/*! 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 28 Block Sticks to Spring A \(700 \... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 28

Block Sticks to Spring A \(700 \mathrm{~g}\) block is released from rest at height \(h_{1}\) above a vertical spring with spring constant \(k=400 \mathrm{~N} / \mathrm{m}\) and negligible mass. The block sticks to the spring and momentarily stops after compressing the spring \(19.0 \mathrm{~cm}\). How much work is done (a) by the block on the spring and (b) by the spring on the block? (c) What is the value of \(h_{1} ?(\mathrm{~d})\) If the block were released from height \(2 h_{1}\) above the spring, what would be the maximum compression of the spring?

Short Answer

Expert verified
a) 7.22 J, b) -7.22 J, c) 0.075 m, d) 0.268 m.

Step by step solution

01

- Determine the mass in kilograms

Convert the mass of the block from grams to kilograms. The mass of the block is given as 700 g. Therefore: \( m = 700 \times 10^{-3} \text{ kg} = 0.7 \text{ kg} \)
02

- Calculate the work done by the block on the spring

The work done by the block on the spring is equal to the kinetic energy loss and potential energy gain. It is given by the elastic potential energy stored in the spring: \( W = \frac{1}{2} k x^2 \), where \( k = 400 \text{ N/m} \) and \( x = 0.19 \text{ m} \). Therefore: \( W = \frac{1}{2} \times 400 \text{ N/m} \times (0.19 \text{ m})^2 = 7.22 \text{ J} \)
03

- Determine the work done by the spring on the block

The work done by the spring on the block is equal and opposite to the work done by the block on the spring, by conservation of energy principles. Therefore, the work done by the spring would be \( -7.22 \text{ J} \).
04

- Find the height \( h_1 \)

The potential energy due to height \( h_1 \) is converted into the elastic potential energy once the block compresses the spring fully. This is given by: \( mgh_1 = \frac{1}{2} k x^2 \). Rearranging for \( h_1 \): \( h_1 = \frac{\frac{1}{2} k x^2}{mg} \). Substitute \( m = 0.7 \text{ kg} \), \( g = 9.8 \text{ m/s}^2 \), \( k = 400 \text{ N/m} \), and \( x = 0.19 \text{ m} \): \( h_1 = \frac{\frac{1}{2} \times 400 \text{ N/m} \times (0.19 \text{ m})^2}{0.7 \text{ kg} \times 9.8 \text{ m/s}^2} = 0.075 \text{ m} \)
05

- Calculate the maximum compression for release from 2h_1

If the block is released from double the height \( 2h_1 = 2 \times 0.075 \text{ m} = 0.15 \text{ m} \), the energy balance equation holds as: \( 2 mgh_1 = \frac{1}{2} k x_{max}^2 \). Solve for \( x_{max} \): \( x_{max} = \frac{\text{2} mgh_1}{\frac{1}{2} k} \). Substituting \( m = 0.7 \text{ kg} \), \( g = 9.8 \text{ m/s}^2 \), \( k = 400 \text{ N/m} \), and \( h_1 = 0.075 \text{ m} \), we get: \( x_{max} = 0.268 \text{ m} \)

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Key Concepts

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

Work Done by Spring
The concept of work done by a spring is tied to the energy stored or released by the spring's deformation. When a block compresses a spring, the work done by the block on the spring is stored as elastic potential energy in the spring. This can be calculated using the formula: Based on the exercise, the work done by the block on the spring when it compresses by 0.19 meters is given by: Therefore, the work done is 7.22 Joules. It is important to note that the spring does an equal and opposite amount of work on the block, which is Using this knowledge, you can solve part (b) by realizing that the work done by the spring on the block will be
Potential Energy
Potential energy in this context refers to the gravitational potential energy of the block. When the block is released from a height, it converts its potential energy into kinetic energy as it falls, which in turn gets converted into elastic potential energy once it compresses the spring. To find the height from which the block was released, we use the principle of conservation of energy. The gravitational potential energy of the block at height The problem states where We solve for Given: This gives us So, This shows how gravitational potential energy converts to elastic potential energy in the spring.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. As the block falls under the influence of gravity, it gains kinetic energy. This kinetic energy is completely transferred to the spring when the block sticks to it and compresses it. Initially, the block has no kinetic energy since it starts from rest. As it falls, its potential energy is converted into kinetic energy. By the time the block hits the spring, all the gravitational potential energy has turned into kinetic energy. And then, as it compresses the spring, this kinetic energy is converted into elastic potential energy stored in the spring. In part (d), the block is released from a height twice that of The increased height leads to more potential energy, which converts to more kinetic energy, allowing the spring to compress further. Using the same conservation of energy principles applied earlier allows us to find the new maximum compression of the spring. Thus, with higher initial potential energy, the spring compresses more, demonstrating the relationship between kinetic energy and potential energy in a spring-block system.

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Most popular questions from this chapter

Block Dropped on a Spring A \(250 \mathrm{~g}\) block is dropped onto a relaxed vertical spring that has a spring constant of \(k=\) 2.5 N/cm (Fig. 10-32). The block becomes attached to the spring and compresses the spring \(12 \mathrm{~cm}\) before momentarily stopping. While the spring is being compressed, what work is done on the block by (a) the gravitational force on it and (b) the spring force? (c) What is the speed of the block just before it hits the spring? (Assume that friction is negligible.) (d) If the speed at impact is doubled, what is the maximum compression of the spring?

Jumping into a Haystack Tom Sawyer wanders out to the barn one fine summer's day. He notices that a haystack has recently been built just outside the barn. The barn has a second-story door into which the hay will be hauled into the barn by a crane. Tom decides it would be a neat idea to jump out of the second-story door onto the haystack. However, he knows from sad experience that if he jumps out of the second-story door onto the ground, that he is likely to break his leg. Knowing lots of physics, Tom decides to estimate whether the haystack will be able to break his fall. He estimates the height of the haystack to be 3 meters. He presses down on top of the stack and discovers that to compress the stack by \(25 \mathrm{~cm}\), he has to exert a force of about \(50 \mathrm{~N}\). The barn door is 6 meters above the ground. Solve the problem by breaking it into pieces as follows: 1\. Model the haystack by a spring. What is its spring constant? 2\. Is the haystack tall enough to bring his speed to zero? (Estimate using conservation of energy.) 3\. If he does come to a stop before he hits the ground, what will the average force exerted on him be?

Rock Slide During a rockslide, a \(520 \mathrm{~kg}\) rock slides from rest down a hillside that is \(500 \mathrm{~m}\) long and \(300 \mathrm{~m}\) high. The coefficient of kinetic friction between the rock and the hill surface is \(0.25 .\) (a) If the gravitational potential energy \(U\) of the rock-Earth system is zero at the bottom of the hill, what is the value of \(U\) just before the slide? (b) How much energy is transferred to thermal energy during the slide? (c) What is the kinetic energy of the rock as it reaches the bottom of the hill? (d) What is its speed then?

Diatomic Molecule The potential energy of a diatomic molecule (a two-atom system like \(\mathrm{H}_{2}\) or \(\mathrm{O}_{2}\) ) is given by $$ U=\frac{A}{r^{12}}-\frac{B}{r^{6}} $$ where \(r\) is the separation of the two atoms of the molecule and \(A\) and \(B\) are positive constants. This potential energy is associated with the force that binds the two atoms together. (a) Find the equilibrium separation-that is, the distance between the atoms at which the force on each atom is zero. Is the force repulsive (the atoms are pushed apart) or attractive (they are pulled together) if their separation is (b) smaller and (c) larger than the equilibrium separation?

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