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Chapter 4: Problem 91

Two blocks are connected by a lightweight, flexible cord that passes over a frictionless pulley. If \(m_{1}=3.6 \mathrm{kg}\) and \(m_{2}=9.2 \mathrm{kg},\) and block 2 is initially at rest \(140 \mathrm{cm}\) above the floor, how long does it take block 2 to reach the floor?

Short Answer

Expert verified
Answer: It takes approximately 0.804 seconds for block 2 to reach the floor.

Step by step solution

01

Identify the forces acting on the system

For this problem, there are two forces acting on the system: the force of gravity on each block and the tension in the rope connecting the blocks.
02

Apply Newton's second law to the system

Designate block 1 as A and block 2 as B. Applying Newton's second law, we get: For block A: \(T - m_a g = m_a a\) (where T is tension and a is acceleration) For block B: \(m_{b} g - T = m_{b} a\)
03

Solve for the acceleration

Add the two equations and we can find the combined acceleration of the system: \(T - m_a g + m_{b} g - T = m_a a + m_{b} a\) The tension cancels out, and the equation simplifies to: \(m_{b} g - m_a g = (m_a + m_{b}) a\) Now, we can solve for the acceleration (a): a = \(\frac{m_{b} g - m_a g}{m_a + m_{b}}\) Plug in the provided values for \(m_a\), \(m_b\), and g (approximate \(g = 9.8 m/s^2\)): a = \(\frac{9.2 \cdot 9.8 - 3.6 \cdot 9.8}{3.6 + 9.2}\) a = \(\frac{55.44}{12.8}\) a = \(4.325 m/s^2\)
04

Calculate the time it takes for block 2 to reach the floor

Now, we apply one of the equations of motion using the distance block 2 falls, the initial velocity of block 2, and the acceleration we found: distance = initial_velocity * time + 0.5 * acceleration * time^2 Since block 2 is initially at rest, its initial velocity is 0. Therefore, we get: distance = 0.5 * acceleration * time^2 The distance is given as 140 cm in the problem, which we need to convert to meters: 140 cm = 1.4 m. Now we can solve for the time (t): 1.4 = 0.5 * 4.325 * time^2 time^2 = \(\frac{1.4}{0.5 \cdot 4.325}\) time^2 = 0.64706 The time is the square root of 0.64706: time = \(\sqrt{0.64706}\) time = 0.804 m/s Therefore, it takes block 2 approximately 0.804 seconds to reach the floor.

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Most popular questions from this chapter

Yolanda, whose mass is \(64.2 \mathrm{kg},\) is riding in an elevator that has an upward acceleration of \(2.13 \mathrm{m} / \mathrm{s}^{2} .\) What force does she exert on the floor of the elevator?
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