A truck of mass 200kg rests on an inclined plane hindered from rolling down the surface by a storing sprint whose force constant is 10^6N/M. neglecting friction, how much work does the truck do on the spring?

Modify the equation fδt=δ (mv) to find the force of friction.

1134567 a jet with mass m = 1.1 ă— 105 kg jet accelerates down the runway for takeoff at 2 m/s2. 1) what is the net horizontal force on the airplane as it accelerates for takeoff? 2.2*10^5 n your submissions: 2.2*10^5 computed value: 220000submitted: saturday, january 26 at 2: 41 pm feedback: correct! 2) what is the net vertical force on the airplane as it accelerates for takeoff? 0 n your submissions: 0 computed value: 0submitted: saturday, january 26 at 2: 41 pm feedback: correct! 3) once off the ground, the plane climbs upward for 20 seconds. during this time, the vertical speed increases from zero to 21 m/s, while the horizontal speed increases from 80 m/s to 95 m/s. what is the net horizontal force on the airplane as it climbs upward? n 4) what is the net vertical force on the airplane as it climbs upward? n 5) after reaching cruising altitude, the plane levels off, keeping the horizontal speed constant, but smoothly reducing the vertical speed to zero, in 13 seconds. what is the net horizontal force on the airplane as it levels off? n 6) what is the net vertical force on the airplane as it levels off?

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).