Suppose that during the launch of a rocket from the surface of the earth, an astronaut onboard the rocket stands on a bathroom scale and measures her weight to be 4.8 times her normal weight. Determine the acceleration of the rocket.

According to newton's second law, when the same force is applied to two objects of different masses, a the object with greater mass will experience a great acceleration, and the object with less mass will experience an even greater acceleration, b. the object with greater mass will experience a smaller acceleration, and the object with less mass will experience a greater acceleration, c. the object with greater mass will experience a greater acceleration, and the object with less mass will experience a smaller acceleration, d. the object with greater mass will experience a small acceleration, and the object with less mass will experience an even smaller acceleration.

Aparticle of mass m is projected with an initial velocity v0 in a direction making an angle α with the horizontal level ground as shown in the figure. the motion of the particle occurs under a uniform gravitational field g pointing downward. (a) write down the lagrangian of the system by using the cartesian coordinates (x, y). (b) is there any cyclic coordinate(s). if so, interpret it (them) physically. (c) find the euler-lagrange equations. find at least one constant of motion. (d) solve the differential equation in part (c) and obtain x and y coordinates of the projectile as a function of time. (e) construct the hamiltonian of the system, h, and write down the hamilton’s equations (canonical equations) of motion.

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).