KCET 2025 Physics Paper

To determine the work done by the net force during the displacement of the body, we need to evaluate the change in kinetic energy ( $\Delta\ \text{KE}$ ) as the object moves from $x=0$ to $x=2\ \text{m}$.The initial velocity $u$ when $x=0$ is zero because $v = k x^{3/2}$ and substituting $x=0$ gives $v=0$.The final kinetic energy at $x=2\ \text{m}$ can be found by calculating the velocity at this point:$v = k \cdot (2)^{3/2}$Given $k=2$ Sl units, the velocity at $x=2$ is:$v = 2 \cdot 2^{3/2}$ The change in kinetic energy is:$\Delta\ \text{KE} = \frac{1}{2} m v^2 - \frac{1}{2} m u^2$Substituting $m=0.25\ \text{kg}$ and $u=0$ : $\Delta\ \text{KE} = \frac{1}{2} \times 0.25 \times (2 \times 2^{3/2})^2 - 0$Calculate $(2 \times 2^{3/2})^2$ :$= (2^1 \times (2^{3/2}))^2 = 2^2 \times (2^3) = 2^5 = 32$So the change in kinetic energy is:$\Delta\ \text{KE} = \frac{1}{2} \times 0.25 \times 32$Finally, solving gives:$=4\ \text{J}$Hence, the work done by the net force during the displacement is 4 Joules.