A horizontal force, F1 = 65 N, and a force, F2 = 12.4 N acting at an angle of θ to the horizontal, are applied to a block of mass m = 3.1 kg. The coefficient of kinetic friction between the block and the surface is μk = 0.2. The block is moving to the right.

Part (a) Solve numerically for the magnitude of the normal force, FN in Newtons, that acts on the block if θ = 30°. t Part (b) Solve numerically for the magnitude of acceleration of the block, a in m/s2, if θ = 30°.

Question

Kamari

Physics

1 May, 19:45

A horizontal force, F1 = 65 N, and a force, F2 = 12.4 N acting at an angle of θ to the horizontal, are applied to a block of mass m = 3.1 kg. The coefficient of kinetic friction between the block and the surface is μk = 0.2. The block is moving to the right.

Part (a) Solve numerically for the magnitude of the normal force, FN in Newtons, that acts on the block if θ = 30°. t Part (b) Solve numerically for the magnitude of acceleration of the block, a in m/s2, if θ = 30°.

+1

Answers (1)

Know the Answer?

Sariah Bernard · Answer 1

(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