MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER Two blocks are connected by a light string that passes over a frictionless pulley as in the figure below. The system is released from rest while m, is on the floor and m, is a distance h above the floor. (a) Assuming m, > m, find an expression for the speed of m, just as it reaches the floor. (Use any variable or symbol stated above along with the following as necessary: g.) (b) Taking m, = 6.9 kg, m, = 3.9 kg, and h= 3,5 m, evaluate your answer to part (a). m/s (c) Find the speed of each block when m, has fallen a distance of 1.2 m. m/s

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Chapter1: Units, Trigonometry. And Vectors
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**Educational Problem Description:**

Two blocks are connected by a light string that passes over a frictionless pulley as shown in the figure. The system is released from rest while \( m_2 \) is on the floor and \( m_1 \) is a distance \( h \) above the floor.

**Diagram Explanation:**
- The diagram shows a pulley system with two blocks, \( m_1 \) and \( m_2 \), connected by a string. 
- Block \( m_1 \) is suspended in the air, while block \( m_2 \) is on the floor. 
- The pulley is positioned above \( m_2 \).
- The distance \( h \) is the initial vertical separation between \( m_1 \) and the floor.

**Questions:**
  
(a) Assuming \( m_1 > m_2 \), find an expression for the speed of \( m_1 \) just as it reaches the floor. (Use any variable or symbol stated above along with the following as necessary: \( g \).)

\[ v = \_\_\_\_\_ \]

(b) Taking \( m_1 = 6.9 \, \text{kg}, \, m_2 = 3.9 \, \text{kg}, \) and \( h = 3.5 \, \text{m}, \) evaluate your answer to part (a).

\[ v = \_\_\_\_\_ \, \text{m/s} \]

(c) Find the speed of each block when \( m_1 \) has fallen a distance of 1.2 m.

\[ v = \_\_\_\_\_ \, \text{m/s} \]
Transcribed Image Text:**Educational Problem Description:** Two blocks are connected by a light string that passes over a frictionless pulley as shown in the figure. The system is released from rest while \( m_2 \) is on the floor and \( m_1 \) is a distance \( h \) above the floor. **Diagram Explanation:** - The diagram shows a pulley system with two blocks, \( m_1 \) and \( m_2 \), connected by a string. - Block \( m_1 \) is suspended in the air, while block \( m_2 \) is on the floor. - The pulley is positioned above \( m_2 \). - The distance \( h \) is the initial vertical separation between \( m_1 \) and the floor. **Questions:** (a) Assuming \( m_1 > m_2 \), find an expression for the speed of \( m_1 \) just as it reaches the floor. (Use any variable or symbol stated above along with the following as necessary: \( g \).) \[ v = \_\_\_\_\_ \] (b) Taking \( m_1 = 6.9 \, \text{kg}, \, m_2 = 3.9 \, \text{kg}, \) and \( h = 3.5 \, \text{m}, \) evaluate your answer to part (a). \[ v = \_\_\_\_\_ \, \text{m/s} \] (c) Find the speed of each block when \( m_1 \) has fallen a distance of 1.2 m. \[ v = \_\_\_\_\_ \, \text{m/s} \]
Expert Solution
Step 1

a) by work energy theorem on system

∆K = Wg + WT

(1/2)(m1+m2) vf- 0 = m1gh - m2gh + 0 

vf2 = 2(m1 - m2)gh / (m1 + m2

vf = √[2(m1 - m2)gh / (m1 + m2)] 

 

b) vf = √[2(6.9-3.9)×9.8×3.5 / (6.9+3.9)] 

vf = 4.36 m/s 

 

c) at h = 1.2 m 

vf = √[2(6.9-3.9)×9.8×1.2 / (6.9+3.9)] 

vf = 2.55 m/s 

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