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

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