/*! 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 71 A small block with a mass of \(0... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 71

A small block with a mass of \(0.0600 \mathrm{~kg}\) is attached to a cord passing through a hole in a frictionless, horizontal surface (Fig. \(\mathrm{P} 6.71\) ). The block is originally revolving at a distance of \(0.40 \mathrm{~m}\) from the hole with a speed of \(0.70 \mathrm{~m} / \mathrm{s}\) The cord is then pulled from below, shortening the radius of the circle in which the block revolves to \(0.10 \mathrm{~m}\). At this new distance, the speed of the block is \(2.80 \mathrm{~m} / \mathrm{s}\). (a) What is the tension in the cord in the original situation, when the block has speed \(v=0.70 \mathrm{~m} / \mathrm{s} ?\) (b) What is the tension in the cord in the final situation. when the block has speed \(v=2.80 \mathrm{~m} / \mathrm{s} ?\) (c) How much work was done by the person who pulled on the cord?

Short Answer

Expert verified
The tension in the cord in the original situation is approximately \(0.0665 \ \mathrm{N}\), in the final situation it's \(5.04 \ \mathrm{N}\), and the work done by the person is approximately \(0.47 \ \mathrm{J}\).

Step by step solution

01

Compute original tension

To solve for the original tension, substitute the given values into the equation \(T=mr\omega^2\). However, since \(\omega=v/r\), the formula can be simplified to \(T=mv^2/r\). With \(m=0.0600 \ \mathrm{kg}\), \(v=0.70 \ \mathrm{m/s}\) and \(r=0.40 \ \mathrm{m}\), the tension becomes \(T=0.0600 \ \mathrm{kg} \times (0.70 \ \mathrm{m/s})^2 / 0.40 \ \mathrm{m} = 0.0665 \ \mathrm{N}\).
02

Compute final tension

Apply the same formula for the final tension as in step 1, but this time with the new values: with \(m=0.0600 \ \mathrm{kg}\), \(v=2.80 \ \mathrm{m/s}\) and \(r=0.10 \ \mathrm{m}\), the tension becomes \(T=0.0600 \ \mathrm{kg} \times (2.80 \ \mathrm{m/s})^2 / 0.10 \ \mathrm{m} = 5.04 \ \mathrm{N}\).
03

Compute work done

The work done pulling on the cord is equal to the change in kinetic energy of the block. The kinetic energy can be calculated using the formula \(KE=0.5mv^2\). Compute the initial kinetic energy with the initial speed and the final kinetic energy with the final speed, then subtract the initial energy from the final energy. The work done is therefore \(W=KE_{final} - KE_{initial} = 0.5 \times 0.0600 \ \mathrm{kg} \times (2.80 \ \mathrm{m/s})^2 - 0.5 \times 0.0600 \ \mathrm{kg} \times (0.70 \ \mathrm{m/s})^2 = 0.47 \ \mathrm{J}\).

