A block with mass m1 = 8.8 kg is on an incline with an angle theta = 40 degree with respect to the horizontal. For the first question there is no friction, but for the rest of this problem the coefficients of friction are: mu_k = 0.38 and mu_s = 0.418. 1) 1) When there is no friction, what is the magnitude of the acceleration of the block? m/s^2 2) Now with friction, what is the magnitude of the acceleration of the block after it begins to slide down the plane? m/s^2 3) 3) To keep the mass from accelerating, a spring is attached. What is the minimum spring constant of the spring to keep the block from sliding if it extends x = 0.13 m from its unstretched length. N/m 4) Now a new block with mass m2 = 16.3 kg is attached to the first block. The new block is made of a different material and has a greater coefficient of static friction. What minimum value for the coefficient of static friction is needed between the new block and the plane to keep the system from accelerating?

Explanation:

1 ) When there is no friction , gravitational force on the block = mgsin40

acceleration = gsin40

= 6.3 m /s²

2 )the acceleration of the block after it begins to slide down the plane

Net force on the block

= mgsin40 - frictional force

mgsin40 - μk mgcosθ

= mg ( sin40 - .38 x cos40)

acceleration

= g ( sin40 - .38 x cos40)

9.8  ( sin40 - .38 x cos40)

= 9.8  ( .6428  - .29)

= 3.46 m /s²

3 ) Net downward force

= 8.8 x 3.46

= 30.448 N

To balance it

restoring force by spring

k x = 30.448

k = 30.448 / .13

= 234.21 N/m

4 )

for equilibrium

mgsin40 = μmgcos40

μ = sin40 /cos40

= tan40

= .84

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).