/*! 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 47 A small glider is placed against... [FREE SOLUTION] | 91Ó°ÊÓ

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A small glider is placed against a compressed spring at the bottom of an air track that slopes upward at an angle of 40.0\(^\circ\) above the horizontal. The glider has mass 0.0900 kg. The spring has \(k\) = 640 N/m and negligible mass. When the spring is released, the glider travels a maximum distance of 1.80 m along the air track before sliding back down. Before reaching this maximum distance, the glider loses contact with the spring. (a) What distance was the spring originally compressed? (b) When the glider has traveled along the air track 0.80 m from its initial position against the compressed spring, is it still in contact with the spring? What is the kinetic energy of the glider at this point?

Short Answer

Expert verified
(a) Spring was compressed approximately 0.260 m. (b) The glider is no longer in contact at 0.80 m, with kinetic energy around zero due to immediate energy conservation.

Step by step solution

01

Identify and Write Down Given Data

List all the given data and constants from the problem:- Mass of the glider, \( m = 0.0900 \) kg- Spring constant, \( k = 640 \) N/m- Maximum distance the glider travels, \( x_\text{max} = 1.80 \) m- Angle of the slope, \( \theta = 40.0^\circ \)- Gravitational acceleration, \( g = 9.81 \) m/s² (assumed)These will be used in subsequent calculations.
02

Apply Energy Conservation at Maximum Distance

Use the conservation of energy principle at the maximum distance:The initial potential energy in the spring transforms into gravitational potential energy at the maximum distance. This can be expressed as:\[ \frac{1}{2} k x^2 = mgh \]Substitute \( h = x_\text{max} \sin(\theta) \) to find the change in vertical height:\[ \frac{1}{2} (640) x^2 = (0.0900)(9.81)(1.80 \sin(40.0^\circ)) \]
03

Solve for Original Spring Compression (Part a)

Calculate the expression derived in step 2:- Calculate \( h = 1.80 \sin(40.0^\circ) \) to find the height.- Substitute \( h \) into the equation and solve for \( x \): \[ \frac{1}{2} (640) x^2 = (0.0900)(9.81) (1.80 \sin(40.0^\circ)) \]Perform the calculations: - \( x^2 = \frac{2 \times 0.0900 \times 9.81 \times 1.159}{640} \) - \( x \approx 0.260 \) m (approximately)
04

Check Contact After 0.80m Travel (Part b)

For the distance of 0.80 m, calculate the kinetic energy (KE) to see if the glider is in contact with the spring:- The spring releases all energy after being compressed by \( x_\text{max} = 0.260 \) m.Calculate the potential energy stored in the spring when compressed by \( 0.260 - 0.80 \):- If \( 0.80m > x \), there's no contact.- KE can be found using the equation:\[ KE = \frac{1}{2} k (x - 0.80)^2 - mgh \]Given \( x - 0.80 \approx -0.54 \text{ m} \), it's negative, showing no spring potential energy.
05

Calculate Kinetic Energy Given 0.80m Displacement

Substitute values into the kinetic energy and potential energy equation: - Since the spring potential energy becomes zero, and at 0.80m it is beyond the spring: \[ KE = mgh - \frac{1}{2} k (0)^2 \] (spring potential is zero at max uncontacted extension)- Calculate: \[ h\_0.8m = 0.80 \sin(40.0^\circ) \]- KE = \( (0.0900 \times 9.81 \times h\_0.8m) \)Substitute and compute Kinetic energy \( \approx- 0.471 \) Joules (approximately lost to gravitational potential, thus kinetic energy is significantly reduced due to h).

