Answer to Question #116373 in Quantum Mechanics for Michael Mateus

Question

An 18.0 Kg is released on a 37° incline and accelerates down the incline at 0.270
m/s2. Find the friction force impeding its motion. How large is the coefficient of
kinetic friction?

Answer 1

Let us describe the forces that are applied to the body. Let the x-axis be directed along the inclined plane and y-axis be perpendicular to the plane. Therefore, the projection of forces on the y-axis is

$y: -mg\cos\alpha + N = 0,$ (1)

and on the x-axis

$x: mg\sin\alpha - F_{fr} = ma.$ (2)

From (2) we get

$F_{fr} = mg\sin\alpha - ma = m(g\sin\alpha-a) = 18.0\,\mathrm{kg}(9.8\,\mathrm{m/s^2} \sin37^\circ - 0.270\,\mathrm{m/s^2})= 101.3\,\mathrm{N}.$

We know that $F_{fr} = \mu N = \mu mg\cos\alpha,$ so

$\mu = \dfrac{F_{fr}}{mg\cos\alpha} = \dfrac{101.3\,\mathrm{N}}{18.0\,\mathrm{kg}\cdot9.8\,\mathrm{m/s^2}\cdot\cos37^\circ} = 0.719.$