7 A stationary gun of mass 4.8kg mounted on wheels is in a battle field. It fires a bullet of mass 85g to the right with a velocity of v ms. The bullet moves horizontally and hits a stationary
wooden block of mass 5.6kg. It gets imbedded in the block. The bullet-block system then moves
a distance 10.78m and comes to rest in a time of 4.5 seconds.
ms
858
5.6kg
7.1 Calculate the combined speed of the bullet block system
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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).

Tall pacific coast redwood trees can reach heights of about 100 m. if air drag is negligibly small, how fast is a sequola cone moving when it reaches the ground f it dropped from the top of a 100 m tree?

How is zig zagging a form of acceleration?