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Key Concepts

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

Spring Force and Compression
The concept of a spring force relates to the energy stored within the spring due to its compression or extension. Springs have a property defined by Hooke's Law, which states that the force exerted by the spring is proportional to its compression or elongation, expressed as:
  • The force, \( F \), is equal to the negative of the spring constant, \( k \), times the displacement, \( x \): \( F = -kx \).
  • The spring constant, \( k \), is a measure of the stiffness of the spring.
Energy stored in a compressed spring is potential energy, calculated as:
  • \( PE = \frac{1}{2} k x^2 \).
When a glider is released from a compressed spring, this potential energy gets converted into kinetic energy as the spring returns to its natural length.
Kinetic Energy
Kinetic Energy (KE) is the energy of motion. It is what a moving object has. When our glider moves along the air track, it has kinetic energy proportional to its mass and the square of its velocity:
  • \( KE = \frac{1}{2} mv^2 \), where \( m \) is the mass, and \( v \) is the velocity.
Kinetic energy changes based on the speed and direction of the moving object. When the spring is released, the stored spring potential energy is transformed into kinetic energy, causing the glider to accelerate along the inclined plane.
Gravitational Potential Energy
Gravitational Potential Energy (GPE) is the energy an object possesses due to its position in a gravitational field, typically relative to the ground. It is affected by:
  • The mass of the object, \( m \).
  • The height above a reference point, \( h \).
  • The gravitational constant, \( g \), commonly \( 9.81 \, \text{m/s}^2 \).
The formula used is:
  • \( GPE = mgh \).
When the glider moves upward along the inclined plane, it converts part of its kinetic energy into gravitational potential energy, slowing down as it gains height.
Mechanical Energy Conservation
The principle of mechanical energy conservation states that in the absence of non-conservative forces like friction, the total mechanical energy of a system remains constant. This total is the sum of kinetic and potential energy:
  • \( E_{total} = KE + PE \).
For the glider on the incline with no friction, we assume all spring potential energy converts into kinetic energy and gravitational potential energy. Therefore:
  • Initial energy stored in the spring equals the energy at the maximum height.
Hence, in this system, as the glider moves up, kinetic energy decreases while gravitational potential energy increases.
Inclined Plane Dynamics
Inclined plane dynamics involve the motion of objects on a slope due to gravitational force. The inclination of the plane affects both the gravitational force component parallel and perpendicular to the slope. The relevant forces include:
  • Gravitational force component parallel to the incline: \( mg \sin(\theta) \).
  • Gravitational force component perpendicular to the incline: \( mg \cos(\theta) \).
As the glider travels upward, its weight's parallel component slows its motion by transforming kinetic energy into gravitational potential energy. Frictionless surfaces simplify calculations, allowing a clear understanding of energy transformations without energy losses due to friction. Knowing these components helps analyze the mechanics of glider motion.

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Most popular questions from this chapter

A mass \(m\) slides down a smooth inclined plane from an initial vertical height \(h\), making an angle \(\alpha\) with the horizontal. (a) The work done by a force is the sum of the work done by the components of the force. Consider the components of gravity parallel and perpendicular to the surface of the plane. Calculate the work done on the mass by each of the components, and use these results to show that the work done by gravity is exactly the same as if the mass had fallen straight down through the air from a height \(h\). (b) Use the work\(-\)energy theorem to prove that the speed of the mass at the bottom of the incline is the same as if the mass had been dropped from height \(h\), independent of the angle \(\alpha\) of the incline. Explain how this speed can be independent of the slope angle. (c) Use the results of part (b) to find the speed of a rock that slides down an icy frictionless hill, starting from rest 15.0 m above the bottom.

A little red wagon with mass 7.00 kg moves in a straight line on a frictionless horizontal surface. It has an initial speed of 4.00 m/s and then is pushed 3.0 m in the direction of the initial velocity by a force with a magnitude of 10.0 N. (a) Use the work\(-\)energy theorem to calculate the wagon's final speed. (b) Calculate the acceleration produced by the force. Use this acceleration in the kinematic relationships of Chapter 2 to calculate the wagon's final speed. Compare this result to that calculated in part (a).

(a) Suppose you cut a massless ideal spring in half. If the full spring had a force constant k, what is the force constant of each half, in terms of \(k\)? (\(Hint\): Think of the original spring as two equal halves, each producing the same force as the entire spring. Do you see why the forces must be equal?) (b) If you cut the spring into three equal segments instead, what is the force constant of each one, in terms of \(k\)?

We usually ignore the kinetic energy of the moving coils of a spring, but let's try to get a reasonable approximation to this. Consider a spring of mass \(M\), equilibrium length \(L_0\), and force constant \(k\). The work done to stretch or compress the spring by a distance \(L\) is \\(\frac{1}{2}\\) \(kX^2\), where \(X = L - L_0\). Consider a spring, as described above, that has one end fixed and the other end moving with speed \(v\). Assume that the speed of points along the length of the spring varies linearly with distance \(l\) from the fixed end. Assume also that the mass \(M\) of the spring is distributed uniformly along the length of the spring. (a) Calculate the kinetic energy of the spring in terms of \(M\) and \(v\). (\(Hint\): Divide the spring into pieces of length \(dl\); find the speed of each piece in terms of \(l\), \(v\), and \(L\); find the mass of each piece in terms of \(dl\), \(M\), and \(L\); and integrate from \(0\) to \(L\). The result is \(not\) \\(\frac{1}{2}\\) \(Mv^2\), since not all of the spring moves with the same speed.) In a spring gun, a spring of mass 0.243 kg and force constant 3200 N/m is compressed 2.50 cm from its unstretched length. When the trigger is pulled, the spring pushes horizontally on a 0.053-kg ball. The work done by friction is negligible. Calculate the ball's speed when the spring reaches its uncompressed length (b) ignoring the mass of the spring and (c) including, using the results of part (a), the mass of the spring. (d) In part (c), what is the final kinetic energy of the ball and of the spring?

A 5.00-kg block is moving at \(\upsilon_0\) \(=\) 6.00 m/s along a frictionless, horizontal surface toward a spring with force constant \(k\) = 500 N/m that is attached to a wall (\(\textbf{Fig. P6.79}\)). The spring has negligible mass. (a) Find the maximum distance the spring will be compressed. (b) If the spring is to compress by no more than 0.150 m, what should be the maximum value of \(\upsilon_0\)?

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