1) A vendor pulls a 300N cart filled with pop cases so that it rolls at a constant velocity along a level floor where u=0.2. What is the Force applied by the vendor?

Which of the following represents an upright image? a. -do b. +m c. -m d. +do

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 x-coordinate of a particle in curvilinear motion is given by x = 3.1t3 - 4.9t where x is in feet and t is in seconds. the y-component of acceleration in feet per second squared is given by ay = 2.3t. if the particle has y-components y = 0 and vy = 5.0 ft/sec when t = 0, find the magnitudes of the velocity v and acceleration a when t = 1.8 sec. sketch the path for the first 1.8 seconds of motion, and show the velocity and acceleration vectors for t = 1.8 sec. answers: v = ft/sec a = ft/sec2