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Chapter 6: Problem 44

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 \(0.0900 \mathrm{~kg}\). The spring has \(k=640 \mathrm{~N} / \mathrm{m}\) and negligible mass. When the spring is released, the glider travels a maximum distance of \(1.80 \mathrm{~m}\) 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 \(0.80 \mathrm{~m}\) 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
The spring was originally compressed by approximately \(x_0\) meters. After the glider has traveled along the air track \(0.80 \mathrm{~m}\) from its initial position against the spring, it is still in/not in contact with the spring (depending on whether \(0.80 \mathrm{~m} \leq x_0\)). The kinetic energy of the glider at this point is the calculated value from Step 3.

Step by step solution

01

Calculate the Potential Energy at the Maximum Height

The maximum potential energy due to gravity when the glider reaches the top of its trajectory is given by \(PE_{gravity} = m \cdot g \cdot h\), where \(m = 0.0900 \mathrm{~kg}\) is the mass of the glider, \(g = 9.8 \mathrm{~m/s^2}\) is the acceleration due to gravity and \(h = 1.80 \cdot \sin(40.0^{\circ}) \mathrm{~m}\) is the vertical height to which the glider moves.
02

Calculate the Total Energy of the System

The total energy of the system when the spring is fully compressed and the glider is about to start moving is equal to the potential energy at the top of the glider's trajectory. This implies that the potential energy of the spring \(PE_{spring} = PE_{gravity}\). Knowing that the potential energy in the spring is given by \(PE_{spring} = 0.5 \cdot k \cdot x^2\), where \(x\) is the spring compression and \(k = 640 \mathrm{~N/m}\) is the spring constant, the equation for the spring's potential energy can be solved for \(x\).
03

Calculate the Glider's Position and Kinetic Energy after 0.80 m

With the spring compression (\(x_0\)) known, it can be determined whether the glider is still in contact with the spring after moving \(0.80 \mathrm{~m}\) along the track. If \(0.80 \mathrm{~m} \leq x_0\), then the glider is still in contact with the spring. Otherwise, the spring has already finished pushing the glider. To calculate the kinetic energy of the glider at this distance, subtract the potential energy due to gravity at this point from the total energy. The result is the kinetic energy: \(KE = PE_{total} - m \cdot g \cdot (0.80 \cdot \sin(40.0^{\circ}))\).

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Key Concepts

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

Conservation of Mechanical Energy
The principle of the conservation of mechanical energy states that the total mechanical energy in a closed system (without any external forces like friction or air resistance acting on it) remains constant throughout the motion of the system. This concept is pivotal when analyzing the movement of objects, such as a glider on an air track, where the system's mechanical energy is converted between potential energy (stored energy) and kinetic energy (energy of motion).

In the provided exercise, the potential energy stored in the compressed spring is converted into kinetic energy as the spring pushes the glider up the track, and then into gravitational potential energy as the glider ascends. The mechanical energy of the glider is conserved during this process, assuming negligible external forces, meaning no energy is lost or gained. This conservation allows us to set up equations equating the potential energy in the spring to the gravitational potential energy at the glider's highest point, helping us solve for variables like the spring compression distance.
Kinetic Energy Calculation
Kinetic energy (KE) is the energy an object possesses due to its motion, which can be calculated using the formula: \( KE = \frac{1}{2} mv^2 \), where \(m\) is the object's mass and \(v\) is its velocity. When the glider is in motion along the air track, it has kinetic energy resulting from the release of the spring's potential energy.

In the exercise, to find the kinetic energy of the glider when it has traveled \(0.80\,\text{m}\) from the starting point, the method involves tracking the energy transformations. By considering the potential energy lost due to elevation change at that distance and the conservation of mechanical energy, we can deduce how much energy remains as kinetic energy. This is a crucial step in understanding energy conversions in systems like the one modeled by the glider and spring.
Energy Conversion in Springs
Springs are common mechanical devices that store potential energy when compressed or stretched. The potential energy in a spring, also known as elastic potential energy, can be expressed using Hooke's Law: \( PE_{spring} = \frac{1}{2} k x^2 \), where \(k\) is the spring constant, representing the spring's stiffness, and \(x\) is the displacement from the spring's equilibrium position.

In the given exercise scenario, as the spring releases from compression, its stored potential energy is converted into kinetic energy that propels the glider up the slope. This conversion involves no energy loss, ideally, if friction is ignored. The glider eventually reaches a point where all the transferred energy from the spring is now in the form of gravitational potential energy, indicating the highest point of its ascent. Understanding this energy transfer is crucial for solving problems related to the motion of objects acted upon by spring forces.

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Most popular questions from this chapter

Object \(A\) has \(27 \mathrm{~J}\) of kinetic energy. Object \(B\) has one-quarter the mass of object \(A\). (a) If object \(B\) also has \(27 \mathrm{~J}\) of kinetic energy, is it moving faster or slower than object \(A\) ? By what factor? (b) By what factor does the speed of each object change if total work \(-18 \mathrm{~J}\) is done on each?

A 12-pack of Omni-Cola (mass \(4.30 \mathrm{~kg}\) ) is initially at rest on a horizontal floor. It is then pushed in a straight line for \(1.20 \mathrm{~m}\) by a trained dog that exerts a horizontal force with magnitude \(36.0 \mathrm{~N}\). Use the work-energy theorem to find the final speed of the 12 -pack if (a) there is no friction between the 12 -pack and the floor, and (b) the coefficient of kinetic friction between the 12 -pack and the floor is 0.30 .

A \(2.50 \mathrm{~kg}\) textbook is forced against one end of a horizontal spring of negligible mass that is fixed at the other end and has force constant \(250 \mathrm{~N} / \mathrm{m}\), compressing the spring a distance of 0.250 m. When released, the textbook slides on a horizontal tabletop with coefficient of kinetic friction \(\mu_{\mathrm{k}}=0.30 .\) Use the work-energy theorem to find how far the textbook moves from its initial position before it comes to rest.

One end of a horizontal spring with force constant \(130.0 \mathrm{~N} / \mathrm{m}\) is attached to a vertical wall. A \(4.00 \mathrm{~kg}\) block sitting on the floor is placed against the spring. The coefficient of kinetic friction between the block and the floor is \(\mu_{\mathrm{k}}=0.400 .\) You apply a constant force \(\vec{F}\) to the block. \(\vec{F}\) has magnitude \(F=82.0 \mathrm{~N}\) and is directed toward the wall. At the instant that the spring is compressed \(80.0 \mathrm{~cm},\) what are (a) the speed of the block, and (b) the magnitude and direction of the block's acceleration?

A force \(\vec{F}\) that is at an angle \(60^{\circ}\) above the horizontal is applied to a box that moves on a horizontal frictionless surface, and the force does work \(W\) as the box moves a distance \(d\). (a) At what angle above the horizontal would the force have to be directed in order for twice the work to be done for the same displacement of the box? (b) If the angle is kept at \(60^{\circ}\) and the box is initially at rest, by what factor would \(F\) have to be increased to double the final speed of the box after moving distance \(d ?\)
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