91Ó°ÊÓ

i) The speed of the block at point A is,$v_{A,1}=2.00\frac{m}{s}$

ii) The speed of the block at point B is,$v_{B,1}=2.60\frac{\mathrm{m}}{\mathrm{s}}$

iii) Second time the speed of the block at point A is,$v_{A,2}=4.00\frac{\mathrm{m}}{\mathrm{s}}$

The problem is based on the principle of conservation of mechanical energy. According to this principle the energy can neither be created nor be destroyed; it can only be internally converted from one form to another if the forces doing work on the system are conservative in nature. change in potential energy. From this, we can find the second time speed of the block at point B.

Formula:

i) The conservation of energy is,$K_A+U_A=K_B+U_B$

ii) The change in kinetic energy is,$\mathrm{Δ}K=K_B-K_A$

The block is sliding down; by conservation of energy, we have

$K_A+U_A=K_B+U_B$

Thus, the change in kinetic energy as the block moves from point A to point B is

$$\begin{gathered} ∆K=K_B-K_A \\ ∆K=-∆U \\ ⇒∆K=-\left(U_B-U_A\right) \end{gathered}$$

Thus, in both circumstances, we have the same potential energy change. Hence,

$\mathrm{Δ}{K}_1=\mathrm{Δ}{K}_2$

The speed of the block at B the second time is given by

$\frac{1}{2}m{v_{B.1}}^2-\frac{1}{2}m{v}_{A,1}^2=\frac{1}{2}m{v}_{B,2}^2-\frac{1}{2}m{v}_{A,2}^2$

Solve for the speed at point B as:

$$\begin{gathered} ⇒v_{B,2}^2=v_{B,2}^2-v_{A,1}^2+v_{A,2}^2 \\ ⇒v_{B,2}=\sqrt{\left(v_{B,1}^2-v_{A,1}^2+v_{A,2}^2\right)} \\ ⇒v_{B,2}^2=\sqrt{\left(2.{60}^2-2.{00}^2+4.00\right)} \\ ⇒v_{B,2}=\sqrt{18.76} \end{gathered}$$

Solve further as:

$⇒v_{B,2}=4.33 \frac{\mathrm{m}}{\mathrm{s}}$