Question

As shown in the figure, a particle of mass $10\text{kg}$ 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 $B$ is $x\text{m/s}$. (Take $g=10{\text{m/s}}^2$) The value of $x$ to the nearest integer is .

Answer 1

Take horizontal surface as reference level for gravitational potential energy.

By energy conservation:
$K_i+U_i=K_f+U_f$

$\Rightarrow 0+10\times 10\times 10=\frac{1}{2}\times 10\times v_B^2+10\times 10\times 5$

$\Rightarrow 1000=5v_B^2+500$

$\Rightarrow v_B^2=\frac{500}{5}=100$

$\Rightarrow v_B=10\text{m/s}$

$\Rightarrow x=10$