N is the magnitude of the normal force; it points down ( - ˆ y direction) Problem 2 (a) 0: Work is given by Work = ´ ~ F · ~ dl ; (i) as the tension always points perpendicular to the direction of motion, ~ T · ~ dl = 0 at all points, so the work done by the tension is zero for any path along this circle. (ii) 0: Gravity is a conservative force, so the work done by gravity depends only on the start and end points and is equal to the negative difference in potential energies Work = ˆ ~ W · ~ dl = ( - mgh final ) - ( - mgh initial ) = - ( U ( h final ) - U ( h initial )) = mg ( h initial - h final ) = mg (0) (b) Motion along a semicircle from the lowest point to the highest point. 0: Again, the tension does no work because it is perpendicular to the direction of motion. -25.1 J: The gravitational work done from the bottom to the top is given by Work = - ( U ( h final ) - U ( h initial )) = - mg h ) = - (0 . 800 kg ) ( 9 . 81 m/s 2 ) (1 . 60 m * 2) = - 25 . 1136 J = - 25 . 1 J 1

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Problem 2’ A 4.80-kg watermelon is dropped from rest from the roof of a 25.0 m tall building. a) calculate the work done by gravity on the watermelon during its displacement from the roof to the ground: Work = - Work = - Δ U = U i - U f = mg ( h i - h f ) = (4 . 80 kg ) ( 9 . 81 m/s 2 ) (25 . 0 m ) = 1177 . 2 J = 1180 J b) What is the kinetic energy and the speed just before it strikes the ground? Work = - Δ U = Δ KE . Initially the watermelon was at rest, so its initial kinetic energy is zero; the final kinetic energy is just the change, and we use KE = 1 2 mv 2 to find the speed. KE = 1180 J = 1 2 mv 2 1180 J = 1 2 (4 . 80 kg ) v 2 v = s 1180 J 1 2 4 . 80 kg = 22 . 18 m/s We could also cut to the chase by starting from the potential energy; this shows that the mass of the object does not factor into the calculations. 1 2 mv 2 = mg ( h i - h f ) v 2 = 2 g ( h i - h f ) v = p 2 gh I = p 2 * 9 . 81 m/s 2 * 25 m = 22 . 15 m/s The difference between these two numbers comes from rounding 1177.2 J to 1180 J for significant figures in the first part.
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