Answer to Question #123766 in Physics for Dr. Horus

Question

A body of mass 2kg lie on a rough plane which is inclined at 23° to the horizontal. A force of 20N is applied to the body parallel to and up the plane. If the body accelerate at 1.5m/s^2. Calculate the coefficient of friction between the body and the plane.

Answer 1

Let's apply the Newton's Second Law of Motion in projections on axis $x$- and $y$:

$\sum F_x = ma_x,$ $\sum F_y = ma_y = 0,$ $F_{appl} - mgsin \theta - F_{fr} = ma, (1)$ $N - mgcos \theta = 0, (2)$ $F_{appl} - mgsin \theta - \mu N = ma,$ $N = mgcos \theta,$ $F_{appl} - mgsin \theta - \mu mgcos \theta = ma.$

From this equation we can find the coefficient of friction between the body and the plane:

$\mu = \dfrac{F_{appl} - mgsin \theta - ma}{mgcos \theta},$ $\mu = \dfrac{20 \ N - 2 \ kg \cdot 9.8 \ \dfrac{m}{s^2} \cdot sin23^{\circ} - 2 \ kg \cdot 1.5 \ \dfrac{m}{s^2}}{2 \ kg \cdot 9.8 \ \dfrac{m}{s^2} \cdot cos23^{\circ}} = 0.52.$

Answer:

$\mu = 0.52.$