/*! 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 10 A horizontal rope is tied to a \... [FREE SOLUTION] | 91Ó°ÊÓ

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A horizontal rope is tied to a \(50 \mathrm{kg}\) box on frictionless ice. What is the tension in the rope if: a. The box is at rest? b. The box moves at a steady \(5.0 \mathrm{m} / \mathrm{s} ?\) c. The box has \(v_{x}=5.0 \mathrm{m} / \mathrm{s}\) and \(a_{x}=5.0 \mathrm{m} / \mathrm{s}^{2} ?\)

Short Answer

Expert verified
The tensions are: a. 0N, b. 0N, c. 250N.

Step by step solution

01

Calculate the tension when the box is at rest

When the box is at rest, its acceleration is \(0 \mathrm{m}/\mathrm{s}^{2}\). By Newton's second law, we have \(F_{net}=ma\). Hence, the tension \(T=ma=m(0)=0 \mathrm{N}\).
02

Calculate the tension when the box moves at a steady speed

When the box is moving at a steady speed, its acceleration is still \(0 \mathrm{m}/\mathrm{s}^{2}\) because the velocity is constant. Hence, using \(T=ma\), we find that the tension is still \(T=ma=m(0)=0 \mathrm{N}\).
03

Calculate the tension when the box has certain acceleration

When the box is accelerating, we need to include the acceleration in our calculation of the tension. The acceleration is given as \(5.0 \mathrm{m}/\mathrm{s}^{2}\). Substituting the mass and the acceleration in \(T=ma\), we get \(T=m(a)=50kg(5.0 \mathrm{m}/\mathrm{s}^{2})=250 \mathrm{N}\).

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Key Concepts

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

Newton's second law
When you're learning physics, one of the most crucial principles you'll encounter is Newton's second law of motion. This law fundamentally links the force applied to an object, its mass, and the acceleration it experiences. The law is elegantly summarized by the equation \( F=ma \), where \( F \) represents the net force acting on the object, \( m \) is the object's mass, and \( a \) is the acceleration.

Newton's second law is pivotal in calculating tension in scenarios like the textbook exercise involving a box on a frictionless surface. When the box is at rest or moving at a constant speed, the acceleration (\( a \)) is zero. Consequently, no net force is acting on the object in the direction of the applied tension, thus resulting in zero tension according to Newton's second law. However, once the box begins to accelerate, there is a net force in the direction of the acceleration, meaning the tension equals the mass of the box multiplied by the given acceleration.
Acceleration
Acceleration refers to the rate at which the velocity of an object changes with time. It's a vector quantity, which means it has both magnitude and direction, expressed in units of \( m/s^2 \). An object accelerates if it speeds up, slows down, or changes direction. In our example, when the box on the ice starts moving from rest, the acceleration is the change in velocity over the time taken to change.

Understanding Acceleration in Tension Problems

When dealing with tension in physics problems, especially ones involving acceleration, you need to factor in any change in velocity. If the acceleration is positive, the object is speeding up in the direction of the force. If the acceleration is zero, as in the case where the box moves at a steady \( 5.0 m/s \), the tension remains zero because the velocity is constant and there's no change occurring in the state of motion.
Frictionless motion
Frictionless motion is an idealized concept often used in physics to simplify problems by neglecting the friction force. In reality, almost all surfaces exert some friction, but assuming a scenario without friction allows us to focus on other forces in play, such as tension. When a surface is frictionless, no force opposes the motion of an object sliding over it.

In the exercise's context, the 'frictionless ice' means that we don't need to consider any resistance the ice might provide against the motion of the box. Therefore, any tension in the rope directly affects the motion of the box without any loss to friction. This assumption makes it much easier to apply Newton's second law to find the tension since the only horizontal force we consider is the tension in the rope.
Constant velocity
A constant velocity indicates that an object is moving at the same speed and in the same direction over time. This is an important concept in motion because an object at constant velocity has no net force acting on it in the direction of its motion and, by definition, no acceleration. From Newton's perspective, no acceleration means no change in the state of motion of the object.

What does this imply for our exercise? For part b, where the box is moving at a steady \( 5.0 m/s \), the zero acceleration means the box maintains its constant velocity without any additional force. As a result, the tension in the rope remains zero. This concept is key to understanding why no tension is calculated even though there is motion since the absence of acceleration signifies a steady state of motion without the influence of a net force.

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Most popular questions from this chapter

In an electricity experiment, a \(1.0 \mathrm{~g}\) plastic ball is suspended on a 60 -cm-long string and given an electric charge. A charged rod brought near the ball exerts a horizontal electrical force \(\vec{F}_{\text {elec }}\) on it, causing the ball to swing out to a \(20^{\circ}\) angle and remain there. a. What is the magnitude of \(\vec{F}_{\text {elec }} ?\) b. What is the tension in the string?

Zach, whose mass is \(80 \mathrm{kg}\), is in an elevator descending at \(10 \mathrm{m} / \mathrm{s} .\) The elevator takes \(3.0 \mathrm{s}\) to brake to a stop at the first floor. a. What is Zach's weight before the elevator starts braking? b. What is Zach's weight while the elevator is braking?

Sam, whose mass is \(75 \mathrm{kg}\), takes off down a 50 -m-high, \(10^{\circ}\) slope on his jet-powered skis. The skis have a thrust of \(200 \mathrm{N}\). Sam's speed at the bottom is \(40 \mathrm{m} / \mathrm{s}\). What is the coefficient of kinetic friction of his skis on snow?

Astronauts in space "weigh" themselves by oscillating on a spring. Suppose the position of an oscillating \(75 \mathrm{kg}\) astronaut is given by \(x=(0.30 \mathrm{m}) \sin ((\pi \mathrm{rad} / \mathrm{s}) \cdot t),\) where \(t\) is in \(\mathrm{s}\). What force does the spring exert on the astronaut at (a) \(t=1.0 \mathrm{s}\) and (b) 1.5 s. Note that the angle of the sine function is in radians.

It's a snowy day and you're pulling a friend along a level road on a sled. You've both been taking physics, so she asks what you think the coefficient of friction between the sled and the snow is. You've been walking at a steady \(1.5 \mathrm{m} / \mathrm{s}\), and the rope pulls up on the sled at a \(30^{\circ}\) angle. You estimate that the mass of the sled, with your friend on it, is \(60 \mathrm{kg}\) and that you're pulling with a force of \(75 \mathrm{N}\). What answer will you give?

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