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Chapter 8: Problem 13

A \(5.0 \mathrm{~g}\) marble is fired vertically upward using a spring gun. The spring must be compressed \(8.0 \mathrm{~cm}\) if the marble is to just reach a target \(20 \mathrm{~m}\) above the marble's position on the compressed spring. (a) What is the change \(\Delta U_{g}\) in the gravitational potential energy of the marble-Earth system during the \(20 \mathrm{~m}\) ascent? (b) What is the change \(\Delta U_{s}\) in the elastic potential energy of the spring during its launch of the marble? (c) What is the spring constant of the spring?

Short Answer

Expert verified
(a) \(0.981 \text{ J}\), (b) \(-0.981 \text{ J}\), (c) \(307.8 \text{ N/m}\).

Step by step solution

01

Identify given values

The mass of the marble is given as \(5.0\, \text{g} = 0.005\, \text{kg}\). The spring compression is \(8.0\, \text{cm} = 0.08\, \text{m}\). The target height is \(20\, \text{m}\) above. The gravitational acceleration \(g\) is \(9.81\, \text{m/s}^2\).
02

Calculate change in gravitational potential energy (鈭哢_g)

The change in gravitational potential energy \(\Delta U_g\) is calculated using the formula \(\Delta U_g = mgh\), where \(m\) is the mass, \(g\) is the gravitational acceleration, and \(h\) is the height. Substituting the values, \(\Delta U_g = 0.005 \times 9.81 \times 20 = 0.981 \text{ J}\).
03

Understand elastic potential energy change (鈭哢_s)

The change in elastic potential energy of the spring \(\Delta U_s\) is equivalent to the gravitational potential energy change since the marble starts at rest and reaches the maximum height with zero velocity. Thus, \(\Delta U_s = -\Delta U_g = -0.981 \text{ J}\).
04

Determine the spring constant (k)

The elastic potential energy is given by \(\Delta U_s = \frac{1}{2} k x^2\). Solving for \(k\), we have \(k = \frac{2 \Delta U_s}{x^2}\). Insert the known values: \(k = \frac{2 \times 0.981}{(0.08)^2} = 307.8 \text{ N/m}\).

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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 the energy an object possesses due to its position in a gravitational field. When you pick up a book and hold it over a table, it has gravitational potential energy because of its position above the ground. This energy depends on three key factors: the mass of the object, the height above the ground, and the gravitational acceleration, which is typically about 9.81 m/s虏 on Earth.

When we calculate the change in gravitational potential energy, we use the formula:
  • \[\Delta U_g = mgh\]
  • "m" is the object's mass (in kilograms),
  • "g" is the acceleration due to gravity,
  • "h" is the change in height (in meters).
In our marble problem, this formula helps us understand how much energy is needed to raise the marble to a height of 20 m. Because the marble has a mass of 5.0 g (0.005 kg), its gravitational potential energy increases as it moves upwards, requiring 0.981 J of energy for the ascent.
Elastic Potential Energy
Elastic Potential Energy is the energy stored in objects that can be stretched or compressed, like a spring or a rubber band. When you compress a spring by pressing on it or when you stretch it by pulling at its ends, you store energy in the spring.

For springs, the formula to calculate elastic potential energy is:
  • \[\Delta U_s = \frac{1}{2} k x^2\]
  • "k" is the spring constant (a measure of the spring's stiffness),
  • "x" is the displacement from its rest position (in meters).
In our exercise, the compressed spring releases 0.981 J of energy as it launches the marble. This energy equals the gravitational potential energy gained by the marble, showing the energy transfer from the spring to the marble as the spring pushes it upwards.
Spring Constant
The Spring Constant is a measure of a spring's stiffness, represented by the symbol "k." A higher spring constant means the spring is stiffer and requires more force to compress or stretch it. It is expressed in newtons per meter (N/m).

To find the spring constant, we rearrange the formula for elastic potential energy:
  • \[k = \frac{2 \Delta U_s}{x^2}\]
  • "\Delta U_s" is the change in elastic potential energy,
  • "x" is the compression or extension distance.
By using the provided values, the spring constant for the marble's spring-gun system is calculated as approximately 307.8 N/m. This means that the spring requires a force of 307.8 N to compress it one meter, signifying the energy needed for it to launch the marble effectively.

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Most popular questions from this chapter

A \(4.0 \mathrm{~kg}\) bundle starts up a \(30^{\circ}\) incline with \(128 \mathrm{~J}\) of kinetic energy. How far will it slide up the incline if the coefficient of kinetic friction between bundle and incline is \(0.30 ?\)

A sprinter who weighs \(670 \mathrm{~N}\) runs the first \(7.0 \mathrm{~m}\) of a race in \(1.6 \mathrm{~s},\) starting from rest and accelerating uniformly. What are the sprinter's (a) speed and (b) kinetic energy at the end of the \(1.6 \mathrm{~s} ?\) (c) What average power does the sprinter generate during the \(1.6 \mathrm{~s}\) interval?

A boy is initially seated on the top of a hemispherical ice mound of radius \(R=13.8 \mathrm{~m} . \mathrm{He}\) begins to slide down the ice, with a negligible initial speed (Fig. 8-47). Approximate the ice as being frictionless. At what height does the boy lose contact with the ice?

Two blocks, of masses \(M=2.0 \mathrm{~kg}\) and \(2 M,\) are connected to a spring of spring constant \(k=200 \mathrm{~N} / \mathrm{m}\) that has one end fixed, as shown in Fig. \(8-69 .\) The horizontal surface and the pulley are frictionless, and the pulley has negligible mass. The blocks are released from rest with the spring relaxed. (a) What is the combined kinetic energy of the two blocks when the hanging block has fallen \(0.090 \mathrm{~m} ?\) (b) What is the kinetic energy of the hanging block when it has fallen that \(0.090 \mathrm{~m} ?\) (c) What maximum distance does the hanging block fall before momentarily stopping?

In Fig. 8 -42, a block of mass \(m=3.20 \mathrm{~kg}\) slides from rest a distance \(d\) down a frictionless incline at angle \(\theta=30.0^{\circ}\) where it runs into a spring of spring constant \(431 \mathrm{~N} / \mathrm{m} .\) When the block momentarily stops, it has compressed the spring by \(21.0 \mathrm{~cm} .\) What are (a) distance \(d\) and (b) the distance between the point of the first block-spring contact and the point where the block's speed is greatest?
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