Aspring of constant 20 n/m has compressed a distance 8 m by a(n) 0.3 kg mass, then released. it skids over a frictional surface of length 2 m with a coefficient of friction 0.16, then compresses the second spring of constant 2 n/m. the acceleration of gravity is 9.8 m/s2.
how far will the second spring compress in order to bring the mass to a stop?

Explanation:

Here we will call:

1. : The energy when the first spring is compress

2. : The energy after the mass is liberated by the spring

3. : The energy before the second string catch the mass

4. : The energy when the second sping compressed

so, the law of the conservations of energy says that:

1.

2.

3.

where is the work of the friction.

1. equation 1 is equal to:

where K is the constant of the spring, x is the distance compressed, M is the mass and the velocity, so:

Solving for velocity, we get:

= 65.319 m/s

2. Now, equation 2 is equal to:

where M is the mass, the velocity in the situation 2, is the velocity in the situation 3, is the coefficient of the friction, N the normal force and d the distance, so:

Volving for , we get:

3. Finally, equation 3 is equal to:

where is the constant of the second spring and is the compress of the second spring, so:

solving for , we get:

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).

v 154.9= 0.9 length length s little over 154.9cm