•What power does a pump develop to lift 35 L of water per minute from a depth of 110 m? (One liter of water has a mass of 1.00 kg)

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).

The slope of a velocity time graph over any interval of time gives the during that interval

1. explain how you calculate the net force in any direction on the box. 2. suppose an upward force of 15n is added to the box. what will be the net vertical force on the box? 3. what force could be applied to the box to make the net force in the horizontal direction zero? explain. 4. suppose a force of 25n to the right is added to the box. what will be the net force to the right?