Question

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

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

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.

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)

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 \(h\): \(h = 1.4\mathrm{m} \cdot \sin(36.9^{\circ})\)

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. \(T \cdot d - m_1g\sin\theta \cdot d + m_2g \cdot h = \frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2}\)

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.