We are given the following data:
The formula for kinetic energy is: \[ KE = \frac{1}{2} m v^2 \]
\[ KE = \frac{1}{2} \times 2 \, \text{kg} \times (5 \, \text{m/s})^2 \]
\[ KE = \frac{1}{2} \times 2 \times 25 = 25 \, \text{J} \]
The kinetic energy of the body is \( 25 \, \text{J} \).
Option 1: 25 J
A block of certain mass is placed on a rough floor. The coefficients of static and kinetic friction between the block and the floor are 0.4 and 0.25 respectively. A constant horizontal force \( F = 20 \, \text{N} \) acts on it so that the velocity of the block varies with time according to the following graph. The mass of the block is nearly (Take \( g = 10 \, \text{m/s}^2 \)): 
