/*! 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 35 In Fig. 8 -42, a block of mass \... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In Fig. 8 -42, a block of mass \(m=3.20 \mathrm{~kg}\) slides from rest a distance \(d\) down a frictionless incline at angle \(\theta=30.0^{\circ}\) where it runs into a spring of spring constant \(431 \mathrm{~N} / \mathrm{m} .\) When the block momentarily stops, it has compressed the spring by \(21.0 \mathrm{~cm} .\) What are (a) distance \(d\) and (b) the distance between the point of the first block-spring contact and the point where the block's speed is greatest?

Short Answer

Expert verified
(a) Distance \(d\) is 0.174 m. (b) The distance for max speed before contact is also 0.174 m.

Step by step solution

01

Understand Energy Conservation

This problem involves mechanical energy conservation. Since the incline is frictionless, the total mechanical energy remains constant throughout the motion of the block. As the block slides down, its gravitational potential energy is converted into kinetic energy and finally into the elastic potential energy of the spring.
02

Set Up Energy Conservation Equation

At the start, the block has only gravitational potential energy, which is given by \(mgh\), where \(h\) is the height. At the point where the spring is compressed, all potential energy is converted to elastic potential energy of the spring, given by \(\frac{1}{2}kx^2\), where \(k\) is the spring constant and \(x\) is the compression in the spring.Equating the gravitational potential energy at the top of the incline to the spring potential energy at maximum compression:\[mgh = \frac{1}{2}kx^2\]\[mgd \sin \theta = \frac{1}{2}kx^2\] since \(h = d \sin \theta\).
03

Solve for Distance d

Rearrange the equation to find \(d\):\[d = \frac{kx^2}{2mg\sin\theta}\]Substitute the known values: \(m = 3.20\text{ kg},\ g = 9.81\text{ m/s}^2,\ k = 431\text{ N/m},\ x = 0.21\text{ m},\ \theta = 30^{\circ}\).\[d = \frac{431 \times (0.21)^2}{2 \times 3.20 \times 9.81 \times \sin(30^{\circ})} \approx 0.174\text{ m}\]
04

Determine Point of Maximum Speed

The point where the block's speed is greatest is when all its potential energy is converted into kinetic energy and none into the spring's compression potential energy. This occurs right before it starts compressing the spring. Hence, the distance between the point of first contact and the maximum speed contact is simply the initial position of the spring relative to the incline.
05

Calculate the Distance Between Contacts

Since the block's speed is greatest just before touching the spring, this means it travels the entire distance \(d\) before compressing it. The calculated distance where the block is at maximum speed from where it first contacts the spring is thus \(d = 0.174\text{ m}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Mechanical Energy
Mechanical energy is a broad term used in physics that represents the sum of both kinetic and potential energy in a system. In our exercise, we're dealing with a frictionless incline, which allows us to focus on the conservation of mechanical energy. Mechanical energy remains consistent if no non-conservative forces, like friction or air resistance, are acting upon the object. Mechanical energy helps us understand how energy shifts forms while still maintaining the total amount as constant. For a block sliding down an incline, the transition of energy goes from gravitational potential energy to kinetic energy, and eventually to elastic potential energy when the block compresses a spring.
  • Total mechanical energy is the sum of kinetic and potential energies.
  • In a frictionless system, the total mechanical energy remains constant.
By appreciating mechanical energy conservation, we can deduce calculations accurately, such as determining the distance a block travels down an incline.
Gravitational Potential Energy
Gravitational Potential Energy (GPE) is the energy stored in an object due to its position in a gravitational field, typically related to its height above the ground. In standard calculations, GPE is represented by the formula \(mgh\), where:
  • \(m\) is mass,
  • \(g\) is the acceleration due to gravity,
  • \(h\) is the height above a reference point.
In our scenario, as the block descends the incline, its height decreases, converting its gravitational potential energy into other energy forms. The critical concept here is recognizing how GPE initially provides the energy needed for motion, subsequently turning into kinetic and then elastic potential energy as it interacts with the spring.Gravitational potential energy changes are essential when calculating distance traveled, \(d\), from the top of an incline to the point of compression into the spring, using the relationship \(h = d \sin \theta\). This highlights how gravitational influence manifests in energy transformations.
Kinetic Energy
Kinetic energy refers to the energy an object possesses due to its movement, with the formula \(\frac{1}{2}mv^2\). It's a key player in our incline problem, as kinetic energy increases as the block slides downward, driven by gravitational forces.Before the block compresses the spring, it reaches its maximum speed and thus its maximum kinetic energy. This is an important point because, at the moment just before the block touches the spring, all gravitational potential energy has turned into kinetic.
  • Kinetic energy is highest at the steepest points of acceleration.
  • It represents transformed gravitational potential energy during the descent.
Acknowledging these energy stages, understanding where kinetic energy reaches its peak and the transition to elastic potential energy allows us to pinpoint moments like maximum speed precisely before the spring interaction.
Elastic Potential Energy
Elastic potential energy is the energy stored in elastic materials, such as springs, when they are compressed or stretched. In our situation, it's described by \(\frac{1}{2}kx^2\), where:
  • \(k\) is the spring constant, which measures the stiffness of the spring.
  • \(x\) is the compression or stretching distance of the spring from its equilibrium position.
As the block slides down the incline and compresses the spring, the kinetic energy is transferred into elastic potential energy, causing the spring to deform. At this maximum compression, kinetic energy is zero because the entire kinetic energy has been converted into stored elastic energy within the spring.This conversion is crucial for calculating the spring's compression distance and understanding the motion dynamics in energy terms. Recognizing how energy is conserved, switching forms but maintaining its total, is fundamental to analyzing similar physical systems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

