Velocity vs Time What is the acceleration of the car at segment C?
O 30 m/s2
0 -30 m/s2
O 40 m/s2
0-40 m/s2
50
40
В
E
30
Velocity (m/s)
N
20
A
10
D
0
0 1 2 3 4 5 6 7 8 9 10
Time (s)

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).

M1 = 2.8 kg, m2 = 6.72 kg, m3 = 11.2 kg, byas in .is ,is .toof m3 it 0.91 m. (in m/s)

A10-kg piece of aluminum sits at the bottom of a lake, right next to a 10-kg piece of lead, which is much denser than aluminum. which one has the greater buoyant force on it? explain. b - the aluminum a) both have the same buoyant force. b) the aluminum c) the lead d) it cannot be determined without knowing their volumes.