Question

4.32 A hanging mass, \(M_{1}=0.50 \mathrm{~kg}\), is attached by a light string that runs over a frictionless pulley to a mass \(M_{2}=1.50 \mathrm{~kg}\) that is initially at rest on a frictionless table. Find the magnitude of the acceleration, \(a,\) of \(M_{2}\)

Short answer

Answer: The magnitude of the acceleration of mass \(M_2\) is approximately \(4.9\, \text{m/s}^2\).

Step-by-step solution

Identify the forces acting on each mass

First, we need to identify the forces acting on each mass. For mass \(M_1\), there is the gravitational force acting downward, which equals \(M_1g\). For mass \(M_2\), there is the force due to the tension in the string, \(T\).

Apply Newton's second law to each mass

Newton's second law states that the net force acting on an object equals its mass times acceleration (\(F_{net} = ma\)). For the vertical direction, the net force acting on mass \(M_1\) is given by the tension in the string minus the gravitational force (\(M_1g\)): \(F_{net1} = T - M_1g\). Similarly, for the horizontal direction, the net force acting on mass \(M_2\) is given by the tension in the string: \(F_{net2} = T\). Note that the two masses have the same acceleration because they are connected by the string. We can write two equations from Newton's second law for the two masses: For mass \(M_1\): \(T - M_1g = M_1a\) For mass \(M_2\): \(T = M_2a\)

Eliminate the unknown tension T

We have two equations and two unknowns, tension \(T\) and acceleration \(a\). We can eliminate the unknown tension by solving the equation for mass \(M_2\) for \(T\) and substituting into the equation for mass \(M_1\): \(T = M_2a\) Substitute into the equation for mass \(M_1\): \(M_2a - M_1g = M_1a\)

Solve for acceleration a

Now we have a single equation and one unknown, acceleration \(a\). Rearrange the equation to solve for \(a\): \(M_2a - M_1a = M_1g\) \((M_2 - M_1)a = M_1g\) \(a = \dfrac{M_1g}{M_2 - M_1}\) Now, we can substitute in the given values for the masses and the gravitational acceleration \(g = 9.81\,\text{m/s}^2\): \(a = \dfrac{(0.50 \,\text{kg})(9.81 \,\text{m/s}^2)}{1.50 \,\text{kg} - 0.50 \,\text{kg}}\) \(a = \dfrac{(0.50 \,\text{kg})(9.81 \,\text{m/s}^2)}{1.00 \,\text{kg}}\) \(a = 4.905 \,\text{m/s}^2\) So, the magnitude of the acceleration of mass \(M_2\) is approximately \(4.9\, \text{m/s}^2\).

Key concepts

Acceleration Calculation

In physics, calculating acceleration is crucial for understanding how objects move. Acceleration, in simple terms, refers to how quickly an object's velocity changes. It's a key concept in Newton's Second Law of Motion, where the net force acting on an object is equal to the product of its mass and acceleration, mathematically expressed as \( F = ma \).

To solve for acceleration in our exercise, we consider two connected masses — one hanging and one laid on a frictionless surface. The forces influencing these masses need to be considered:

• For the hanging mass \( M_1 = 0.50 \, ext{kg} \), gravity acts downward with force \( M_1g \), where \( g = 9.81 \, ext{m/s}^2 \).
• For the mass on the table \( M_2 = 1.50 \, ext{kg} \), tension in the string acts horizontally.

By setting up the equations based on these forces, as shown in the exercise, we can solve for acceleration \( a \). The calculated acceleration is approximately \( 4.9 \, ext{m/s}^2 \), demonstrating how the interplay of these forces influences motion.

Tension in Strings

When you think of tension in strings, you're essentially thinking about the pulling force transmitted along a string or rope. In our scenario, the string transmits force between the hanging mass and the mass on the table.

The tension \( T \) in the string is crucial as it directly affects both masses. For the mass on the table, the tension \( T \) is what pulls it horizontally, resulting in acceleration \( a \). Conversely, for the hanging mass, tension works against the gravitational pull:

• For the table mass, net force is \( F_{net2} = T \) equating to \( T = M_2a \).
• For the hanging mass, net force is \( F_{net1} = T - M_1g \) equating to \( T - M_1g = M_1a \).

By manipulating these equations, specifically solving for \( T \) and substituting into one another, we eliminate the tension to simplify the acceleration calculation. This process highlights how tension balances, yet couples the forces across different directions.

Frictionless Pulley System

A frictionless pulley system simplifies the study of physics by removing resistive forces that could complicate calculations, like friction. In this idealized setup, a small pulley alters the direction of a pulling force without adding extra resistance.

This scenario implies that the only forces we are working with are those we actively account for — namely, tension and gravitational force. The absence of friction means:

• The pulley does not slow down or oppose the motion.
• Tension remains consistent throughout the string.
• The system faithfully follows Newton's Second Law.

By focusing on these simplifications, students can hone in on understanding how forces interact without considering additional variables. Mastering these basics in frictionless conditions supports learning when confronting more complex scenarios with friction and additional forces.