Question

In Fig. 8-51, a block is sent sliding down a frictionless ramp. Its speeds at points A and B are$\mathbf{2}\mathbf{.}\mathbf{00}\frac{\mathbf{m}}{\mathbf{s}}$and$\mathbf{2}\mathbf{.}\mathbf{60}\frac{\mathbf{m}}{\mathbf{s}}$, respectively. Next, it is again sent sliding down the ramp, but this time its speed at point A is$\mathbf{4}\mathbf{.}\mathbf{00}\frac{\mathbf{m}}{\mathbf{s}}$. What then is its speed at point B?

Short answer

The second-time speed at point B is,$v_{B,2}=4.33\frac{\mathrm{m}}{\mathrm{s}}$

Step-by-step solution

Given

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,$v_{A,2}=4.00\frac{\mathrm{m}}{\mathrm{s}}$

Determine the concept and the formulas

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,$\Delta K=K_B-K_A$

Calculate the block’s speed at point B

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

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

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

$\Delta K_1=\Delta K_2$

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:

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

Solve further as:

$\Rightarrow v_{B,2}=4.33\hspace{0.33em}\frac{\mathrm{m}}{\mathrm{s}}$