/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 34 A crate of mass \(m_{1}\) on a f... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 34

A crate of mass \(m_{1}\) on a frictionless inclined plane is attached to another crate of mass \(m_{2}\) by a massless rope. The rope passes over an ideal pulley so the mass \(m_{2}\) is suspended in air. The plane is inclined at an angle \(\theta=36.9^{\circ} .\) Use conservation of energy to find how fast crate \(m_{2}\) is moving after \(m_{1}\) has traveled a distance of $1.4 \mathrm{m}\( along the incline, starting from rest. The mass of \)m_{1}$ is \(12.4 \mathrm{kg}\) and the mass of \(m_{2}\) is \(16.3 \mathrm{kg}\).

Short Answer

Expert verified
Answer: To find the final velocity of crate \(m_2\), we can apply the work-energy theorem on the system and use the given values for masses, angle, and distance. After calculating the height crate \(m_2\) rises and solving for the velocity, we find that the final velocity of crate \(m_2\) is equal to the final velocity of crate \(m_1\). By plugging in the given values and solving the equation, we can find the magnitude of the final velocity for both crates.

Step by step solution

01

Represent the system diagrammatically

Draw a diagram with crate \(m_1\) on the inclined plane at an angle \(\theta\) with respect to the horizontal, and crate \(m_2\) hanging vertically. The massless rope connects both crates over an ideal pulley.
02

Identify the forces and work done on the system

There are three forces acting on the system: tension force \(T\) in the rope, gravitational force \(m_1g\sin\theta\) acting parallel to the inclined plane on \(m_1\), and gravitational force \(m_2g\) acting on \(m_2\). As we are to use work-energy theorem, let's calculate the work done by each force: 1. Work done by tension, \(W_T = T \cdot d\) (where \(d\) is the distance traveled by \(m_1\)) 2. Work done by the gravitational force on \(m_1\), \(W_{g1} = -m_1g\sin\theta \cdot d\) (negative as it acts opposite to the motion) 3. Work done by the gravitational force on \(m_2\), \(W_{g2} = m_2g \cdot h\) (where \(h\) is the height \(m_2\) rises)
03

Calculate height using trigonometry

We need to find the height \(h\) that \(m_2\) rises when \(m_1\) travels a distance of \(1.4\mathrm{m}\) along the incline. Using trigonometry, \(h = d\sin\theta\). Plug in the given values to find \(h\): \(h = 1.4\mathrm{m} \cdot \sin(36.9^{\circ})\)
04

Apply work-energy theorem to the system

According to the work-energy theorem, the total work done on the system is equal to the change in kinetic energy of the system. Mathematically, \(W_T + W_{g1} + W_{g2} = \Delta KE\). Plug in the expressions for work done and set the initial kinetic energy as zero because both crates start from rest. \(T \cdot d - m_1g\sin\theta \cdot d + m_2g \cdot h = \frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2}\)
05

Use given values and solve for \(v_2\)

Now we have all values necessary to solve for the final velocity of crate \(m_2\). Use the given values for \(m_1\), \(m_2\), \(d\), and calculate \(h\) and solve for \(v_2\). Plug in the values to find \(v_2\): \(12.4\mathrm{kg} \cdot 9.81\mathrm{m/s^2} \cdot \sin(36.9^{\circ}) \cdot 1.4\mathrm{m} + 16.3\mathrm{kg} \cdot 9.81\mathrm{m/s^2} \cdot 1.4\mathrm{m} \cdot \sin(36.9^{\circ}) = \frac{1}{2} 12.4\mathrm{kg} \cdot v_{1}^{2} + \frac{1}{2} 16.3\mathrm{kg} \cdot v_{2}^{2}\) Note that since the rope is massless and the pulley is ideal, both crates have equal magnitudes of velocities at any given time, i.e., \(v_1 = v_2\). Now, solve for \(v_1\) (and, subsequently, \(v_2\)) from the equation above.

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