Astudent was trying to find the relationship between mass and force. he placed four different masses on a table and pulled them using a spring scale. the table shows the different masses used in the experiment and the force required to pull each mass. the student concluded that more force was required to pull heavier objects. what comment would you make regarding his conclusion? a) no clear relation can be observed between mass and force from the data. b) there is a direct proportion between the mass and force listed in the table. c) gravity should have been taken into account while performing the experiment. d) there is an inverse proportion between the mass and force listed in the table.

mass (kg) force (n)

5 25

10 50

15 75

20 100

Apendulum has a mass of 1.5 kg and starts at a height of 0.4 m. if it is released from rest, how fast is it going when it reaches the lowest point of its path? acceleration due to gravity is g = 9.8 m/s2. a. 2.8 m/s b. 0 m/s c. 5.9 m/s d. 4.3 m/s

A1000-kg weather rocket is launched straight up. the rocket motor provides a constant acceleration for 16-s, then the motor stops. the rocket altitude 20-s after launch is 510-m. you can ignore any effects of air resistance. a. what was the rocket’s acceleration during the first 16-s? b. what is the rocket’s speed as it passes through a cloud 5100-m above the ground?

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).