Question

As shown in the figure, a particle of mass $10kg$ is placed at point $A$. When the particle is slightly displaced to its right, it starts moving and reaches point $B$. The speed of the particle at $B$ is $xm{s}^{-1}$. (Take $g=10m{s}^{-1}$ ) The value of $\text{'}x\text{'}$to the nearest integer is $\_\_\_\_\_\_\_.$

Answer 1

Step 1: Given data

Mass of the particle, $m=10kg$

Height of particle at point $A,h_A=10m$

The velocity of particle at point $A,v_A=0m/s$

Height of particle at point $B,h_B=5m$

Step 2: Find the velocity of the particle at point $B$

Let the velocity of the particle at point $B=v_B$

Applying conservation of energy,

Total energy at point $A=$Total energy at the point $B$

$$\begin{gathered} {\left(K.E\right)}_A+{\left(P.E\right)}_A={(K.E)}_B+{(P.E)}_B \\ 0+m\times g\times h_A=\frac{1}{2}m{\left(v_B\right)}^2+m\times g\times h_B \\ 10\times 10\times 10=\frac{1}{2}\times 10\times {\left(v_B\right)}^2+10\times 10\times 5[As,g=10m/s^2] \\ 1000=5{\left(v_B\right)}^2+500 \\ {\left(v_B\right)}^2=100 \\ {v}_B=10m/s \end{gathered}$$

Hence, The value of $\text{'}x\text{'}$ to the nearest integer is $10$.