/*! 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 32 4.32 A hanging mass, \(M_{1}=0.5... [FREE SOLUTION] | 91Ó°ÊÓ

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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

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

Step by step solution

01

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, \(T\).
02

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 \(M_2\): \(T = M_2a\)
03

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 \(M_1\): \(M_2a - M_1g = M_1a\)
04

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 \(4.9\, \text{m/s}^2\).

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Key Concepts

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

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.

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Most popular questions from this chapter

4.1 A car of mass M travels in a straight line at constant speed along a level road with a coefficient of friction between the tires and the road of \(\mu\) and a drag force of \(D\). The magnitude of the net force on the car is a) \(\mu M g\). c) \(\sqrt{(\mu M g)^{2}+D^{2}}\) b) \(\mu M g+D\)

offee filters behave Ince small parachutes, with a drag force that is proportional to the velocity squared, \(F_{\text {drag }}=K v^{2}\). A single coffee filter, when dropped from a height of \(2.0 \mathrm{~m}\), reaches the ground in a time of \(3.0 \mathrm{~s}\). When a second coffee filter is nestled within the first, the drag force remains the same, but the weight is doubled. Find the time for the combined filters to reach the ground. (Neglect the brief period when the filters are accelerating up to their terminal speed.)

A wedge of mass \(m=36.1 \mathrm{~kg}\) is located on a plane that is inclined by an angle \(\theta=21.3^{\circ}\) with respect to the horizontal. A force \(F=302.3 \mathrm{~N}\) in the horizontal direction pushes on the wedge, as shown in the figure. The coefficient of kinetic friction between the wedge and the plane is 0.159 What is the acceleration of the wedge along the plane?

A box of books is initially at rest a distance \(D=0.540 \mathrm{~m}\) from the end of a wooden board. The coefficient of static friction between the box and the board is \(\mu_{s}=0.320\), and the coefficient of kinetic friction is \(\mu_{k}=0.250 .\) The angle of the board is increased slowly, until the box just begins to slide; then the board is held at this angle. Find the speed of the box as it reaches the end of the board. -4.55 A block of mass \(M_{1}=0.640 \mathrm{~kg}\) is initially at rest on a cart of mass \(M_{2}=0.320 \mathrm{~kg}\) with the cart initially at rest on a level air track. The coefficient of static friction between the block and the cart is \(\mu_{s}=0.620\), but there is essentially no friction between the air track and the cart. The cart is accelerated by a force of magnitude \(F\) parallel to the air track. Find the maximum value of \(F\) that allows the block to accelerate with the cart, without sliding on top of the cart.

-4.44 A mass \(m_{1}=20.0 \mathrm{~kg}\) on a frictionless ramp is attached to a light string. The string passes over a frictionless pulley and is attached to a hanging mass \(m_{2}\). The ramp is at an angle of \(\theta=30.0^{\circ}\) above the horizontal. \(m_{1}\) moves up the ramp uniformly (at constant speed). Find the value of \(m_{2}\)

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