Question

A block 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 ?

Answer 1

Answer:

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 ,

$N = mg cos \theta + F sin \theta$

$N = 7.5\times 9.81\times cos25^0 + F sin 25^0$

N = 66.68+ .423 F

for the mass

$F cos \theta - mg sin \theta - \mu_k\ N = ma$

$F cos 25^0 - 7.5\times 9.81\times sin 25^0 - 0.312\times ( 66.68+ .423 F) = 7.5\times 3$

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