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Bharat Somnath Bakshi
Asked: 2022-11-07T07:54:12+05:30 In:

A block of mass `m=1kg` moving on a horizontal surface with speed `v_(i)=2ms^(-1)` enters a rough patch ranging from `x0.10m to x=2.01m`. The retarding force `F_(r)` on the block in this range ins inversely proportional to x over this range
`F_(r)=-(k)/(x) fo r 0.1 lt xlt 2.01m`
`=0` for `lt 0.1m` and `x gt 2.01m` where `k=0.5J`. What is the final K.E. and speed `v_(f)` of the block as it crosses the patch?

A block of mass `m=1kg` moving on a horizontal surface with speed `v_(i)=2ms^(-1)` enters a rough patch ranging from `x0.10m to x=2.01m`. The retarding force `F_(r)` on the block in this range ins inversely proportional to x over this range
`F_(r)=-(k)/(x) fo r 0.1 lt xlt 2.01m`
`=0` for `lt 0.1m` and `x gt 2.01m` where `k=0.5J`. What is the final K.E. and speed `v_(f)` of the block as it crosses the patch?
Parabola
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  1. Here, `m=1kg, v_(i)=2ms^(-1),k=0.5J`
    Initial K.E., `K_(i)=(1)/(2)mv_(i)^(2)=(1)/(2)xx1(2)^(2)=2J`
    Work done against friction
    `W=int_(x=0.1m)^(x=2.01m)F_(r), dx=int_(x=0.1m)^(x=2.01m)-(k)/(x)dx`
    `=-0.5[log_(e)x]_(x=01m)^(x=2.01m)`
    `=-0.5″log”_(e)(2.01)/(0.1)`
    `W=-0.5xx2.303log_(10)20.10`
    `=-0.5xx2.303xx1.303=-1.5J`
    `:.` Final K.E., `K_(f)=K_(i)+W=2.0-1.5J`
    `=0.5J`
    `v_(f)=sqrt((2k_(f))/(m))=sqrt((2xx0.5)/(1))=1ms^(-1)`

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