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An ingenious bricklayer builds a device for shooting bricks up to the top of the wall where he is working. He places a brick on a vertical compressed spring with force constant \(k = 450\) N/m and negligible mass. When the spring is released, the brick is propelled upward. If the brick has mass 1.80 kg and is to reach a maximum height of 3.6 m above its initial position on the compressed spring, what distance must the bricklayer compress the spring initially? (The brick loses contact with the spring when the spring returns to its uncompressed length. Why?)

Short Answer

Expert verified
The spring must be compressed by approximately 0.5315 meters.

Step by step solution

01

Identify Known Values

First, let's list the given information from the problem. We have the spring constant \(k = 450\) N/m, the mass of the brick \(m = 1.80\) kg, and the desired maximum height \(h = 3.6\) m.
02

Calculate Potential Energy at Maximum Height

The potential energy at the maximum height is given by the gravitational potential energy formula. The formula is \(U = mgh\), where \(g\) is the acceleration due to gravity (9.8 m/s\(^2\)). Substitute the known values: \(U = 1.80 \times 9.8 \times 3.6\).
03

Solve for Potential Energy

Perform the calculation: \(U = 1.80 \times 9.8 \times 3.6 = 63.504\) Joules. This is the energy needed to raise the brick to the maximum height of 3.6 m.
04

Relate Spring Compression to Energy

The potential energy stored in a compressed spring is given by \(PE_{spring} = \frac{1}{2}kx^2\), where \(x\) is the spring compression. We want this energy to equal the gravitational potential energy at the max height, so set \(\frac{1}{2}kx^2 = 63.504\).
05

Solve for Spring Compression

Now solve \(\frac{1}{2}kx^2 = 63.504\) for \(x\). Substitute \(k = 450\): \(\frac{1}{2} \times 450 \times x^2 = 63.504\). Simplify to get \(225x^2 = 63.504\).
06

Calculate Distance of Compression

Divide both sides by 225: \(x^2 = \frac{63.504}{225}\). Simplify the division: \(x^2 = 0.28224\). Take the square root to find \(x\): \(x = \sqrt{0.28224}\).
07

Result Interpretation

Calculate the square root: \(x \approx 0.5315\) meters. Therefore, the bricklayer must compress the spring by approximately 0.5315 meters to achieve the desired height.

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Key Concepts

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

Gravitational Potential Energy
Gravitational potential energy is a type of energy possessed by an object due to its position relative to the Earth. It's an essential concept in physics, especially when dealing with the motion of objects under the influence of gravity. This energy is calculated using the formula:
  • \( U = mgh \)
In this equation:
  • \( U \) represents the gravitational potential energy, measured in joules.
  • \( m \) is the mass of the object, in kilograms.
  • \( g \) is the acceleration due to gravity, approximately \( 9.8 \text{ m/s}^2 \) on Earth.
  • \( h \) is the height above the reference point, in meters.
In practical terms, when you lift an object to a height, you are storing energy in it, which can be released if the object falls. This energy plays a crucial role in determining how much energy is required to move the brick to a specified height using the spring mechanism.
Spring Constant
The spring constant, denoted as \( k \), is a measure of the stiffness of a spring. It quantifies the force required to compress or extend a spring by a unit length. The spring constant is expressed in newtons per meter (N/m).To understand this concept, consider Hooke's Law, which states:
  • \( F = -kx \)
Here, \( F \) is the force applied to the spring, \( k \) is the spring constant, and \( x \) is the displacement from the spring's equilibrium position. A higher spring constant means a stiffer spring, which requires more force to compress or extend.In our problem, the spring constant \( k \) is 450 N/m. This value influences how much the spring must be compressed to store the necessary energy to propel the brick to the desired height. The stiffness of the spring directly affects its ability to convert potential energy into kinetic energy and vice versa.
Energy Conservation
Energy conservation is a fundamental principle in physics, stating that energy cannot be created or destroyed, only transformed from one form to another. In the context of our problem, the energy conservation principle is applied to the system consisting of the brick and the spring.Initially, energy is stored in the compressed spring as elastic potential energy. When the spring is released, this energy transforms into kinetic energy as the spring expands and later into gravitational potential energy as the brick reaches its peak height.The equation that represents energy conservation in this context is:
  • \( \frac{1}{2}kx^2 = mgh \)
This equation states that the elastic potential energy stored in the spring must equal the gravitational potential energy when the brick is at its maximum height. This transformation is why understanding energy conservation is vital in solving such physics problems.
Physics Problem Solving
Physics problem solving involves a systematic approach to understanding and calculating physical phenomena. It's not just about crunching numbers; it's about comprehending the underlying principles and applying them skillfully. Here are some crucial steps to enhance physics problem-solving skills:
  • Identify and list all known values, like mass, spring constant, and height in our scenario.
  • Recognize the applicable physical laws, such as energy conservation.
  • Derive equations that relate the known values to the desired quantity—in our case, spring compression.
  • Carefully solve the equations, step by step, ensuring unit consistency and logical flow.
  • Interpret results in the context of the physical problem.
This structured approach helps break down complex problems into more manageable parts, making it easier to reach a solution logically and efficiently.

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Most popular questions from this chapter

A block of ice with mass 2.00 kg slides 1.35 m down an inclined plane that slopes downward at an angle of 36.9\(^\circ\) below the horizontal. If the block of ice starts from rest, what is its final speed? Ignore friction.

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A 30.0-kg crate is initially moving with a velocity that has magnitude 3.90 m/s in a direction 37.0\(^\circ\) west of north. How much work must be done on the crate to change its velocity to 5.62 m/s in a direction 63.0\(^\circ\) south of east?

A large crate sits on the floor of a warehouse. Paul and Bob apply constant horizontal forces to the crate. The force applied by Paul has magnitude 48.0 N and direction 61.0\(^\circ\) south of west. How much work does Paul's force do during a displacement of the crate that is 12.0 m in the direction 22.0\(^\circ\) east of north?

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