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

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Unformatted text preview: Physics 112 Homework #6 Solutions, Spring 2008 March 3, 2008 Problem 1 At any point in the path of the remote control cars 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 (- 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 = ~ 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 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 =- U = U i- U f = mg ( h i- h f ) = (4 . 80 kg ) ( 9 . 81 m/s 2 ) (25 . m ) = 1177 . 2 J = 1180 J b) What is the kinetic energy and the speed just before it strikes the ground? Work =- U = 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....
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This note was uploaded on 04/06/2008 for the course PHYS 1112 taught by Professor Leclair,a during the Spring '07 term at Cornell University (Engineering School).

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

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