Question:
As shown in the figure, a particle of mass 10 $\mathrm{kg}$ is placed at a point A. When the particle is slightly displaced to its right, it starts moving and reaches the point $B$. The speed of the particle at $\mathrm{B}$ is $\times \mathrm{m} / \mathrm{s}$. (Take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$ )
The value of ' $x$ ' to the nearest integer is_________.
Solution:
Using work energy theorem,
$\mathrm{W}_{\mathrm{g}}=\Delta \mathrm{K} . \mathrm{E}$
$(10)(g)(5)=\frac{1}{2}(10) v^{2}-0$
$\mathrm{v}=10 \mathrm{~m} / \mathrm{s}$