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Chapter 7: Problem 19

A 130-g arrow is shot vertically from a bow whose effective spring constant is 400 N/m. If the bow is drawn 85 cm before shooting, to what height does the arrow rise?

Short Answer

Expert verified
Solving the above steps, you will find that the arrow rises to approximately 140.8 meters.

Step by step solution

01

Calculate Initial Potential Energy of the arrow due to the spring

The potential energy stored in a spring is given by the equation \(PE = 0.5 * k * x^2\) where k is the spring constant and x is the amount the spring is drawn. So substitute the given values \(PE = 0.5 * 400N/m * (0.85m)^2 \) to get PE.
02

Calculate Gravitational Potential Energy at maximum height

The gravitational potential energy is given by the equation \(PE = m * g * h\) where m is the mass, g is the gravitational pull and h is the height. In this case, kinetic energy is zero and all the energy is potential due to gravity, so from energy conservation, we equate initial potential energy due to the spring to the gravitational energy at the top, so we have \( 0.5 * 400N/m * (0.85m)^2 = 0.13kg * 9.8m/s^2 * h\)
03

Solve the equation to find the height

Solving the equation from step 2 will provide the height to which the arrow rises.

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Key Concepts

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

Potential Energy
Potential energy is a type of energy that is stored in an object due to its position, configuration, or state. In our context, when the arrow is drawn back on the bow, energy is stored in the form of elastic potential energy within the bowstring. This stored energy has the potential to be converted into kinetic energy when the arrow is released. This concept is crucial when analyzing systems where energy transformation occurs, such as with springs or elastic materials.
  • The formula for the potential energy in a spring is given by: \( PE = 0.5 \times k \times x^2 \).
  • Here, \( k \) is the spring constant and \( x \) is the displacement from the equilibrium position.
  • This form of energy is always ready to be transformed into other types like kinetic energy.
Understanding potential energy helps us predict how the arrow behaves once it is released from the bow.
Spring Constant
The spring constant, denoted as \( k \), plays a central role in determining how a spring behaves when force is applied. It reflects the stiffness of a spring with higher values indicating a stiffer spring. In our example, a spring constant of 400 N/m means that every meter the spring is compressed or stretched, 400 newtons of force are applied.
  • The spring constant is fundamental to calculating the energy stored in a spring, influencing how much potential energy is available to be converted into kinetic energy.
  • A large spring constant makes the bow harder to pull back but results in more stored potential energy, resulting in a faster arrow.
Knowing the spring constant helps in designing or using systems where controlled energy storage and release are necessary.
Gravitational Potential Energy
Gravitational potential energy (GPE) is the energy an object possesses due to its position relative to the Earth. It depends on the object's mass, the height at which it is positioned, and the gravitational acceleration. As the arrow rises to its maximum height, all the initial energy from the spring transfers into gravitational potential energy, ignoring air resistance and other forces.
  • The formula for gravitational potential energy is: \( PE = m \times g \times h \).
  • \( m \) represents the mass, \( g \) is the acceleration due to gravity (approximately \( 9.8 m/s^2 \)), and \( h \) is the height.
Gravitational potential energy gives insight into how high the arrow can rise, assuming energy conservation within a closed system.
Kinetic Energy
Kinetic energy is the energy of motion. As the arrow is released from the drawn bowstring, all the stored potential energy converts into kinetic energy. It then projects the arrow upwards until the kinetic energy decreases to zero, at which point it reaches its maximum height.
  • The equation for kinetic energy is given by: \( KE = 0.5 \times m \times v^2 \).
  • Here, \( m \) stands for mass and \( v \) is the velocity of the object.
  • During the arrow's ascent, kinetic energy gradually converts into gravitational potential energy.
By understanding kinetic energy, we can measure how quickly and forcefully the arrow moves through space after being released from the bow.
Energy Conservation Law
The law of energy conservation is a fundamental principle of physics stating that energy cannot be created or destroyed; it can only transform from one form to another. In our exercise, this law explains how energy transitions from the potential energy in the bowstring to the kinetic energy of the flying arrow, and ultimately into gravitational potential energy as the arrow reaches the maximum height.
  • When the arrow is released, the stored potential energy becomes kinetic energy, moving the arrow upwards.
  • At the peak of its trajectory, all kinetic energy is converted into gravitational potential energy.
  • Conservation of energy helps us calculate the maximum height the arrow will achieve, knowing that all initial energy in the bow is transferred to the arrow鈥檚 rise.
Energy conservation is key to understanding and predicting how energy behaves in any isolated system.

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Most popular questions from this chapter

A climbing rope exerts a force given by F = - kx - cx2 . Find an expression for c such that when the rope is stretched a distance d its potential energy is twice what it would be if the rope were an ideal spring with F = - kx.

Your engineering department is asked to evaluate the performance of a new 460-hp sports car. You know that 29% of the engine鈥檚 power can be converted to kinetic energy of the 1400-kg car and that the power delivered is independent of the car鈥檚 velocity. What do you report for the time it will take to accelerate from rest to 100 km/h on a level road?

In a railroad switchyard, a rail car of mass 28,600 kg starts from rest and rolls down an incline and onto a level stretch of track. It then hits a spring bumper at the end of the track. If the spring constant is 1.88 MN/m and if the spring compresses a maximum of 1.03 m, what鈥檚 the height at which the car started? Neglect friction.

The reservoir at Northfield Mountain Pumped Storage Project is 214 m above the pump/generators and holds 2.1*1010 kg of water (see Application on p. 133). The generators can produce electrical energy at the rate of 1.08 GW. Find (a) the gravitational potential energy stored, taking zero potential energy at the generators, and (b) the length of time the station can generate power before the reservoir is drained.

A particle of mass \(m\) is subject to a force \(\vec{F}=(a \sqrt{x}) \hat{l}\), where \(a\) is a constant. The particle is initially at rest at the origin and is given a slight nudge in the positive \(x\)-direction. Find an expression for its speed as a function of position \(x\).
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