/*! 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 16 A \(700 \mathrm{~g}\) block is r... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 16

A \(700 \mathrm{~g}\) block is released from rest at height \(h_{0}\) 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 \(h_{0} ?(\mathrm{~d})\) If the block were released from height \(2.00 h_{0}\) 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) 1.05 m, (d) 0.38 m.

Step by step solution

01

Calculate the Work Done by the Block on the Spring

The work done by the block on the spring is equal to the potential energy stored in the spring at maximum compression. Use the formula for spring potential energy, \( W = \frac{1}{2} k x^2 \), where \( k = 400 \, \mathrm{N/m} \) and \( x = 0.19 \, \mathrm{m} \).Substitute the values into the formula:\[ W = \frac{1}{2} \times 400 \, \mathrm{N/m} \times (0.19 \, \mathrm{m})^2 = 7.22 \, \mathrm{J} \]
02

Calculate the Work Done by the Spring on the Block

The work done by the spring on the block is equal in magnitude but opposite in direction to the work done by the block on the spring. Therefore, the work done by the spring on the block is \(-7.22 \, \mathrm{J}\).
03

Use Energy Conservation to Find Initial Height (h鈧�)

Apply conservation of energy from the initial height to the point of maximum compression. The potential energy at height \( h_0 \) is transformed into spring energy:\( mgh_0 = \frac{1}{2} k x^2 \).Re-arranging gives:\[ h_0 = \frac{1}{2} \times \frac{400 \, \mathrm{N/m} \times (0.19 \, \mathrm{m})^2}{0.7 \, \mathrm{kg} \times 9.8 \, \mathrm{m/s^2}} \]Calculate to find:\[ h_0 = 1.05 \, \mathrm{m} \]
04

Determine Maximum Compression for a Release Height of 2h鈧�

Using energy conservation from initial height \( 2h_0 \), the gravitational potential energy equates to the new spring energy at maximum compression \( X \):\( mg(2h_0) = \frac{1}{2} k X^2 \).Substitute known quantities:\( 0.7 \, \mathrm{kg} \times 9.8 \, \mathrm{m/s^2} \times 2.1 \, \mathrm{m} = \frac{1}{2} \times 400 \, \mathrm{N/m} \times X^2 \).Solving for \( X \):\( X^2 = 0.01449 \; \Rightarrow \; X = 0.38 \, \mathrm{m} \).

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Key Concepts

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

Spring Potential Energy
When a spring is compressed or stretched, it stores energy in the form of potential energy, known as **spring potential energy**. This energy can be calculated using the formula:
  • The formula is: \( W = \frac{1}{2} k x^2 \)
  • In this equation, \( k \) is the spring constant, which measures the stiffness of the spring.
  • The variable \( x \) is the displacement from the spring鈥檚 equilibrium position.
In our exercise, when the block compresses the spring by \( 0.19 \space \text{m} \), it stores \( 7.22 \space \text{J} \) of energy in the spring. This energy is initially imparted by the block as it comes to rest. Understanding the concept of spring potential energy helps explain how objects interact with springs.
Conservation of Energy
The principle of **conservation of energy** is fundamental in physics. It states that energy cannot be created or destroyed, only transformed from one form to another. In this problem:
  • The initial potential energy when the block is released transforms into spring energy when it compresses the spring.
  • Mathematically, this relationship is expressed as: \( mgh_0 = \frac{1}{2} k x^2 \)
  • Here, \( mgh_0 \) is the gravitational potential energy where \( m \) is mass, \( g \) is acceleration due to gravity, and \( h_0 \) is the height above the spring.
  • On maximum compression, the gravitational energy fully transforms into the spring potential energy.
In the exercise, the block moves from height \( h_0 \) to zero height upon full compression, demonstrating energy conservation as measured by its potential energy transformation into spring energy.
Kinetic and Potential Energy Transformations
Energy transformation between kinetic and potential forms is crucial to analyzing the motion of objects. **Kinetic energy** is the energy of motion, given by \( K = \frac{1}{2} mv^2 \), whereas **potential energy** relates to the position of an object.
In the exercise, as the block is initially at rest at height \( h_0 \), all the energy is potential. When it is released, it starts to move downward, converting potential energy into kinetic energy.
  • During descent, energy shifts from potential to kinetic, until the block makes contact with the spring.
  • The spring then resists the motion, storing energy as potential, halting the block.
  • This full transformation completes when all kinetic energy becomes spring potential energy, marking the compression.
Ultimately, understanding how energy morphs from one type to another is essential for predicting and analyzing physical systems effectively.

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Most popular questions from this chapter

An automobile with passengers has weight \(16400 \mathrm{~N}\) and is moving at \(113 \mathrm{~km} / \mathrm{h}\) when the driver brakes, sliding to a stop. The frictional force on the wheels from the road has a magnitude of \(8230 \mathrm{~N}\). Find the stopping distance.

A \(70 \mathrm{~kg}\) firefighter slides, from rest, \(4.3 \mathrm{~m}\) down a vertical pole. (a) If the firefighter holds onto the pole lightly, so that the frictional force of the pole on her is negligible, what is her speed just before reaching the ground floor? (b) If the firefighter grasps the pole more firmly as she slides, so that the average frictional force of the pole on her is \(500 \mathrm{~N}\) upward, what is her speed just before reaching the ground floor?

A \(5.0 \mathrm{~kg}\) block is projected at \(5.0 \mathrm{~m} / \mathrm{s}\) up a plane that is inclined at \(30^{\circ}\) with the horizontal. How far up along the plane does the block go (a) if the plane is frictionless and (b) if the coefficient of kinetic friction between the block and the plane is \(0.40 ?(\mathrm{c})\) In the latter case, what is the increase in thermal energy of block and plane during the block's ascent? (d) If the block then slides back down against the frictional force, what is the block's speed when it reaches the original projection point?

A pendulum consists of a \(2.0 \mathrm{~kg}\) stone swinging on a 4.0 \(\mathrm{m}\) string of negligible mass. The stone has a speed of \(8.0 \mathrm{~m} / \mathrm{s}\) when it passes its lowest point. (a) What is the speed when the string is at \(60^{\circ}\) to the vertical? (b) What is the greatest angle with the vertical that the string will reach during the stone's motion? (c) If the potential energy of the pendulum-Earth system is taken to be zero at the stone's lowest point, what is the total mechanical energy of the system?

In a circus act, a \(60 \mathrm{~kg}\) clown is shot from a cannon with an initial velocity of \(16 \mathrm{~m} / \mathrm{s}\) at some unknown angle above the horizontal. A short time later the clown lands in a net that is \(3.9 \mathrm{~m}\) vertically above the clown's initial position. Disregard air drag. What is the kinetic energy of the clown as he lands in the net?
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