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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

A locomotive pulls a train of 10 identical cars, on a track that runs east- west, with a force of \(2.0 \times 10^{6} \mathrm{N}\) directed east. What is the force with which the last car to the west pulls on the rest of the train?
A model rocket is fired vertically from rest. It has a net acceleration of \(17.5 \mathrm{m} / \mathrm{s}^{2} .\) After \(1.5 \mathrm{s}\), its fuel is exhausted and its only acceleration is that due to gravity. (a) Ignoring air resistance, how high does the rocket travel? (b) How long after liftoff does the rocket return to the ground?
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