Question

In a children-park an inclined plane is constructed with an angle of incline 45° in the middle part (figure 6−E1). Find the acceleration of boy sliding on it if the friction coefficient between the cloth of the boy and the incline is 0.6 and g = 19 m/s².

Figure

Answer 1

Let m be the mass of the boy.



From the above diagram:
R − mg cos 45° = 0
$R=mg\cos 45°=\frac{mg}{\sqrt{2}}\left(1\right)$
Net force acting on the boy, making him slide down
= mg sin 45° − μR
= mg sin 45° − μmg cos 45°
$$\begin{gathered} =m\times 10\times \left(\frac{1}{\sqrt{2}}\right)-0.6\times m\times 10\times \left(\frac{1}{\sqrt{2}}\right) \\ =m\left(5\sqrt{2}-3\sqrt{2}\right)=m\times 2\times \sqrt{2} \end{gathered}$$
The acceleration of the boy $=\frac{\mathrm{Force}}{\mathrm{Mass}}$
$$\begin{gathered} =\frac{m\left(2\sqrt{2}\right)}{m} \\ =2\sqrt{2}\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$