/*! 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 136 A spring with spring constant \(... [FREE SOLUTION] | 91Ó°ÊÓ

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A spring with spring constant \(k=620 \mathrm{~N} / \mathrm{m}\) is placed in a vertical orientation with its lower end supported by a horizontal surface. The upper end is depressed \(25 \mathrm{~cm}\), and a block with a weight of \(50 \mathrm{~N}\) is placed (unattached) on the depressed spring. The system is then released from rest. Assume that the gravitational potential energy \(U_{g}\) of the block is zero at the release point \((y=0)\) and calculate the kinetic energy \(K\) of the block for \(y\) equal to (a) 0 . (b) \(0.050 \mathrm{~m},(\mathrm{c}) 0.10 \mathrm{~m},(\mathrm{~d}) 0.15 \mathrm{~m}\), and \((\mathrm{e}) 0.20 \mathrm{~m}\). Also, \((\mathrm{f})\) how far above its point of release does the block rise?

Short Answer

Expert verified
Kinetic energies: (a) 0 J, (b) 4.5 J, (c) 7 J, (d) 7.5 J, (e) 6 J. Maximum height: 0.212 m.

Step by step solution

01

Understand the problem

The problem involves calculating the kinetic energy of a block on a spring as it moves upwards. We use the conservation of energy principle to determine the kinetic energy at different positions and find how high the block rises. Given values include the spring constant, initial depression, and weight of the block.
02

Calculate initial energy ( U_s ) in the system

The initial potential energy stored in the spring is the energy at the moment just before release. This is because the spring is compressed by 0.25 m from its natural state (where spring force would be zero): \[ U_s = \frac{1}{2} k x^2 = \frac{1}{2} \times 620 \times (0.25)^2 = 19.375 \, \text{J} \]
03

Calculate gravitational potential energy at different positions

For each position given by y (0, 0.050 m, 0.10 m, 0.15 m, 0.20 m), calculate the gravitational potential energy (U_g) of the block: \[ U_g = mg y = 50 \times y \] This gives U_g values: (a) 0 J, (b) 2.5 J, (c) 5 J, (d) 7.5 J, (e) 10 J.
04

Calculate spring potential energy at each position

As the block moves upwards, the spring's potential energy changes since the compression changes. For each position y, the spring compression position is (0.25-y): \[ U_s(y) = \frac{1}{2} k (0.25-y)^2 \]Calculate this for each given y: \[ \text{(a) } U_s = 19.375 \text{ J} \] \[ \text{(b) } U_s = 12.375 \text{ J} \] \[ \text{(c) } U_s = 7.375 \text{ J} \] \[ \text{(d) } U_s = 4.375 \text{ J} \] \[ \text{(e) } U_s = 3.375 \text{ J} \]
05

Apply conservation of energy to find kinetic energy (K)

Using the conservation of energy, where total initial energy equals total energy at any point:\[ U_s(0) = K + U_g(y) + U_s(y) \] Solve for K (kinetic energy) at each position: \[ \text{(a) } K = 19.375 - 0 - 19.375 = 0 \text{ J} \] \[ \text{(b) } K = 19.375 - 2.5 - 12.375 = 4.5 \text{ J} \] \[ \text{(c) } K = 19.375 - 5 - 7.375 = 7 \text{ J} \] \[ \text{(d) } K = 19.375 - 7.5 - 4.375 = 7.5 \text{ J} \] \[ \text{(e) } K = 19.375 - 10 - 3.375 = 6 \text{ J} \]
06

Calculate maximum height above release point

The block rises until its kinetic energy is zero (at maximum height), meaning all initial energy is converted to gravitational and spring potential energy. Solving for y when K=0 gives us: \[ U_s(0) = U_g + U_s(y_{\text{max}}) \]This simplifies to finding y where \[ 19.375 = 50y + \frac{1}{2} \times 620 \times (0.25-y)^2 \]. Solving for y iteratively or using a solver method gives\( y = 0.212 \text{ m}\).
07

Conclusion

The kinetic energies at positions a-e are 0 J, 4.5 J, 7 J, 7.5 J, 6 J, respectively. The block rises to a maximum height of 0.212 m above the release point.

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Key Concepts

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

Kinetic Energy
Kinetic Energy is the energy a body possesses due to its motion. In our scenario with the block and spring, kinetic energy helps us understand how much energy the block has as it moves after being released from the depressed spring.
Kinetic energy (\(K\)) can be determined using the equation:\[ K = \frac{1}{2} m v^2\]For this problem, however, we use the conservation of energy principle to directly calculate kinetic energy without needing to evaluate velocity.
  • Initially, the spring is compressed, meaning the block's kinetic energy at \(y = 0\) is zero, as it's just being released.
  • As the block moves upwards, it gains kinetic energy when the spring's potential energy and gravitational potential energy decrease.
The conservation of energy equation in the exercise shows how kinetic energy is calculated by balancing the spring's potential energy decrease with the increase in gravitational potential energy. This approach simplifies the calculation immensely, as it directly relates to changes in potential energies.
Gravitational Potential Energy
Gravitational Potential Energy (GPE) refers to the energy stored by an object's position relative to a height in a gravitational field.
In our problem, gravitational potential energy (\(U_g\)) varies as the block rises from its initial release point. It is calculated using:\[ U_g = m g y \]where \(m\) is the mass of the object, \(g\) is the acceleration due to gravity (often \(9.81 \text{ m/s}^2\) for calculations, though in this exercise weight provides a direct relation), and \(y\) is the height above a defined reference point.
  • When \(y = 0\), GPE is zero since this is our reference point.
  • As the block ascends to different heights like 0.050 m or 0.10 m, its GPE increases, showing the block stores more energy due to its elevated position.
  • This incremental addition of potential energy is crucial, as the energy from the spring transforms partly into this gravitational form.
Through each step of the height change, GPE is calculated to track the energy's transformation within the system, illustrating how energy conserves altogether.
Spring Potential Energy
Spring Potential Energy is the energy stored in a compressed or stretched spring. This energy is crucial in our problem as the primary source of energy that allows the block to move upwards after being released.
The potential energy stored in a spring (\(U_s\)) is calculated by:\[ U_s = \frac{1}{2} k x^2 \]where \(k\) is the spring constant, and \(x\) is the distance from the spring's natural length (zero force position).
If the spring is initially compressed by 25 cm:
  • At the starting point (\(y = 0\)), the potential energy is 19.375 J, since all energy is stored in the compressed spring.
  • As the block moves upwards and the spring decompresses, this stored energy decreases, becoming kinetic and gravitational potential energy. For \(0.25 - y\) deformations, \(U_s\) constantly changes.
Understanding these energy conversions helps in applying the conservation of energy principles effectively, and explains why the block has kinetic energy as the system evolves.

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Most popular questions from this chapter

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