Ablock with mass m = 7.50 kg is initially moving up the incline with speed v 0 and is increasing speed with acceleration a = 3 m/s 2 . the applied force f is horizontal. the coefficients of friction between the block and incline are µ s = 0.443 and µ k = 0.312. the angle of the incline is 25.0 degrees. what is the magnitude of the force f ?

F = 96.11 N

friction force =33.48 N

Explanation:

given,

mass of block M = 7.50 kg

acceleration = 3 m/s²

µ s = 0.443 and µ k = 0.312

angle with the horizontal = 25°

the component of force along the incline is F cos theta

normal reaction ,

N = 66.68+ .423 F

for the mass

0.906 F - 0.132 F - 31.09 - 20.8 = 22.5

0.774F = 74.39

F = 96.11 N

normal force ,

N = 66.68 + .423×96.11

  =107.33 N

friction force = .312 × N

                     = 0.312× 107.33

friction force =33.48 N

b.

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energy can be converted from one form to another, but it cannot be created or destroyed in ordinary chemical or physical means. do not understand what you are asking but hope this