Question

A 120 kg piano is at the top of a ramp that makes an angle of 30° with the level ground. As the piano slides down the ramp it experiences a force of friction. The coefficient of kinetic friction is 0.10.

Answer 1

$\textsf{From the diagram,}\\ F_r=\textsf{Kinetic force of friction}\\ \mu=\textsf{coefficient of kinetic friction}\\ N_r=\textsf{Normal force on the piano}\\ m=\textsf{mass of the piano}\\ W=\textsf{weight of the piano}\\ g=\textsf{acceleration due to gravity}\\ \theta=\textsf{angle of inclination of the ramp}\\$

(ii) The normal force,

$\hspace{0.4cm}\begin{aligned}N_r&=mgcos\theta\\ &=120kg × 9.81ms^{-2} × cos(30°)\\ &=1019.49N\end{aligned}$ (iii) The kinetic force of friction,

$\hspace{0.4cm}\begin{aligned}F_r&=\mu N_r\\ &=0.1×1019.49N\\ &=101.95N\end{aligned}$

(iii) According to Newton's second law,

$\hspace{0.4cm}\begin{aligned}&F-F_r=ma\\ &mgsin\theta-F_r=ma\\ &120kg×9.81ms^{-2}×sin(30°)-101.95N=120kg×a\\ &a=\frac{486.65N}{120kg}\\ &a=4.06ms^{-2}\end{aligned}\\$