When an inclined plane is used, a. the mechanical advantage increases as the height increases. b. a smaller force is exerted over a longer distance. c. a negative mechanical advantage can result. d. a greater force is exerted over a shorter distance

Part f - example: finding two forces (part i) two dimensional dynamics often involves solving for two unknown quantities in two separate equations describing the total force. the block in (figure 1) has a mass m=10kg and is being pulled by a force f on a table with coefficient of static friction îľs=0.3. four forces act on it: the applied force f (directed î¸=30â above the horizontal). the force of gravity fg=mg (directly down, where g=9.8m/s2). the normal force n (directly up). the force of static friction fs (directly left, opposing any potential motion). if we want to find the size of the force necessary to just barely overcome static friction (in which case fs=îľsn), we use the condition that the sum of the forces in both directions must be 0. using some basic trigonometry, we can write this condition out for the forces in both the horizontal and vertical directions, respectively, as: fcosî¸â’îľsn=0 fsinî¸+nâ’mg=0 in order to find the magnitude of force f, we have to solve a system of two equations with both f and the normal force n unknown. use the methods we have learned to find an expression for f in terms of m, g, î¸, and îľs (no n).

Awater slide of length l has a vertical drop of h. abby's mass is m. an average friction force of magnitude f opposes her motion. she starts down the slide at initial speed vi. use work-energy ideas to develop an expression for her speed at the bottom of the slide. then evaluate your result using unit analysis and limiting case analysis. express your answer in terms of the variables h, m, l, vi, f and appropriate constants. vf v f

Abasketball star covers 2.65 m horizontally in a jump to dunk the ball. his motion through space can be modeled precisely as that of a particle at his center of mass. his center of mass is at elevation 1.02 m when he leaves the floor. it reaches a maximum height of 1.90 m above the floor and is at elevation 0.910 m when he touches down again. (a) determine his time of flight (his "hang time"). (b) determine his horizontal velocity at the instant of takeoff. (c) determine his vertical velocity at the instant of takeoff. (d) determine his takeoff angle. (e) for comparison, determine the hang time of a whitetail deer making a jump with center-of-mass elevations yi = 1.20 m, ymax = 2.45 m, and yf = 0.750 m.