/*! 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 81 A child's pogo stick (Fig. P5.81... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A child's pogo stick (Fig. P5.81) stores energy in a spring \((k=\) \(\left.2.50 \times 10^{4} \mathrm{~N} / \mathrm{m}\right)\). At position (A) \(\left(x_{1}=-0.100 \mathrm{~m}\right)\), the spring compression is a maximum and the child is momentarily at rest. At position (B) \((x=0)\), the spring is relaxed and the child is moving upward. At position (C), the child is again momentarily at rest at the top of the jump. Assuming that Figure p \(5.81\) the combined mass of child and pogo stick is \(25.0 \mathrm{~kg}\), (a) calculate the total energy of the system if both potential energies are zero at \(x=0\), (b) determine \(x_{2}\), (c) calculate the speed of the child at \(x=0\), (d) determine the value of \(x\) for which the kinetic energy of the system is a maximum, and (c) obtain the child's maximum upward speed.

Short Answer

Expert verified
The total energy of the system on the pogo stick remains constant and is equal to the initial potential energy of the spring. \(x_2\) is calculated using the relationship between gravitational potential energy and total energy. The speed of the child at \(x=0\) can be calculated using conservation of total energy. The kinetic energy of the system is maximum at \(x=0\), and the child's maximum upward speed is found by using the conservation of energy principle.

Step by step solution

01

Calculate total energy of the system at position A

The child is momentarily at rest at position A so kinetic energy\( KE = 0 \). The child is at maximum compression of the spring, so there is spring potential energy \( U_{s} = \frac{1}{2} k x_{1}^{2} \), where k is the spring constant and \(x_{1}\) is displacement from the equilibrium position of the spring. There is no gravitational potential energy here as we are taking potential energies to be zero at x=0. So, total energy of the system will be the spring potential energy, \( E_{total} = U_{s} \). Substituting the given values, we will get value for total energy of the system.
02

Calculate displacement \(x_{2}\) at position C

At the top of the jump at position C, the child is momentarily at rest, so \(KE = 0\). Here, the spring is in its relaxed state i.e., \( U_{s} = 0 \). Therefore, all the total energy will now be in the form of gravitational potential energy \( U_{g} = m g x_{2}\), where m is mass of the child, g is acceleration due to gravity and \(x_{2}\) is displacement from the equilibrium position. Total energy of the system will be equal to the gravitational potential energy at this position \( E_{total} = U_{g} \). Solve the equation to get value of \(x_{2}\).
03

Calculate speed of the child at \(x=0\)

At position \(x = 0\), the spring is in its relaxed state i.e., \( U_{s} = 0 \). The child is above ground so there would be gravitational potential energy at this point. The mechanical energy of the system would be both kinetic and potential energy \( E_{total} = KE + U_{g} \). However, we know that the total energy is equal to the kinetic energy (as there's no potential energy) at \(x=0\), so by substituting \( E_{total} = \frac{1}{2} m v^{2}\) we can solve for v.
04

Calculate the value of \(x\) for which kinetic energy of the system is maximum

The kinetic energy of the system is at a maximum when potential energy is at a minimum. By considering the equations for potential energy for both the spring and gravity, we find that at \(x=0\), the potential energy is at it's minimum. Therefore the maximum kinetic energy is at \(x=0\), which is the total energy of the system, as computed already in step 1.
05

Calculate the child's maximum upward speed

The child's maximum upward speed is going to be at the point where the all the potential energy has been converted back into kinetic energy. Using the principle of energy conservation, we can state that total energy at maximum speed is \( E_{total} = KE + U_{g} \). As maximum speed is at \(x=0\) and at \(x=0\) the gravitational potential energy is \( U_{g} = 0 , so E_{total} = KE = \frac{1}{2} m V_{max}^{2} \). Solving the equation for \(V_{max}\), we find the maximum speed of the child.

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.

