Problem 7-62 Energy cons. with friction - Part 7

A pen of mass \(0.057 \;kg\) is sliding across a horizontal desk. In sliding \(25 \;cm\), its speed decreases to \(5.7 \;cm/s\). What was its initial speed if the force of kinetic friction exerted on the pen by the desk is \(0.15\; N\) in magnitude? Use conservation of energy.

Accumulated Solution

\(E_K = (1/2)mv^2 \\ \text{Work against friction} = F_K \Delta x \\ (1/2)mv_0{^2} = (1/2)mv^2 + F_K \Delta x \)

\((1/2)mv_0{^2} = (1/2)mv^2 + F_K \Delta x \\ v_0 =(v^2 + 2F_K \Delta x/m)^{1/2} \\ = [(0.057 m/s)^2 + 2((0.15 \;N)(0.25 \;N)/(0.057\; kg)]^{1/2} = 1.1\; m/s\)

You have completed this problem.