Assuming weightless pulleys and 100% efficiency, what is the minimum input force required to lift a 120 n weight using a single fixed pulley?

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).

In all trials, the magnitude of the final velocity for g1 + g2 was less than the magnitude of any initial velocity. as mass increased, what happened to the velocity? the velocity decreased. the velocity increased. the velocity of g1 + g2 could not be measured. the velocity was not affected by the mass increase.

Ihave a density of 1.61g/cm^3 and a mass of 28g. find the missing value