Respuesta :
Answer:
(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
The solution values are ;
(a) The magnitude of the normal is 25.211 N
(b) The acceleration of the block, a ≈ 22.81 m/s²
The reason the above values are correct is given as follows:
The given parameter are;
The magnitude of the horizonal force, F₁ = 65 N
The magnitude of the force, F₂ = 12.4 N
The direction of F₂ = θ
The mass of the block, m = 3.1 kg
The coefficient of friction between the block and the surface, μk = 0.2
The direction of motion of the block = Right
Part (a) The normal force, [tex]F_N[/tex] in Newtons is given as follows;
The weight of the block, W = mg = 3.1 kg × 9.81 m/s² = 31.411 N
Given that the block is moving right, we have;
The normal reaction, [tex]F_N[/tex] = W - F₂·sin(θ)
When θ = 30, we have;
[tex]F_N[/tex] = 31.411 - 12.4 × sin(30°) = 25.211
The magnitude of the normal, [tex]F_N[/tex] = 25.211 N
(b) The frictional force acting on the block, [tex]F_f[/tex] = [tex]F_N[/tex] × μ
∴ [tex]F_f[/tex] = 25.211 × 0.2=5.0422
The sum of the horizontal forces, Fₓ = F₁ + F₂ × sin(θ) - [tex]F_f[/tex]
∴ Fₓ = 65 + 12.4 × cos(30°) - 5.0422 ≈ 70.7
Fₓ ≈ 70.7N
Force = Mass × Acceleration
[tex]Acceleration, \ a = \dfrac{Force, \, F }{Mass, \, m}[/tex]
Therefore;
[tex]a_x = \dfrac{F_x}{m} = \dfrac{70.7 \ N}{3.1 \ kg} \approx 22.81[/tex]
The acceleration of the block, a ≈ 22.81 m/s²
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