A horizontal force, F1 = 95 N, and a force, F2 = 18.9 N acting at an angle of θ to the horizontal, are applied to a block of mass m = 2.9 kg. The coefficient of kinetic friction between the block and the surface is μk = 0.2. The block is moving to the right. Calculate the normal force when the angle is 30. and acceleration when the angle is 30.

(a) FN = 24.18 N

(b) a = 22.87 m/s²

Explanation:

Newton's second law of the  block:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Forces acting on the box

We define the x-axis in the direction parallel to the movement of the block on the surface   and the y-axis in the direction perpendicular to it.

F₁ : Horizontal force

F₂ : acting at an angle of θ to the horizontal,

W: Weight of the block  : In vertical direction

FN : Normal force : perpendicular to the direction the surface

fk : Friction force: parallel to the direction to the surface

Known data

m =3.1 kg : mass of the  block

F₁ = 65 N,  horizontal force

F₂ = 12.4 N acting at an angle of θ to the horizontal

θ = 30° angle θ of F₂ with respect to the horizontal

μk = 0.2 : coefficient of kinetic friction between the block and the surface

g = 9.8 m/s² : acceleration due to gravity

Calculated of the weight  of the block

W= m*g  = (3.1 kg)*(9.8 m/s²) = 30.38 N

x-y F₂ components

F₂x = F₂cos θ= (12.4)*cos(30)° = 10.74 N

F₂y = F₂sin θ= (12.4)*sin(30)° = 6.2 N

a)Calculated of the Normal force  (FN)

We apply the formula (1)

∑Fy = m*ay    ay = 0

FN+6.2-30.38 = 0

FN = -6.2+30.38

FN = 24.18 N

Calculated of the Friction force:

fk=μk*N=  0.2* 24.18 N = 4.836 N

b) We apply the formula (1) to calculated acceleration of the block:

∑Fx = m*ax ,  ax= a  : acceleration of the block

F₁ + F₂x -fk = ( m)*a

65 N + 10.74 -4.836 = ( 3.1)*a

70.904 = ( 3.1)*a

a = (70.904 ) / ( 3.1)

a = 22.87 m/s²

does anyone know or what?