A1.60 kg , horizontal, uniform tray is attached to a vertical ideal spring of force constant 200 n/m and a 285 g metal ball is in the tray. the spring is below the tray, so it can oscillate up-and-down. the tray is then pushed down 14.8 cm below its equilibrium point (call this point a) and released from rest.

Two horizontal plates with infinite length and width are separated by a distance h in the z direction. the bottom plate is moving at a velocity u. the incompressible fluid trapped between the plates is moving in the positive x-direction with the bottom plate. align gravity with positive z. assume that the flow is fully-developed and laminar. if the systems operates at steady state and the pressure gradient in x-direction can be ignored, do the following: 1. sketch your system 2. identify the coordinate system to be used. 3. show your coordinates and origin point on the sketch. list all your assumptions. 5. apply the continuity equation to your system. nts of navier stokes equations of choice to your system 7. solve the resulting differential equation to obtain the velocity profile within the system make sure to list your boundary conditions. check units of velocity 8. describe the velocity profile you obtain using engineering terminology. sketch that on the same sketch you provided in (1). 9. obtain the equation that describes the volumetric flow rate in the system. check the units.

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).

What is the momentum of a 2 kg object moving at 15 meters per second