/*! 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 0.0... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 71

A small block with a mass of 0.0600 kg is attached to a cord passing through a hole in a frictionless, horizontal surface (\(\textbf{Fig. P6.71}\)). The block is originally revolving at a distance of 0.40 m from the hole with a speed of 0.70 m/s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.10 m. At this new distance, the speed of the block is 2.80 m/s. (a) What is the tension in the cord in the original situation, when the block has speed \(\upsilon = 0.70\) m/s? (b) What is the tension in the cord in the final situation, when the block has speed \(\upsilon = 2.80\) m/s? (c) How much work was done by the person who pulled on the cord?

Short Answer

Expert verified
(a) 0.0735 N (b) 4.704 N (c) 0.2205 J

Step by step solution

01

Understanding the Initial Tension Calculation

To find the tension in the cord when the block is revolving at a speed of 0.70 m/s and radius 0.40 m, we use the centripetal force formula. The tension force provides the necessary centripetal force, so it equals \( T = \frac{m v^2}{r} \). Here, \( m = 0.0600 \text{ kg} \), \( v = 0.70 \text{ m/s} \), and \( r = 0.40 \text{ m} \).
02

Calculating the Initial Tension

Substitute the known values into the centripetal force formula: \[T = \frac{0.0600 \times (0.70)^2}{0.40}\]Calculate to find:\[T = \frac{0.0600 \times 0.49}{0.40} = \frac{0.0294}{0.40} = 0.0735 \text{ N}\]So, the tension in the cord initially is 0.0735 N.
03

Understanding the Final Tension Calculation

For the final scenario, we again use the formula for centripetal force with the new speed and radius of the revolution. The tension in the cord when \( v = 2.80 \text{ m/s} \) and \( r = 0.10 \text{ m} \) is given by\[T = \frac{m v^2}{r}.\]
04

Calculating the Final Tension

Substitute the new values into the centripetal force formula:\[T = \frac{0.0600 \times (2.80)^2}{0.10}\]Calculate to find:\[T = \frac{0.0600 \times 7.84}{0.10} = \frac{0.4704}{0.10} = 4.704 \text{ N}\]Thus, the tension in the cord in the final situation is 4.704 N.
05

Understanding Work Done by the Force

Work done by the person pulling the cord can be found using the work-energy principle, which states that the work done is the change in kinetic energy. Calculate the initial and final kinetic energies and find their difference. Initial kinetic energy:\[ K_i = \frac{1}{2} m v_i^2 \]where \( v_i = 0.70 \text{ m/s} \).Final kinetic energy:\[ K_f = \frac{1}{2} m v_f^2 \]where \( v_f = 2.80 \text{ m/s} \).
06

Calculating Initial and Final Kinetic Energies

Calculate the initial kinetic energy:\[K_i = \frac{1}{2} \times 0.0600 \times (0.70)^2 = \frac{1}{2} \times 0.0600 \times 0.49 = 0.0147 \text{ J}\]Calculate the final kinetic energy:\[K_f = \frac{1}{2} \times 0.0600 \times (2.80)^2 = \frac{1}{2} \times 0.0600 \times 7.84 = 0.2352 \text{ J}\]
07

Calculating the Work Done

The work done (W) by pulling the cord is the change in kinetic energy:\[W = K_f - K_i = 0.2352 \text{ J} - 0.0147 \text{ J} = 0.2205 \text{ J}\]So, the work done by the person is 0.2205 J.

