Answer:
The magnitude of the kinetic frictional force acting on the child is 162.93 N
Explanation:
Given;
mass of the child, m = 35 kg
height of the slide, h = 3.8 m
length of the slide, d = 8.0 m
The change in thermal energy associated with the kinetic frictional force is calculated as follows;
![\Delta E_{th} + \Delta K.E + \Delta U = 0\\\\\Delta E_{th} + (\frac{1}{2} mv_f^2 - \frac{1}{2} mv_i^2) + (mgh_f - mgh_i) =0\\\\since \ the \ speed \ is \ constant, \ v_f = v_i \ and \ \Delta K.E = 0\\\\Also, \ final \ height \ , h _f= 0\\\\\Delta E_{th} - mgh_i = 0\\\\\Delta E_{th} = mgh_i\\\\\Delta E_{th} = 35 \times9.8 \times 3.8\\\\\Delta E_{th} = 1303.4 \ J](https://tex.z-dn.net/?f=%5CDelta%20E_%7Bth%7D%20%2B%20%5CDelta%20K.E%20%2B%20%5CDelta%20U%20%3D%200%5C%5C%5C%5C%5CDelta%20E_%7Bth%7D%20%2B%20%28%5Cfrac%7B1%7D%7B2%7D%20mv_f%5E2%20-%20%5Cfrac%7B1%7D%7B2%7D%20mv_i%5E2%29%20%2B%20%28mgh_f%20-%20mgh_i%29%20%3D0%5C%5C%5C%5Csince%20%5C%20the%20%5C%20speed%20%5C%20is%20%5C%20constant%2C%20%5C%20v_f%20%3D%20v_i%20%5C%20and%20%5C%20%5CDelta%20K.E%20%3D%200%5C%5C%5C%5CAlso%2C%20%5C%20final%20%5C%20height%20%5C%20%2C%20h%20_f%3D%200%5C%5C%5C%5C%5CDelta%20E_%7Bth%7D%20-%20mgh_i%20%3D%200%5C%5C%5C%5C%5CDelta%20E_%7Bth%7D%20%3D%20mgh_i%5C%5C%5C%5C%5CDelta%20E_%7Bth%7D%20%3D%2035%20%5Ctimes9.8%20%5Ctimes%203.8%5C%5C%5C%5C%5CDelta%20E_%7Bth%7D%20%3D%201303.4%20%5C%20J)
The magnitude of the kinetic frictional force that produced this thermal energy is calculated from the work done by frictional force;
![\Delta E_{th} = F \times d\\\\F = \frac{\Delta E_{th} }{d} \\\\F = \frac{1303.4}{8} \\\\F = 162.93 \ N](https://tex.z-dn.net/?f=%5CDelta%20E_%7Bth%7D%20%3D%20F%20%5Ctimes%20d%5C%5C%5C%5CF%20%3D%20%5Cfrac%7B%5CDelta%20E_%7Bth%7D%20%7D%7Bd%7D%20%5C%5C%5C%5CF%20%3D%20%5Cfrac%7B1303.4%7D%7B8%7D%20%5C%5C%5C%5CF%20%3D%20162.93%20%5C%20N)
Therefore, the magnitude of the kinetic frictional force acting on the child is 162.93 N