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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).

If a rock is thrown upward on the planet mars with a velocity of 12 m/s, its height (in meters) after t seconds is given by h = 12t − 1.86t2. (a) find the velocity of the rock after two seconds. m/s (b) find the velocity of the rock when t = a. 12−3.72a m/s (c) when will the rock hit the surface? (round your answer to one decimal place.) t = s (d) with what velocity will the rock hit the surface? m/s

You throw a small rock straight up from the edge of a highway bridge that crosses a river. the rock passes you on its way down, 7.00 s after it was thrown. what is the speed of the rock just before it reaches the water 28.0 m below the point where the rock left your hand? ignore air resistance.