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

depends on what you mean by "group." for order, it would be coleoptera (beetles). for class, it would be insecta (insects). for phylum, it would be the arthropoda (animals with jointed exoskeletons).

explanation: