In a children's park, an inclined plane is built with an angle of incline of 45^ in its middle section, as shown in the figure. Determine the acceleration of a boy sliding on this part if the friction coefficient between the boy's clothing and the incline is 0.6 . Take g = 10 m/s ^2 .

Step-by-step solution

The forces acting on the boy along the inclined plane are the component of gravity downwards and the frictional force upwards. The equation of motion for the boy sliding down the incline is given by: ma = mg - f_k where f_k = N is the force of kinetic friction and N = mg is the normal reaction. Substituting f_k into the equation, we get: ma = mg - mg a = g( - ) Given values are g = 10 m/s ^2 , = 45^ , and = 0.6 . Substituting these values: a = 10 ( 45^ - 0.6 45^ ) a = 10 ( 1 2 - 0.6 1 2 ) a = 10 2 (1 - 0.6) a = 10