Ahorizontal 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°.

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, 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, = W - F₂·sin(θ)

When θ = 30, we have;

= 31.411 - 12.4 × sin(30°) = 25.211

The magnitude of the normal, = 25.211 N

(b) The frictional force acting on the block, = × μ

∴ = 25.211 × 0.2=5.0422

The sum of the horizontal forces, Fₓ = F₁ + F₂ × sin(θ) -

∴ Fₓ = 65 + 12.4 × cos(30°) - 5.0422 ≈ 70.7

Fₓ ≈ 70.7N

Force = Mass × Acceleration

Therefore;

The acceleration of the block, a ≈ 22.81 m/s²

Learn more about friction here:

link

link

link