A 26.0-kg block is initially at rest on a horizontal surface. A horizontal force of 72.0 N is required to set the block in motion, after which a horizontal force of 58.0 N is required to keep the block moving with constant speed. (a) Find the coefficient of static friction between the block and the surface. (b) Find the coefficient of kinetic friction between the block and the surface.

a)  μ = 0.283 , b)    μ = 0.228

Explanation:

For this exercise we must use the translational equilibrium conditions in each exercise

Y Axis  

       N- W = 0

       N = W = mg

X axis

       F -fr = 0

       F = fr

friction force has an expression

        Fr = μ N

 we substitute

       F = μ mg

        μ = F / mg

  let's calculate

a) μ = 72.0 / 26.0 9.8      static

      μ = 0.283

b) μ = 58/26 9.8             dinamicic

      μ = 0.228

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.

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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).

the current is 0.6

voltage ÷ resistance = current

so 12 ÷ 20 = 0.6