Answer to Question #229301 in Mechanics | Relativity for Anand

Question

A child of mass 30 kg slides from rest down a wooden pole by a distance of 3.0 m. Calculate the average force of friction acting on the child if her speed is 2 m/s.
when she hits the ground.
Take g = 10m/s

Answer 1

start by calculating the acceleration using the third equation of motion

$v^2=u^2+2as$

from rest u=0

$2^2=0^2+2(a)(3)$

$4=6a, a=0.67m/s^2$

now $a=\frac{F_{net}}{m}=\frac{mg-friction force}{m}$

$ma=mg-friction force$

$friction force=ma-mg=30\times 0.67-30\times 10=-279.9N$

so the average friction force which is acting in the opposite direction is 279.9N