Answer:
a) 12.23 N
b) 2.2 m/s²
Explanation:
m = mass of the block = 3.9 kg
F = applied force = 27 N
θ = angle of the applied force with the horizontal = 40°
μ = Coefficient of kinetic friction = 0.22
[tex]F_{n}[/tex] = normal force
[tex]F_{g}[/tex] = weight of the block = mg
Along the vertical direction, force equation is given as
[tex]F_{n}[/tex] = F Sinθ + [tex]F_{g}[/tex]
[tex]F_{n}[/tex] = F Sinθ + mg
Kinetic frictional force is given as
f = μ [tex]F_{n}[/tex]
f = μ (F Sinθ + mg)
f = (0.22) (27 Sin40 + (3.9)(9.8))
f = 12.23 N
b)
Force equation along the horizontal direction is given as
F Cosθ - f = ma
27 Cos40 - 12.23 = 3.9 a
a = 2.2 m/s²