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Key Concepts

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

Tension in Cord
Tension in a cord refers to the force transmitted through a string, cable, or cord when it is pulled tight by forces acting from opposite ends. It is important to understand the concept of tension, especially when dealing with objects in circular motion, like in the original exercise.
In circular motion, tension acts as the centripetal force needed to keep an object moving in a circle. The formula for tension in the cord when an object is revolving in a circle can be expressed as:
  • \( T = \frac{mv^2}{r} \)
Here, \( T \) represents the tension, \( m \) is the mass of the object, \( v \) is the velocity, and \( r \) is the radius of the circle.
This formula helps determine how the tension changes when any of these variables鈥攎ass, speed, and radius鈥攁re altered. For the given exercise, as the block's speed increases and the radius decreases, the tension significantly rises from 0.0665 N to 5.04 N, illustrating how dynamically they are interconnected.
Work Done
Work done refers to the process of energy transfer from one object to another, usually resulting in some form of mechanical change. In the context of the problem, the work is done by the person pulling the cord and influences the motion of the block significantly.
Mathematically, work done \( W \) by moving an object through distance can be expressed by:
  • \( W = F \cdot d \cdot \cos(\theta) \)
Here, \( F \) is the force applied, \( d \) is the distance over which the force is applied, and \( \theta \) is the angle between the force and direction of motion. In circular motion, work done is more commonly discussed in terms of changes in kinetic energy.
In this exercise, the work done comes from the change in kinetic energy as the radius of the block's path reduces and its speed increases. The change in kinetic energy can be calculated by finding the difference between the initial and final kinetic energy of the block, which results in the work done being 0.47 J. This represents the amount of energy that has been added to the system to change the motion of the block.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. It plays a crucial role in understanding how changing conditions like speed and radius affect the movement of an object in circular motion.
Kinetic energy \( KE \) is quantified by the equation:
  • \( KE = \frac{1}{2}mv^2 \)
Here, \( m \) is the mass of the object and \( v \) is its velocity.
In this particular exercise, the kinetic energy of the block is what changes as the cord is shortened. As the radius is reduced from 0.40 m to 0.10 m, the speed of the block increases from 0.70 m/s to 2.80 m/s, resulting in an increase in kinetic energy.
The initial kinetic energy, when the speed was 0.70 m/s, is calculated to be 0.0147 J. With the speed increase to 2.80 m/s, the final kinetic energy becomes 0.504 J. The significant increase in kinetic energy is a key factor in the analysis of the work done and the dynamics of tension in the exercise.

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Most popular questions from this chapter

A 4.80 kg watermelon is dropped from rest from the roof of an 18.0 m-tall building and feels no appreciable air resistance. (a) Calculate the work done by gravity on the watermelon during its displacement from the roof to the ground. (b) Just before it strikes the ground, what are the watermelon鈥檚 (i) kinetic energy and (ii) speed? (c) Which of the answers in parts (a) and (b) would be different if there were appreciable air resistance?

A luggage handler pulls a \(20.0 \mathrm{~kg}\) suitcase up a ramp inclined at \(32.0^{\circ}\) above the horizontal by a force \(\vec{F}\) of magnitude \(160 \mathrm{~N}\) that acts parallel to the ramp. The coefficient of kinetic friction between the ramp and the incline is \(\mu_{\mathrm{k}}=0.300 .\) If the suitcase travels \(3.80 \mathrm{~m}\) along the ramp, calculate (a) the work done on the suitcase by \(\overrightarrow{\boldsymbol{F}} ;\) (b) the work done on the suitcase by the gravitational force; (c) the work done on the suitcase by the normal force; (d) the work done on the suitcase by the friction force; (e) the total work done on the suitcase. (f) If the speed of the suitcase is zero at the bottom of the ramp, what is its speed after it has traveled \(3.80 \mathrm{~m}\) along the ramp?

A \(75.0 \mathrm{~kg}\) painter climbs a ladder that is \(2.75 \mathrm{~m}\) long and leans against a vertical wall. The ladder makes a \(30.0^{\circ}\) angle with the wall. (a) How much work does gravity do on the painter? (b) Does the answer to part (a) depend on whether the painter climbs at constant speed or accelerates up the ladder?

A spring of force constant \(300.0 \mathrm{~N} / \mathrm{m}\) and unstretched length \(0.240 \mathrm{~m}\) is stretched by two forces, pulling in opposite directions at opposite ends of the spring, that increase to \(15.0 \mathrm{~N}\). How long will the spring now be, and how much work was required to stretch it that distance?

A \(5.00 \mathrm{~kg}\) block is moving at \(v_{0}=6.00 \mathrm{~m} / \mathrm{s}\) along a frictionless, horizontal surface toward a spring with force constant \(k=500 \mathrm{~N} / \mathrm{m}\) that is attached to a wall (Fig. P6.79). The spring has negligible mass. (a) Find the maximum distance the spring will be compressed. (b) If the spring is to compress by no more than \(0.150 \mathrm{~m},\) what should be the maximum value of \(v_{n} ?\)
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