When a click beetle is upside down on its back, it jumps upward by suddenly arching its back, transferring energy stored in a muscle to mechanical energy. This launching mechanism produces an audible click, giving the beetle its name. Videotape of a certain clickbeetle jump shows that a beetle of mass \(m=4.0 \times 10^{-6} \mathrm{~kg}\) moved directly upward by \(0.77 \mathrm{~mm}\) during the launch and then to a maximum height of \(h=0.30 \mathrm{~m}\). During the launch, what are the average magnitudes of (a) the external force on the beetle's back from the floor and (b) the acceleration of the beetle in terms of \(g ?\)

Next, the particle is released from rest at \(x=0 .\) What are (l) its kinetic energy at \(x=5.0 \mathrm{~m}\) and \((\mathrm{m})\) the maximum positive position \(x_{\max }\) it reaches? (n) What does the particle do after it reaches \(x_{\max } ?\) $$ \begin{array}{lc} {\text { Range }} & \text { Force } \\ \hline 0 \text { to } 2.00 \mathrm{~m} & \vec{F}_{1}=+(3.00 \mathrm{~N}) \hat{\mathrm{i}} \\ 2.00 \mathrm{~m} \text { to } 3.00 \mathrm{~m} & \vec{F}_{2}=+(5.00 \mathrm{~N}) \hat{\mathrm{i}} \\ 3.00 \mathrm{~m} \text { to } 8.00 \mathrm{~m} & F=0 \\ 8.00 \mathrm{~m} \text { to } 11.0 \mathrm{~m} & \vec{F}_{3}=-(4.00 \mathrm{~N}) \hat{\mathrm{i}} \\ 11.0 \mathrm{~m} \text { to } 12.0 \mathrm{~m} & \vec{F}_{4}=-(1.00 \mathrm{~N}) \hat{\mathrm{i}} \\ 12.0 \mathrm{~m} \text { to } 15.0 \mathrm{~m} & F=0 \end{array} $$ For the arrangement of forces in Problem \(81,\) a \(2.00 \mathrm{~kg}\) particle is released at \(x=5.00 \mathrm{~m}\) with an initial velocity of \(3.45 \mathrm{~m} / \mathrm{s}\) in the negative direction of the \(x\) axis. (a) If the particle can reach \(x=0 \mathrm{~m},\) what is its speed there, and if it cannot, what is its turning point? Suppose, instead, the particle is headed in the positive \(x\) direction when it is released at \(x=5.00 \mathrm{~m}\) at speed \(3.45 \mathrm{~m} / \mathrm{s} .\) (b) If the particle can reach \(x=13.0 \mathrm{~m},\) what is its speed there, and if it cannot, what is its turning point?

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?

A \(30 \mathrm{~g}\) bullet moving a horizontal velocity of \(500 \mathrm{~m} / \mathrm{s}\) comes to a stop \(12 \mathrm{~cm}\) within a solid wall. (a) What is the change in the bullet's mechanical energy? (b) What is the magnitude of the average force from the wall stopping it?

Tarzan, who weighs 688 N, swings from a cliff at the end of a vine \(18 \mathrm{~m}\) long \((\) Fig. \(8-40) .\) From the top of the cliff to the bottom of the swing, he descends by \(3.2 \mathrm{~m}\). The vine will break if the force on it exceeds \(950 \mathrm{~N}\). (a) Does the vine break? (b) If no, what is the greatest force on it during the swing? If yes, at what angle with the vertical does it break?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.