A sphere of mass m = 2.5 kg can move across a horizontal, frictionless surface. Attached to the sphere is an ideal spring with spring constant k = 27 N/m. At time t = 0 the sphere is pulled aside from the equilibrium position, x=0, a distance d = 12 cm in the positive direction and released from rest. After this time, the system oscillates between x = ± d.

Determine the magnitude of the force, in newtons, required to initially displace the sphere d=11 cm from equilibrium

F = 2.97 N

Explanation:

Since the surface is frictionless, the mass of the sphere has no effect on the force required to pull the sphere. The force that the spring applies is as follows

Therefore, the magnitude of the force required to initially displace the sphere d = 11 cm is

most voltage drop calculators want the distance between the source and destination. then internally, they double it so both the source and return wires are calculated. but you need to know which distance the calculator uses so you're not half of what you should be or double what you should be.

most circuit breakers will not take a #6 directly, so the electrician will have to splice on a smaller pigtail. you also have to upsize the grounding wire when you upsized the ungrounded (hot) wire (see note in 250.122). this will preclude using most cable assemblies, as a cable with #6 copper will most likely only have a #10 grounding conductor. you need a #6 grounding conductor if the circuit is 20a. a #8 cable is no better, as it will have a #10 grounding conductor too. but most 20a breakers will take a #8 directly.

in your related post, you didn't say what type of load this is. if it is a light or heater, voltage drop won't be a big issue. if this is a motor, the current needs are huge at startup and the voltage drop can pose a problem. if this is a motor, knowing if the motor is under much load during startup will answer the voltage drop effect (a saw is easier to start than a compressor).