Question

A block of mass 2 Kg moving on a horizontal surface with speed v = 2 m/s, enters a rough patch ranging from x =
0.10 m to x= 2.0 m. The retarding force F on the block is F = -k/x for 0.1 kinetic energy?

Answer 1

For 0.1<x<2.0

$\sf F=\frac{-k}{x}$

$\sf \implies \: ma = \frac{ - k}{x}$

$\sf \implies mv \frac{dv}{dx} = \frac{ - k}{x}$

$\sf \implies \huge[ \small \frac{m}{ - k} \huge] \small\int _{v_t}^{v_f} \small \: \: v.dv = \int_{0.1}^{2.0} \frac{ - 1}{x}$

$\sf \implies \huge[ \small \frac{m}{ - k} \huge] \huge[ \small \frac{ {v}^{2} }{2} \huge] \small_{v_t}^{v_f} = [ln \: x]_{0.1}^{2.0}$

$\sf2.0 \implies \: \frac{m {v}^{2}_{f} }{2} - \frac{m {v}^{2}_{t} }{2} = [ - k \: ]ln \: (2.0)$

$\\ \\ \sf \: \frac{m {v}^{2} _f}{2} = [ - k]ln \: (2.0) + \frac{m {v}^{2}_t }{2}$

$\sf \implies \: K.E_f = ( - 0.5)ln(2.0) + \frac{1}{2} {g}^{2}$

$\implies \: k.e \: \approx \: 4.9 \: j$