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
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
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Question
![**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.
\[ v = \_\_\_\_\_ \, \text{m/s} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd9f8dd60-2822-4162-bc66-228f0a545829%2Fe846ec0c-c7a0-41c8-94d5-78e68c73791b%2Fz612q4_processed.jpeg&w=3840&q=75)
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
![](/static/compass_v2/shared-icons/check-mark.png)
Step 1
a) by work energy theorem on system
∆K = Wg + WT
(1/2)(m1+m2) vf2 - 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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