Pogo Stick Physics
Pogo sticks, a playful symbol of childhood fun, also offer an engaging example of how physics principles like energy conservation operate. When a child jumps on a pogo stick, they are actively converting between different forms of energy. Think of a pogo stick as a dynamic dance between spring potential energy, gravitational potential energy, and kinetic energy.
Understanding this energy conversion is crucial to grasp how pogos work. When the spring is compressed (at position A in our original problem), energy is stored in the form of spring potential energy. As the pogo stick releases this energy, the spring pushes, propelling the child upwards towards position C, transforming stored energy into kinetic energy and some gravitational potential energy.
  • The cycle is fascinating because it continuously transforms energy between potential and kinetic states.
  • Real-world pogo stick movements serve as a practical manifestation of mechanical energy conservation principles.
Comprehending pogo stick physics provides an appreciation of energy transitions in everyday toys, shedding light on core physics concepts.
Spring Potential Energy
Spring potential energy is a fascinating concept that explains how springs store energy. In the context of the pogo stick, spring potential energy is pivotal when the spring is either compressed or extended.
At position A, the pogo stick's spring is maximally compressed. Spring potential energy here can be calculated using the formula, \( U_s = \frac{1}{2} k x^2 \),
where:
  • \(k\) is the spring constant, indicating the spring's stiffness.
  • \(x\) is the compression distance from the spring's equilibrium position.
It's essential to note how energy stored in a compressed spring can be transferred effectively into mechanical actions, as seen in our example. Mechanical energy transformation from spring potential to kinetic energy forms the backbone of the pogo stick's movement. This conversion exemplifies a broader physics principle: the conservation of mechanical energy from potential to kinetic states and vice versa.
Kinetic Energy Calculations
Kinetic energy is the energy of motion. For the pogo stick in our example, once the spring is released at position A, the child propelled by the spring begins to move upward, converting stored spring potential energy into kinetic energy.
The kinetic energy of a system is calculated with the formula, \( KE = \frac{1}{2} m v^2 \), where:
  • \(m\) is the mass of the object in motion.
  • \(v\) is the velocity of the object.
By applying the conservation of total mechanical energy, we find that when all spring potential energy is converted to kinetic energy, the child achieves maximum speed. This concept is pivotal in understanding objects in motion, highlighting the interplay of forces and energy.
At position B, where the child's velocity is maximum, it's crucial to equate the original spring potential energy to the kinetic energy at this point to find this peak speed.
Gravitational Potential Energy
Gravitational potential energy (GPE) is a crucial aspect of the pogo stick problem. At position C, when the child is momentarily at rest at the peak of the jump, the energy transformed into gravitational potential energy.
The amount of gravitational potential energy at this point is represented by:\( U_g = m g h \), where:
  • \(m\) is the mass of the child and pogo stick.
  • \(g\) is the acceleration due to gravity (approximately \(9.81 \text{ m/s}^2\) on Earth).
  • \(h\) is the height from the reference point, typically the ground or the relaxed spring state (at \(x=0\)).
In the case of the pogo stick, understanding how kinetic energy transforms into gravitational potential energy as the child moves upwards is crucial. Once again, the principle of energy conservation plays a vital role. As the child descends, GPE converts back to kinetic energy and, upon spring compression, to spring potential energy, illustrating the continuous energy transformation cycle.

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

In 1990 Walter Arfeuille of Belgium lifted a \(281.5-\mathrm{kg}\) object through a distance of \(17.1 \mathrm{~cm}\) using only his teeth. (a) How much work did Arfeuille do on the object? (b) What magnitude force did he exert on the object during the lift, assuming the force was constant?

A \(7.80-\mathrm{g}\) bullet moving at \(575 \mathrm{~m} / \mathrm{s}\) penetrates a tree trunk to a depth of \(5.50 \mathrm{~cm}\). (a) Use work and energy considerations to find the average frictional force that stops the bullet. (b) Assuming the frictional force is constant, determine how much time elapses between the moment the bullet enters the tree and the moment it stops moving.

A toy gun uses a spring to project a \(5.3-g\) soft rubber sphere horizontally. The spring constant is \(8.0 \mathrm{~N} / \mathrm{m}\), the barrel of the gun is \(15 \mathrm{~cm}\) long, and a constant frictional force of \(0.032 \mathrm{~N}\) exists between barrel and projectile. With what speed does the projectile leave the barrel if the spring was compressed \(5.0 \mathrm{~cm}\) for this launch?

Bio In a needle biopsy, a narrow strip of tissue is extracted from a patient with a hollow needle. Rather than being pushed by hand, to ensure a clean cut the needle can be fired into the patient's body by a springAssume the needle has mass \(5.60 \mathrm{~g}\), the light spring has force constant \(375 \mathrm{~N} / \mathrm{m}\), and the spring is originally compressed \(8.10 \mathrm{~cm}\) to project the needle horizontally without friction. The tip of the needle then moves through \(2.40 \mathrm{~cm}\) of skin and soft tissue, which exerts a resistive force of \(7.60 \mathrm{~N}\) on it. Next, the needle cuts \(3.50 \mathrm{~cm}\) into an organ, which exerts a backward force of \(9.20 \mathrm{~N}\) on it. Find (a) the maximum speed of the needle and (b) the speed at which a flange on the back end of the needle runs into a stop, set to limit the penetration to \(5.90 \mathrm{~cm}\).

Tarzan swings on a \(30.0\)-m-long vine initially inclined at an angle of \(37.0^{\circ}\) with the vertical. What is his speed at the bottom of the swing (a) if he starts from rest? (b) If he pushes off with a speed of \(4.00 \mathrm{~m} / \mathrm{s}\) ?

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.