Question

In a children - park an inclined plane is constructed with angle of incline ${45}^{\circ }$ in the middle part (figure 6-E1). Find the acceleration of a boy sliding on it, if the friction coefficient between the cloth of the o and the incline is 0.6 and g $=10m{s}^2$

Answer 1

Let m = mass of child, R- mg cos ${45}^{\circ }=0$
$R=mg\cos {45}^{\circ }=\frac{mg}{\sqrt{2}}$
Net force acting on the boy due to which it slides down
$=mgsin{45}^{\circ }-\mu R=mg\cos {45}^{\circ }-\mu mg\cos {45}^{\circ }$

$=m\times 10\times \left(\frac{1}{\sqrt{2}}\right)-0.6\times m\times 10\times \left(\frac{1}{\sqrt{2}}\right)=m(5\sqrt{2}-3\sqrt{2})=m\times 2\times \sqrt{2}\text{Acceleration}=\frac{\text{Force}}{mass}=\frac{m\left(\left(2\sqrt{2}\right)\right)}{m}=2\sqrt{2}m/s^2$