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

radiation warms things up temp rises convection currents air circ windy weather

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.