/*! 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 17 A light rope is attached to a bl... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A light rope is attached to a block with mass \(4.00 \mathrm{~kg}\) that rests on a friction less, horizontal surface. The horizontal rope passes over a friction less, massless pulley, and a block with mass \(m\) is suspended from the other end. When the blocks are released, the tension in the rope is \(15.0 \mathrm{~N}\). (a) Draw two free-body diagrams: one for each block. (b) What is the acceleration of either block? (c) Find \(m\). (d) How does the tension compare to the weight of the hanging block?

Short Answer

Expert verified
The acceleration of either block is \(a = \frac{T}{m_1}= \frac{15N}{4kg}\). The mass of the hanging block \(m_2 = \frac{T}{g + a}\). The tension does compare to the weight of the hanging block depending on whether \(T > W_2\), \(T < W_2\), or \(T = W_2\).

Step by step solution

01

Draw free-body diagrams

The first block on the frictionless surface has two forces acting on it: the tension force \(T\) directed to the right and the weight \(W_1\) of the block directed downwards. For the hanging block, the forces acting on it are the weight \(W_2\) pointing downward and the tension force \(T\) directed upward.
02

Applying Newton's second law for horizontal block

As there is no friction, the net force on the horizontal block is equal to the tension force \(T\). So according to the Newton's second law \(F = ma\), we have \(T = m_1*a\), where \(m_1\) is the mass of the horizontal block and \(a\) is the acceleration. We know \(T= 15N\) and \(m_1 = 4kg\), substitute these values we get \(a = \frac{T}{m_1} = \frac{15N}{4kg}\).
03

Finding the mass of the hanging block

For the hanging block, the net force is \(T - W_2\) (tension force - weight of the hanging block). Again applying Newton's second law, \(T - W_2 = m_2*a\), where \(m_2\) is the mass of the hanging block and \(W_2 = m_2*g\) (weight is mass times gravity). Replace \(W_2\) with \(m_2*g\) we get, \(T - m_2*g = m_2 * a\). Solve for \(m_2\), we have \(m_2 = \frac{T}{g + a}\). Use \(g = 9.8m/s^2\), \(T = 15N\) and \(a\) from step 2 to calculate \(m_2\).
04

Compare the tension to the weight of the hanging block

To compare the tension \(T\) with the weight of the hanging block \(W_2 = m_2*g\), substitute the value of \(m_2\) from step 3 into \(W_2 = m_2*g\) and compare with \(T\). If \(T > W_2\), the tension is greater than the weight. If \(T < W_2\), the weight is greater than the tension. If \(T = W_2\), the tension is equal to the weight.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Newton's Second Law

Understanding Newtons's second law of motion is crucial for analyzing how objects behave when forces are applied. Simply put, this law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The law is represented mathematically by the equation: \[\begin{equation} F = ma \end{equation}\] \where \(F\) is the net force, \(m\) is the mass of the object, and \(a\) is the acceleration.


When considering the exercise mentioned, we apply Newton's second law to find the acceleration of a block moving on a frictionless surface when a tension force is applied. Since there are no opposing forces (like friction), the tension force in the rope directly translates to the acceleration of the block with mass \(m_1\), as shown in the step-by-step solution. This principle is broadly applicable in physics, from the swing of a pendulum to the orbits of planets.


As an exercise improvement advice, it could be beneficial for students to practice identifying and calculating the net force in various scenarios using Newton's second law, reinforcing their understanding of how acceleration depends on both force and mass.

Frictionless Motion

The term frictionless motion refers to an idealized situation in which no frictional forces oppose the movement of an object. Friction is the resistive force that occurs when two surfaces slide against one another, and it is not present in this hypothetical environment. In such cases, the only forces considered are those directly applied to the object and normal forces, such as gravity or tension.