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Key Concepts

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

Kinetic Energy
Kinetic energy is an essential concept in understanding how energy moves with objects as they speed up or slow down. It reflects the energy an object possesses due to its motion. For a given object, the kinetic energy is calculated using the formula:\[ K = \frac{1}{2} mv^2 \]where:- *m* is the mass of the object (in kilograms),- *v* is the velocity of the object (in meters per second).In our scenario, a small block moves at varying speeds depending on how far it is from the hole it's revolving around. Initially, the block has a kinetic energy of 0.0147 J when revolving at 0.70 m/s.When the radius decreases and its speed increases to 2.80 m/s, the block's kinetic energy spikes to 0.2352 J. This showcases how significantly kinetic energy enhances as speed multiplies, demonstrating that kinetic energy is very sensitive to changes in speed due to the squared factor of velocity.
Tension
Tension is a force exerted by a string or rope when it is pulled tight by forces acting from opposite ends. In circular motion, such as that of a block attached to a cord, tension provides the necessary centripetal force for the block to maintain its circular path.For each scenario, tension is computed as:\[ T = \frac{mv^2}{r} \]where:- *m* is the mass,- *v* is the velocity,- *r* is the radius of rotation.In the initial setting, tension is calculated to be 0.0735 N at a speed of 0.70 m/s with a radius of 0.40 m. However, when the radius is reduced to 0.10 m and speed increases to 2.80 m/s, tension drastically rises to 4.704 N. This shows how tension, directly related to speed and inversely related to radius, increases sharply with a decrease in radius and increase in velocity.
Work-Energy Principle
The work-energy principle bridges the concept of work done on an object and its kinetic energy change. It states that the work done by forces on an object results in a change in its kinetic energy. The work done can be calculated by:\[ W = \Delta K = K_f - K_i \]where:- \( K_f \) is the final kinetic energy,- \( K_i \) is the initial kinetic energy.In the exercise involving the rotating block, the work done by pulling the cord from below alters its kinetic energy from 0.0147 J to 0.2352 J. As a result, the work done is measured as 0.2205 J. This perfectly illustrates how an increase in kinetic energy, due to increasing speed, reflects the work applied on the block, thereby making the block move faster along a tighter circle.

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Most popular questions from this chapter

A net force along the \(x\)-axis that has \(x\)-component \(F{_x}= -12.0\mathrm{N} +(0.300\mathrm{N/m{^2}})x{^2}\) is applied to a 5.00-kg object that is initially at the origin and moving in the \(-x\)-direction with a speed of 6.00 m/s. What is the speed of the object when it reaches the point \(x = 5.00\) m?

A luggage handler pulls a 20.0-kg suitcase up a ramp inclined at 32.0\(^\circ\) above the horizontal by a force \(\overrightarrow{F}\) of magnitude 160 N that acts parallel to the ramp. The coefficient of kinetic friction between the ramp and the incline is \(\mu_k = 0.300\). If the suitcase travels 3.80 m along the ramp, calculate (a) the work done on the suitcase by \(\overrightarrow{F}\); (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 m along the ramp?

A box is sliding with a speed of 4.50 m/s on a horizontal surface when, at point \(P\), it encounters a rough section. The coefficient of friction there is not constant; it starts at 0.100 at \(P\) and increases linearly with distance past \(P\), reaching a value of 0.600 at 12.5 m past point \(P\). (a) Use the work\(-\)energy theorem to find how far this box slides before stopping. (b) What is the coefficient of friction at the stopping point? (c) How far would the box have slid if the friction coefficient didn't increase but instead had the constant value of 0.100?

Consider a spring that does not obey Hooke's law very faithfully. One end of the spring is fixed. To keep the spring stretched or compressed an amount \(x\), a force along the \(x\)-axis with \(x\)-component \(F_x = kx - bx^2 + cx^3\) must be applied to the free end. Here \(k = 100 \, \mathrm {N/m}\), \(b = 700 \, \mathrm {N/m{^2}}\), and \(c = 12,000 \, \mathrm{N/m}^3\). Note that \(x > 0\) when the spring is stretched and \(x< 0\) when it is compressed. How much work must be done (a) to stretch this spring by 0.050 m from its unstretched length? (b) To \(compress\) this spring by 0.050 m from its unstretched length? (c) Is it easier to stretch or compress this spring? Explain why in terms of the dependence of \(F_x\) on \(x\). (Many real springs behave qualitatively in the same way.)

A 128.0-N carton is pulled up a frictionless baggage ramp inclined at 30.0\(^\circ\) above the horizontal by a rope exerting a 72.0-N pull parallel to the ramp's surface. If the carton travels 5.20 m along the surface of the ramp, calculate the work done on it by (a) the rope, (b) gravity, and (c) the normal force of the ramp. (d) What is the net work done on the carton? (e) Suppose that the rope is angled at 50.0\(^\circ\) above the horizontal, instead of being parallel to the ramp's surface. How much work does the rope do on the carton in this case?
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