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 1 2.0 ms when she hits the ground. Take   2 g 10 ms . 

Answer 1

Explanations & Calculations

• By considering the conservation of mechanical energy of the child from the top of the rod to the bottom of it and taking the average frictional force to be $\small f$, the average frictional force could be calculated.

$\qquad\qquad \begin{aligned} \small E_p+E_k &=\small E_{p1}+E_{k1}+E_{external-work}\\ \small mgh+0&=\small 0+\frac{1}{2}mv^2+fs\\ \small f&=\small \frac{mgh-0.5mv^2}{s}\\ &=\small \frac{30kg.10ms^{-2}.3m-0.5.30kg.(2ms^{-1})^2}{3m}\\ &=\small \bold{280\,N} \end{aligned}$

• To understand in simple terms, the energy she had at the top is expended to give that speed at the bottom (in terms of kinetic energy) & to overcome the friction imposed on her by the pole.