Chapter 6, Problem 8Q

A hand exerts a constant horizontal force on a block that is free to slide on a frictionless surface (Fig. 6-30). The block starts from rest at point A, and by the time it has traveled a distance d to point B it is traveling with speed v_B. When the block has traveled another distance d to point C, will its speed be greater than, less than, or equal to 2v_e? Explain your reasoning.

To determine

Whether the speed of block at point C greater than, less than or equal to $2v_B$ .

Answer to Problem 8Q

Solution:The speed of block at point C is greater than the speed at point B. The exact value is $v_c=\sqrt{2}{v}_B$

Explanation of Solution

Given: The force applied $\overrightarrow{F}$ is constant and free to slide on a frictionless surface. The speed $v_{\text{A}}$ of block at point A is zero. The speed of block at point Bis $v_{\text{B}}$ .

Formula used: According to Work-Energy Principle we have:

  $W=\Delta E_{\text{C}}=K_f-K_i$  (1)

Where:

• $W$ is the total work on the system.
• $\Delta E_c$ is the change in kinetic energy.
• $K_f$ is the final kinetic energy, which can be defined as:

  $K_f=\frac{1}{2}m{\left(v_f\right)}^2$  (2)

• $K_i$ is the initial kinetic energy:

  $K_i=\frac{1}{2}m{\left(v_i\right)}^2$   (3)

Therefore, equation (1) can be written as:

  $W=\frac{1}{2}m{\left(v_f\right)}^2-\frac{1}{2}m{\left(v_i\right)}^2$  (4)

Calculation:

The force applied on the block is constant at the same direction as displacement, therefore equation (4) can be written as:

  $F\cdot x=\frac{1}{2}m{\left(v_f\right)}^2-\frac{1}{2}m{\left(v_i\right)}^2$  (5)

Now, according to figure given, the work done between points A and B is:

  $F\cdot d=\frac{1}{2}m{\left(v_B\right)}^2-\frac{1}{2}m{\left(v_A\right)}^2$  (6)

But:

  $v_A=0$  (7)

Therefore, equation (6) can be written as:

  $F.d=\frac{1}{2}m{v}_B^2$  (8)

Isolating $v_B$ from equation (8):

  $v_B=\sqrt{\frac{2Fd}{m}}$  (9)

Similarly, between points B and C the work done is:

  $F.d=\frac{1}{2}m{v}_c^2-\frac{1}{2}m{v}_B^2$  (10)

Replacing (9) into equation (10):

  $F.d=\frac{1}{2}m{v}_C^2-F\cdot d$  (11)

Isolating $v_c$ from equation (12):

  $v_c=\sqrt{\frac{4Fd}{m}}$  (11)

But, equation (12) can be written as:

  $v_c=\sqrt{2}\sqrt{\frac{2Fd}{m}}$  (13)

Replacing (9) into equation (13):

  $v_c=\sqrt{2}{v}_B$  (14)

Conclusion: According to equation (14), the speed of block at point C is greater than the speed at point B. The exact value is $v_c=\sqrt{2}{v}_B$ .