/*! 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 11 A block of mass \(0.250 \mathrm{... [FREE SOLUTION] | 91Ó°ÊÓ

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A block of mass \(0.250 \mathrm{kg}\) is placed on top of a light vertical spring of force constant \(5000 \mathrm{N} / \mathrm{m}\) and pushed downward so that the spring is compressed by \(0.100 \mathrm{m}\). After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise?

Short Answer

Expert verified
The maximum height the block will reach above the point of release is approximately 10.2 m.

Step by step solution

01

Calculate the spring's potential energy

First, calculate the potential energy in the spring when the block is released. This can be found using the formula for elastic potential energy \(U = 0.5 * k * x^2\), where \(k\) is the spring constant and \(x\) is the displacement (compression). Plugging the given values into the formula, we find \(U = 0.5 * 5000 N/m * (0.100 m)^2 = 25 J\).
02

Calculate the maximum height

Next, the maximum height can be found by setting the elastic potential energy equal to the gravitational potential energy at the maximum height. The formula for gravitational potential energy is \(U = mgh\), where \(m\) is the mass, \(g\) is the acceleration due to gravity, and \(h\) is the height. We can solve for \(h\) as follows: \(25 J = 0.250 kg * 9.81 m/s^2 * h\). Solving for \(h\) gives \(h = 25 J / (0.250 kg * 9.81 m/s^2) = 10.2 m\).

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Key Concepts

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

Spring Potential Energy
Spring potential energy is the energy stored in a spring when it is either compressed or stretched from its equilibrium position. This is a kind of mechanical energy, and it is known as elastic potential energy. When a spring is compressed or stretched, it can perform work, such as moving an object attached to it. The amount of energy stored depends on two factors:
1. The spring constant (k) - This represents the stiffness of the spring. A higher spring constant means the spring is stiffer and requires more force to compress or stretch.
2. The displacement (x) - This is the distance the spring is compressed or stretched.

The formula to calculate spring potential energy (U) is given by:
\[ U = \frac{1}{2} k x^2 \]
This equation tells us that the potential energy is directly proportional to both the spring constant and the square of the displacement. Thus, even a small change in the displacement can greatly increase the energy stored in the spring, making springs efficient at storing and releasing energy.
Gravitational Potential Energy
Gravitational potential energy refers to the energy an object possesses due to its position in a gravitational field. It is a form of potential energy associated with the height of an object above a reference point, such as the ground.
The amount of gravitational potential energy depends on:
1. Mass of the object (m) - More massive objects have higher potential energy for the same height.
2. Height (h) - The higher the object is from the reference point, the more potential energy it has.
3. Gravitational acceleration (g) - On Earth, this is approximately \(9.81 \ m/s^2\).

The formula for gravitational potential energy (U) is:
\[ U = mgh \]
In the given exercise, you can calculate how much potential energy the block possesses at its maximum height by using its mass, the height it reaches, and the gravitational acceleration. This concept is crucial in analyzing how energy transforms in mechanical systems.
Energy Conservation
The principle of energy conservation states that energy cannot be created or destroyed; it can only be transformed from one form to another. In mechanics, it often involves the conversion of potential energy to kinetic energy and vice versa. When dealing with springs and gravity, you often witness these transformations.
In the exercise, when the spring is compressed, it has maximum spring potential energy. Once released, this energy is converted to kinetic energy as the block moves, and eventually to gravitational potential energy when the block reaches its maximum height above the point of release.
This transformation is seamless because of energy conservation:
- Initial spring potential energy = Final gravitational potential energy
This equation ensures that the total mechanical energy of the system remains constant, assuming no energy is lost to friction or air resistance.
Spring Force Constant
The spring force constant, often represented as k, is a crucial parameter in determining how a spring behaves when compressed or stretched. It is measured in Newtons per meter (N/m) and quantifies the stiffness of a spring. A spring with a high force constant is stiffer and harder to compress.
This parameter appears in Hooke's Law, which states that the force needed to compress or stretch a spring by a certain distance (x) is directly proportional to that distance:
\[ F = kx \]
Where F is the force applied to the spring. The same constant, k, is used in the spring potential energy formula, illustrating its influence in both force dynamics and energy storage.
This constant is essential in calculating how much work is needed to compress the spring and how much energy is stored when the spring is compressed by a certain amount, like in our exercise.

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Most popular questions from this chapter

A block of mass \(M\) rests on a table. It is fastened to the lower end of a light vertical spring. The upper end of the spring is fastened to a block of mass \(m .\) The upper block is pushed down by an additional force \(3 m g\), so the spring compression is \(4 \mathrm{mg} / \mathrm{k}\). In this configuration the upper block is released from rest. The spring lifts the lower block off the table. In terms of \(m,\) what is the greatest possible value for \(M ?\)

A light rigid rod is \(77.0 \mathrm{cm}\) long. Its top end is pivoted on a low- friction horizontal axle. The rod hangs straight down at rest with a small massive ball attached to its bottom end. You strike the ball, suddenly giving it a horizontal velocity so that it swings around in a full circle. What minimum speed at the bottom is required to make the ball go over the top of the circle?

A single conservative force acts on a \(5.00-\mathrm{kg}\) particle. The equation \(F_{x}=(2 x+4)\) N describes the force, where \(x\) is in meters. As the particle moves along the \(x\) axis from \(x=1.00 \mathrm{m}\) to \(x=5.00 \mathrm{m},\) calculate (a) the work done by this force, (b) the change in the potential energy of the system, and \((\mathrm{c})\) the kinetic energy of the particle at \(x=5.00 \mathrm{m}\) if its speed is \(3.00 \mathrm{m} / \mathrm{s}\) at \(x=1.00 \mathrm{m}\).

A simple pendulum, which we will consider in detail in Chapter \(15,\) consists of an object suspended by a string. The object is assumed to be a particle. The string, with its top end fixed, has negligible mass and does not stretch. In the absence of air friction, the system oscillates by swinging back and forth in a vertical plane. If the string is \(2.00 \mathrm{m}\) long and makes an initial angle of \(30.0^{\circ}\) with the vertical, calculate the speed of the particle (a) at the lowest point in its trajectory and (b) when the angle is \(15.0^{\circ}\).

Air moving at \(11.0 \mathrm{m} / \mathrm{s}\) in a steady wind encounters a windmill of diameter \(2.30 \mathrm{m}\) and having an efficiency of \(27.5 \% .\) The energy generated by the windmill is used to pump water from a well \(35.0 \mathrm{m}\) deep into a tank \(2.30 \mathrm{m}\) above the ground. At what rate in liters per minute can water be pumped into the tank?

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