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?

Explanation

To solve the problem step by step, we will follow the principles of work, energy, and the equations of motion.

### Step 1: Understand the forces acting on the block
The block of mass \( m = 1 \, \text{kg} \) is moving with an initial speed \( v_i = 2 \, \text{m/s} \) and enters a rough patch where the retarding force \( F_r \) is given by:
\[ F_r = -\frac{k}{x} \]
for \( 0.1 < x < 2.01 \, \text{m} \), where \( k = 0.5 \, \text{J} \). Outside this range, the force is zero.
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