Two blocks weighing 200 N and 300 N are hung to the ends of a rope passing over an ideal pulley. The velocity of the system at time t = 0 is 3 m/s and velocity of the system at time t becomes 5 m/s. The block weighing 200 N shows an upward displacement of h. Acceleration due to gravity is 10 m/s2. According to work - energy method, the tensions in the rope on its sides and the value of h are

Concept:

To solve this problem, we can use the work-energy principle and Newton's second law. The system consists of two blocks connected by a rope over an ideal pulley, with one block moving upward and the other downward. The change in kinetic energy will be related to the work done by gravity and tension.

Calculation:

Given:

• Weight of Block 1 (\( W_1 \)) = 200 N (upward displacement)
• Weight of Block 2 (\( W_2 \)) = 300 N (downward displacement)
• Initial velocity (\( v_1 \)) = 3 m/s
• Final velocity (\( v_2 \)) = 5 m/s
• Acceleration due to gravity (\( g \)) = 10 m/s²

1. Finding the Acceleration:

The change in velocity is from 3 m/s to 5 m/s. We need to find the acceleration (\( a \)):

\( a = \frac{v_2^2 - v_1^2}{2h} = \frac{25 - 9}{2h} = \frac{16}{2h} = \frac{8}{h} \)

2. Equations of Motion for Each Block:

Let \( T \) be the tension in the rope:

• For Block 1 (200 N, moving upward):
  • \( T - W_1 = \frac{200}{10} \cdot a \)
  • \( T - 200 = 20a \quad ...(1) \)
• For Block 2 (300 N, moving downward):
  • \( W_2 - T = \frac{300}{10} \cdot a \)
  • \( 300 - T = 30a \quad ...(2) \)

3. Solving the Equations:

Adding equations (1) and (2) to eliminate \( T \):

\( (T - 200) + (300 - T) = 20a + 30a \)

\( 100 = 50a \)

\( a = 2 \, \text{m/s}^2 \)

4. Finding the Tension \( T \):

Substitute \( a = 2 \) m/s² into equation (1):

\( T - 200 = 20 \cdot 2 \)

\( T = 240 \, \text{N} \)

The tension is the same for both blocks.

5. Calculating the Displacement \( h \):

Using the kinematic equation:

\( v_2^2 = v_1^2 + 2ah \)

\( 25 = 9 + 4h \)

\( 4h = 16 \)

\( h = 4 \, \text{m} \)