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A block of mass m = 1kg, moving on a hor...

A block of mass m = 1kg, moving on a horizontal surface with speed `v_(1) = 2ms^(-1)` enters a rough patch ranging from `x=0.10m to x=2.01m` The retarding force `F_(r)` on the block in the range is inversely proportional to x over the range.
`F_(r) = -K/x` for 0.1 `ltxlt2.01`m =0 for `xlt0.1m` and `xgt2.01m` where k=0.5 J. What is the final kinetic energy and speed `v_(f)` of the block as it crosses this patch?

Text Solution

Verified by Experts

From work energy therorem `K_(f)-K_(i)=int Fdx`
`rArr K_(f)=K_(i)+ int_(0.1)^(2.01)((-k))/(x)dx =(1)/(2)mv_(i)^(2)-k1n(x):|_(0.1)^(2.01)`
`=(1)/(2)mv_(i)^(2)-k1n(2.01//0.1)=2-0.5 log_(e )(20.1)`
`=2-1.5=0.5 J " " v_(f)=sqrt(2K_(f )//m)=1ms^(-1)`
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