In the context of the problem from the textbook, a frictionless surface allows us to assume that the horizontal block's acceleration is solely due to the tension in the rope. The lack of friction means that the tension is the net force causing the block to accelerate. This concept helps simplify problems and focus on the basic principles of mechanics, like mass, acceleration, and force, without the added complexity that friction would bring.


For a more comprehensive learning experience, students could compare and contrast the outcomes of scenarios with and without friction to deepen their understanding of how significantly friction can affect motion.

Tension in Physics

Tension is a force within a rope, string, or other continuous object that arises when it's pulled tight by forces acting from opposite ends. In physics problems, tension is typically treated as a pulling force transmitted through the object, and it is always directed along the length of the rope or wire. Tension can vary from one end of the rope to the other if other forces are acting on the rope (like gravity or friction), but in many idealized problems, like the one we're discussing, it is constant throughout.


Importance of Tension in the Exercise

In our exercise, tension acts as the primary force causing both the block on the frictionless surface and the hanging block to accelerate. The rope transmits the same tension force to both blocks, which is a characteristic of a light, inextensible rope in physics problems. This assumption is critical because it simplifies the calculations by ensuring the acceleration is the same for both blocks, despite their different masses.


To enhance understanding, students should try creating their own problems with varying tension forces, perhaps introducing other complicating factors such as an angle in the rope's path or additional forces. This will help them to better grasp how tension interacts with other physical forces in different contexts.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A physics major is working to pay her college tuition by performing in a traveling carnival. She rides a motorcycle inside a hollow, transparent plastic sphere. After gaining sufficient speed, she travels in a vertical circle with radius \(13.0 \mathrm{~m}\). She has mass \(70.0 \mathrm{~kg},\) and her motorcycle has mass \(40.0 \mathrm{~kg} .\) (a) What minimum speed must she have at the top of the circle for the motorcycle tires to remain in contact with the sphere? (b) At the bottom of the circle, her speed is twice the value calculated in part (a). What is the magnitude of the normal force exerted on the motorcycle by the sphere at this point?

A steel ball with mass \(m\) is suspended from the ceiling at the bottom end of a light, 15.0 -m-long rope. The ball swings back and forth like a pendulum. When the ball is at its lowest point and the rope is vertical, the tension in the rope is three times the weight of the ball, so \(T=3 m g .\) (a) What is the speed of the ball as it swings through this point? (b) What is the speed of the ball if \(T=m g\) at this point, where the rope is vertical?

A \(75.0 \mathrm{~kg}\) wrecking ball hangs from a uniform, heavy-duty chain of mass \(26.0 \mathrm{~kg}\). (a) Find the maximum and minimum tensions in the chain. (b) What is the tension at a point three-fourths of the way up from the bottom of the chain?

BIO Stay Awake! An astronaut is inside a \(2.25 \times 10^{6} \mathrm{~kg}\) rocket that is blasting off vertically from the launch pad. You want this rocket to reach the speed of sound \((331 \mathrm{~m} / \mathrm{s})\) as quickly as possible, but astronauts are in danger of blacking out at an acceleration greater than \(4 g\). (a) What is the maximum initial thrust this rocket's engines can have but just barely avoid blackout? Start with a free-body diagram of the rocket. (b) What force, in terms of the astronaut's weight \(w\), does the rocket exert on her? Start with a free-body diagram of the astronaut. (c) What is the shortest time it can take the rocket to reach the speed of sound?

Friction in an Elevator. You are riding in an elevator on the way to the 18 th floor of your dormitory. The elevator is accelerating upward with \(a=1.90 \mathrm{~m} / \mathrm{s}^{2} .\) Beside you is the box containing your new computer; the box and its contents have a total mass of \(36.0 \mathrm{~kg} .\) While the elevator is accelerating upward, you push horizontally on the box to slide it at constant speed toward the elevator door. If the coefficient of kinetic friction between the box and the elevator floor is \(\mu_{\mathrm{k}}=0.32,\) what magnitude of force must you apply?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.