Two blocks are attached by a thin inextensible string over a frictionless, massless pulley. there is a frictional force between the block on the incline and the incline. the acceleration of gravity is 9.8 m/s 2 . 15 kg 14 kg µ t 1.2 m/s 0 m/s 2 34◦ calculate the magnitude of the frictional force acting on the block on the incline. answer in units of n.

The force of friction fk = 146.24 N

Explanation:

a = 9.8 m/s²

m₁ = 15 kg, m₂ = 14 kg

μ = 1.2  Ф = 34 °  

Fₓ = T - m₁ * sin (Ф ) - f = m₁ * a

Fₙ = N - m₁ * g * cos (Ф) = m₁ * 0

N - m₁ * g * cos (Ф) = 0

N = m₁ * g * cos (Ф) = 15 kg * 9.8 m/s² * cos (34°)

N = 121.86 N

Now to find the force ofk = 146.24 N f friction using the coefficient of μ

fk = N * μ

fk = 121.86 * 1.2

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