Aball of mass 0.35 kg hangs straight down on a string of length 13 cm. it is then swung upward, keeping the string taut, until the string makes an angle of 70° with respect to the vertical. find the change in the gravitational potential energy of the earth-ball system.

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).

How much force was applied to an object if was moved 2 meters and the work done on the object was 8 joules? a. 0.25 n b. 4 n c. 6 n d. 16 n

When a bicycle coasts uphill, it moves slower and slower as it climbs. why? a. its kinetic energy is transforming into heat energy. b. its potential energy is transforming into kinetic energy. c. its kinetic energy is transforming into potential energy and heat energy. d. its potential energy is transforming into kinetic energy and heat energy.