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GP A horizontal spring attached to a wall has a force constant of \(850 \mathrm{~N} / \mathrm{m}\). A block of mass \(1.00 \mathrm{~kg}\) is attached to the spring and oscillates freely on a horizontal, frictionless surface as in Figure \(5.20\). The initial goal of this problem is to find the velocity at the equilibrium point after the block is released. (a) What objects constitute the system, and through what forces do they interact? (b) What are the two points of interest? (c) Find the energy stored in the spring when the mass is stretched \(6.00 \mathrm{~cm}\) from equilibrium and again when the mass passes through equilibrium after being released from rest. (d) Write the conservation of energy equation for this situation and solve it for the speed of the mass as it passes equilibrium. Substitute to obtain a numerical value. (c) What is the speed at the halfway point? Why isn't it half the speed at equilibrium?

Step by step solution

01

Identify the system and the interacting forces

The system consists of the spring-block system. They interact through the spring force, which is the restoring force that tends to bring the block to the equilibrium position. The force can be described using Hooke's law, F_spring = -k*x, where k is the spring constant and x is the displacement from the equilibrium position.

02

Identify the two points of interest

The two points of interest are when the spring is stretched \6.00 \mathrm{~cm}\) from its equilibrium position and when the block is at the equilibrium position after being released.

03

Calculate the energy stored in the spring

The energy stored in a spring when it is displaced from its equilibrium position can be given by the elastic potential energy formula, U = 1/2 * k * x², where k is the spring constant and x is the displacement from equilibrium. When the spring is at \6.00 \mathrm{~cm}\), or \0.06 \mathrm{~m}\), the energy stored it in is U = 1/2 * 850 N/m * (0.06 m)² = 1.53 Joules. When the block is at the equilibrium, the displacement x is zero, hence the potential energy is also zero.

04

Derive the conservation of energy equation and solve

The principle of conservation of energy can be written for this system as Kinetic energy at equilibrium = Elastic potential energy at stretch. This gives, 1/2 * m * v² = 1/2 * k * x². From this, v can be calculated as \(v = \sqrt{k/m} * x\). Substitute the given values to get the speed \(v = \sqrt{850 \mathrm{N/m}/1.00 \mathrm{kg}} * 0.06 \mathrm{m} = 0.70 \mathrm{m/s}\).

05

Calculate the speed at the halfway point

At the halfway point, i.e., when the block is \3.00 \mathrm{~cm}\) from equilibrium, we apply kinetic plus potential energy equality. K_halfway + U_halfway = 1/2 * m * v_halfway² + 1/2 * k * (x/2)² = 1/2 * m * v². This gives \(v_halfway = \sqrt{850 \mathrm{N/m}/1.00 \mathrm{kg}} * 0.06 \mathrm{m} * 1/\sqrt{2} = 0.50 \mathrm{m/s}\). This is not half the speed at equilibrium due to conservation of energy and its redistribution among kinetic and potential energy forms.

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Key Concepts

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

Spring Constant

The spring constant, often denoted as k, is a measure of a spring's stiffness. It's revealed in Hooke’s law, which states the force exerted by a spring is directly proportional to the displacement from its equilibrium position. Mathematically, it's expressed as F = -kx. The negative sign indicates that the force is a restoring force acting in the opposite direction of the displacement.

For a stiffer spring, the spring constant is higher, indicating a larger force is required to stretch or compress it. In this exercise, the spring constant is provided as 850 N/m, which allows us to calculate the elastic potential energy stored in the spring when it is displaced. This value is essential in understanding the behavior of the spring as it affects both the potential energy stored and the resulting movement of the mass attached to it when released.

Conservation of Energy

The conservation of energy is a fundamental principle stating that energy cannot be created or destroyed, only transformed from one form to another. In the context of our spring-block system, the energy initially stored in the spring as elastic potential energy is fully transformed into kinetic energy when the block passes through the equilibrium position.

The solution exploits this principle by equating the elastic potential energy at the spring's stretched position to the kinetic energy of the block at equilibrium, allowing us to solve for the block's velocity. Understanding this conservation helps explain the motion and energy transformations of the system from the point of release to the equilibrium and beyond.

Hooke's Law

Hooke’s law provides a basis for understanding how springs exert force. The law posits that the force needed to extend or compress a spring by some distance is proportional to that distance. In formula terms, this is F = -kx, where F is the force applied by the spring, k is the spring constant, and x is the stretch or compression distance from the spring's position at rest.

The negative sign suggests the restoring nature of the spring force: it acts to bring the spring back to its equilibrium state. Hooke's law, a crucial piece of our exercise, is most accurate for small deformations: the larger the displacement, the more likely we are to encounter non-linear behavior where the force is no longer directly proportional to displacement.

Kinetic Energy

Kinetic energy refers to the energy an object possesses due to its motion. It's directly proportional to the mass of the moving object and to the square of its velocity (KE = 1/2 mv²). In our spring-block problem, when the block is at the equilibrium point, all the energy stored in the spring as elastic potential energy has been converted to kinetic energy, allowing us to determine the block's speed.

It's worth noting that the kinetic energy is not at its peak at the halfway point between the equilibrium and maximum compression/stretch, even though one might initially assume the velocity would be half of that at the equilibrium. The kinetic energy, while still obeying the law of conservation, is now shared with the potential energy the spring retains at this new halfway position.