A body of mass `2kg` is resting on rough horizontal surface. A force `20N` is applied to it for `10s` parallel to the surface. If coefficient of kinetic friction between the surfaces in contact in `0.2.` Caluculate (i) work done by applied force in `10s`. (ii) change in `KE` of object in `10s`.
Take `g=10m//s^(2)`

Explanation

To solve the problem step by step, we will break it down into two parts: (i) calculating the work done by the applied force in 10 seconds, and (ii) calculating the change in kinetic energy of the object in 10 seconds.

### Given Data:
- Mass of the body, \( m = 2 \, \text{kg} \)
- Applied force, \( F = 20 \, \text{N} \)
- Time, \( t = 10 \, \text{s} \)
- Coefficient of kinetic friction, \( \mu = 0.2 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)
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