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A 1.20 \(\mathrm{kg}\) piece of cheese is placed on a vertical spring of negligible mass and force constant \(k=1800 \mathrm{N} / \mathrm{m}\) that is compressed 15.0 \(\mathrm{cm}\) . When the spring is released, how high does the cheese rise from this initial position? (The cheese and the spring are not attached.)

Short Answer

Expert verified
The cheese rises approximately 1.72 meters from the initial position.

Step by step solution

01

Understand the problem

We need to find how high the cheese rises after being released from a compressed spring. The cheese gains potential energy when the spring releases and converts it into kinetic energy and eventually gravitational potential energy. The energy conservation principle will guide the solution.
02

Write the energy conservation equation

Initially, the spring has potential energy stored due to its compression, given by the formula: \[ PE_{ ext{spring}} = \frac{1}{2} k x^2 \]where \(k = 1800 \, \text{N/m}\) is the spring constant and \(x = 0.15 \, \text{m}\) is the compression distance (remember to convert cm to m). This energy converts fully to gravitational potential energy \( PE_g \) when the cheese reaches the highest point. The equation representing this energy transformation is:\[ \frac{1}{2} k x^2 = mgh \]where \(m = 1.20 \, \text{kg}\) is the mass of the cheese, \(g = 9.81 \, \text{m/s}^2\) is the acceleration due to gravity, and \(h\) is the height we need to find.
03

Solve for height \(h\)

Rearrange the equation from Step 2 to solve for \(h\):\[ h = \frac{\frac{1}{2} k x^2}{mg} \]Substitute the known values into the equation:\[ h = \frac{\frac{1}{2} \times 1800 \, \text{N/m} \times (0.15 \, \text{m})^2}{1.20 \, \text{kg} \times 9.81 \, \text{m/s}^2} \]Calculate \(h\): \[ h = \frac{\frac{1}{2} \times 1800 \times 0.0225}{1.20 \times 9.81} = \frac{20.25}{11.772} \approx 1.72 \, \text{m} \]So, the cheese rises approximately 1.72 meters from the initial compressed position.

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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 compressed or stretched. This energy is calculated using the formula \[ PE_{\text{spring}} = \frac{1}{2} k x^2 \]where:
  • k is the spring constant, which tells us how stiff the spring is. A larger value of k means the spring is stiffer.
  • x is the distance the spring is compressed or stretched from its usual length.
When discussing Spring Potential Energy, remember that this energy is potential, meaning it is stored and has the potential to do work. In our example of the cheese on the spring, when the spring is compressed by 0.15 meters, energy is stored within it. Once released, this stored energy can be transformed into other forms of energy, like kinetic or gravitational potential energy, allowing the cheese to rise. Understanding how Spring Potential Energy works is crucial for problems involving mechanical springs.
Gravitational Potential Energy
Gravitational Potential Energy (GPE) is the energy an object possesses because of its position in a gravitational field. It's expressed by the formula:\[ PE_g = mgh \]Here:
  • m stands for the mass of the object.
  • g is the acceleration due to gravity, about 9.81 m/s^2 on Earth.
  • h is the height above the reference point, such as the ground.
In the problem statement, as the cheese is launched upwards by the energy from the spring, it gains gravitational potential energy as it ascends. The higher the cheese rises, the greater its gravitational potential energy. This concept explains why objects gain energy that can be released as they fall back down. For instance, at the peak of its movement, the cheese has maximum gravitational potential and minimum kinetic energy. This intricate dance of energy forms is key to solving energy conservation problems.
Kinetic Energy Transformation
Kinetic Energy is the energy of motion. When an object is moving, it has kinetic energy, calculated by the formula:\[ KE = \frac{1}{2} mv^2 \]where:
  • m indicates the mass of the object.
  • v is the velocity of the object.
In the scenario of the cheese on the spring, the energy transformation process beautifully demonstrates the principle of energy conservation. Initially, the energy is stored within the compressed spring. Upon release, this potential energy transforms into kinetic energy as the cheese is propelled upwards. As the cheese continues to rise, its velocity decreases until it momentarily stops at the peak, where kinetic energy is entirely transformed into gravitational potential energy. Understanding this transformation cycle helps us grasp how energy shifts between states to facilitate physical movements. Energy transformation principles are vital for solving mechanical energy problems and designing efficient systems.

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Most popular questions from this chapter

Gravity in One Dimension. Two point masses, \(m_{1}\) and \(m_{2}\) , lie on the \(x\) -axis, with \(m_{1}\) held in place at the origin and \(m_{2}\) at position \(x\) and free to move. The gravitational potential energy of these masses is found to be \(U(x)=-G m_{1} m_{2} / x,\) where \(G\) is a constant (called the gravitational constant). You'll learn more about gravitation in Chapter 12 . Find the \(x\) -component of the force acting on \(m_{2}\) due to \(m_{1} .\) Is this force attractive or repulsive? How do you know?

A wooden block with mass 1.50 \(\mathrm{kg}\) is placed against a compressed spring at the bottom of an incline of slope \(30.0^{\circ}\) (point \(A\) ). When the spring is released, it projects the block up the incline. At point \(B,\) a distance of 6.00 \(\mathrm{mup}\) the incline from \(A\) , the block is moving up the incline at 7.00 \(\mathrm{m} / \mathrm{s}\) and is no longer in contact with the spring. The coefficient of kinetic friction between the block and the incline is \(\mu_{\mathrm{k}}=0.50 .\) The mass of the spring is negligible. Calculate the amount of potential energy that was initially stored in the spring.

A machine part of mass \(m\) is attached to a horizontal ideal spring of force constant \(k\) that is attached to the edge of a friction-free horizontal surface. The part is pushed against the spring compressing it a distance \(x_{0},\) and then released from rest. Find the maximum (a) speed and (b) acceleration of the machine part. (c) Where in the motion do the maxima in parts (a) and (b) occur? (d) What will be the maximum extension of the spring? (e) Describe the subsequent motion of this machine part. Will it ever stop permanently?

A \(2.00-\mathrm{kg}\) block is pushed against a spring with negligible mass and force constant \(k=400 \mathrm{N} / \mathrm{m}\) , compressing it 0.220 \(\mathrm{m}\) . When the block is released, it moves along a frictionless, horizontal surface and then up a frictionless incline with slope \(37.0^{\circ}\) (Fig. 7.30\()\) . (a) What is the speed of the block as it slides along the horizontal surface after having left the spring? (b) How far does the block travel up the incline before starting to slide back down?

A certain spring is found not to obey Hooke's law; it exerts a restoring force \(F_{x}(x)=-\alpha x-\beta x^{2}\) if it is stretched or compressed, where \(\alpha=60.0 \mathrm{N} / \mathrm{m}\) and \(\beta=18.0 \mathrm{N} / \mathrm{m}^{2} .\) The mass of the spring is negligible. (a) Calculate the potentinl-energy function \(U(x)\) for this spring. Let \(U=0\) when \(x=0\) (b) An object with mass 0.900 \(\mathrm{kg}\) on a frictionless, horizontal surface is attached to this spring, pulled a distance 1.00 \(\mathrm{m}\) to the right (the \(+x\) -direction) to stretch the spring, and released. What is the speed of the object when it is 0.50 \(\mathrm{m}\) to the right of the \(x=0\) equilibrium position?

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