Two masses, \(m_1\) and \(m_2\), are connected by a string of negligible mass, which passes over a massless pulley, as shown in the figure. Assume \(m_1>m_2\). What is the acceleration \(a\) of \(m_1\) ? (A) \(a=\left(m_1-m_2\right) g /\left(m_1+m_2\right)\) (B) \(a=\left(m_1+m_2\right) g /\left(m_1-m_2\right)\) (C) \(a=\left(m_1 m_2\right) g /\left(m_1+m_2\right)\) (D) \(a=\left(m_2-m_1\right) g /\left(m_1+m_2\right)\) (E) \(a=\left(m_1-m_2\right) g /\left(m_1 m_2\right)\)

Short Answer

Expert verified
Question: Calculate the acceleration of mass \(m_1\) in a two-mass system, given that \(m_1>m_2\) and the two masses are connected by an inextensible string passing over a frictionless pulley. Answer: The acceleration of mass \(m_1\) is given by \(a=\left(m_1-m_2\right) g /\left(m_1+m_2\right)\), where \(g\) is the acceleration due to gravity.

Step by step solution

01

Identify the Forces acting on each mass

For each mass, we will have two forces acting on them: their weight (gravitational force) and the tension force from the string. Let's identify the forces acting on each mass: - On \(m_1\): - Weight: \(m_1g\) (downwards) - Tension: \(T\) (upwards) - On \(m_2\): - Weight: \(m_2g\) (downwards) - Tension: \(T\) (upwards)
02

Apply Newton's second law of motion

For each mass, we need to apply Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration (\(F=ma\)). So, for each mass, we have: - For mass \(m_1\) (moving downwards): \begin{equation} \sum F_1 = m_1a_1 \Rightarrow m_1g - T = m_1a_1 \end{equation} - For mass \(m_2\) (moving upwards): \begin{equation} \sum F_2 = m_2a_2 \Rightarrow T - m_2g = m_2a_2 \end{equation} Since the string is assumed to be inextensible, the two masses will have the same magnitude of acceleration, i.e., \(|a_1|=|a_2|=a\).
03

Solve the system of equations

Now, we will solve the system of equations formed in Step 2. First, we can find tension force, \(T\), from equation (2) and substitute it into equation (1) to solve for acceleration, \(a\): From equation (2), we have: \begin{equation} T = m_2a_2 + m_2g \Rightarrow T = m_2a + m_2g \end{equation} Substituting (3) into equation (1), we get: \begin{equation} m_1g - (m_2a + m_2g) = m_1a \end{equation}
04

Solve for acceleration \(a\)

Now, we will solve equation (4) for acceleration, \(a\). Simplify the equation: \begin{equation} m_1a + m_2a = m_1g - m_2g \end{equation} Factor out the common term \(a\): \begin{equation} a (m_1 + m_2) = g (m_1 - m_2) \end{equation} Finally, solve for \(a\): \begin{equation} a = \frac{(m_1-m_2)g}{(m_1+m_2)} \end{equation} Thus, the acceleration of mass \(m_1\) is given by Option (A): \[a=\left(m_1-m_2\right) g /\left(m_1+m_2\right)\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Newton's Second Law of Motion
Understanding Newton's second law of motion is crucial when tackling physics acceleration problems. This law is often succinctly stated as F = ma, which means the force (F) acting on an object is equal to the mass (m) of the object multiplied by its acceleration (a). In other words, how much an object accelerates depends on the total force applied and the object's mass.

When multiple forces are acting on an object, such as gravitational force and tension as in our textbook exercise, the net force is what influences the acceleration. This is the force that results after all other forces have been combined, considering both their magnitudes and directions. For the masses in the pulley system, we consider the downwards force of gravity and the upwards force of tension. By calculating the net force for each mass and applying Newton's second law, we can determine the acceleration of each mass with ease. The beauty of this law lies in its ability to simplify complex interactions into manageable equations that can provide clear and accurate predictions about an object's motion.
Pulley System Mechanics
Pulley systems are a staple in physics problems due to their practical applications in lifting objects and their ability to demonstrate fundamental mechanics principles. In our example, a pulley system is used to connect two masses with a string. Mechanics of such a system involve understanding the forces in action, including tension in the string and gravitational forces acting on the masses.

A key characteristic to remember is that a pulley, assumed to be frictionless and massless, only changes the direction of the tension force and does not alter its magnitude. Thus, if the string is inextensible, any acceleration of one mass must be identical in magnitude, though opposite in direction, to the acceleration of the other mass. Essentially, what happens to one mass will directly affect the other due to the constraints of the system. Through the lens of pulley system mechanics, we're equipped to analyze complex setups by reducing them to interactions between tension and gravity, leading to the solution for the acceleration of connected bodies.
System of Equations in Physics
In physics, particularly in dynamics, we frequently employ systems of equations to solve for multiple unknowns. This approach is necessary when a single equation is not enough to unravel the variables at play. Our textbook exercise exemplifies this when dealing with two masses connected by a pulley system.

By applying Newton's second law separately to each mass, we end up with two equations, which together represent a system that must be solved simultaneously. The elegance of using a system of equations lies in its ability to cross-reference multiple relations and constraints, leading to a complete solution that individual equations cannot provide alone. Solving such a system, as demonstrated in our textbook solution, might involve substitution or elimination methods to find the value of unknowns such as the tensions in the strings and the common acceleration of the masses. Mastery of this mathematical technique unlocks the power to tackle a vast array of physics problems, enhancing our overall conceptual toolkit.

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Most popular questions from this chapter

Pick all correct responses. Air bags in cars protect occupants in a crash by (A) increasing the time needed for the occupant to come to a stop, thereby decreasing the force of impact. (B) increasing the momentum of the occupant. (C) decreasing the impulse of the occupant. (D) decreasing the amount of kinetic energy lost during the collision. (E) None of the above

An object of mass \(m_1\) and an object of mass \(m_2\) are initially attached to each other and at rest. At the instant \(t=0\) they are pushed apart by a spring that has been compressed between them. After \(t=0\), \(v_1=2 \mathrm{~m} / \mathrm{s}\) and \(v_2=-3 \mathrm{~m} / \mathrm{s}\). One can conclude that (A) \(m_1 / m_2=3 / 2\) (B) \(m_1=2 \mathrm{~kg}\) and \(m_2=3 \mathrm{~kg}\) (C) \(m_1=3 \mathrm{~kg}\) and \(m_2=2 \mathrm{~kg}\) (D) \(m_1 / m_2=2 / 3\) (E) \(m_1 / m_2=-3 / 2\)

Two planets, Gamma and Delta, with masses \(m_G\) and \(m_D\) and radii \(R_G\) and \(R_D\) are initially at rest a large distance \(d\) apart from each other. Khan the Conqueror suddenly allows the two planets to approach under their mutual gravitational attraction until they collide. To determine the velocity at which Gamma and Delta collide you could (A) use conservation of energy alone. (B) use conservation of momentum alone. (C) use both conservation of energy and momentum. (D) The problem cannot be solved using energy and momentum conservation. (E) The problem cannot be solved unless you know how long it takes the two planets to collide.

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