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II A slingshot obeying Hooke's law is used to launch pebbles vertically into the air. You observe that if you pull a pebble back \(20.0 \mathrm{~cm}\) against the elastic band, the pebble goes \(6.0 \mathrm{~m}\) high. (a) Assuming that air drag is negligible, how high will the pebble go if you pull it back \(40.0 \mathrm{~cm}\) instead? (b) How far must you pull it back so it will reach \(12.0 \mathrm{~m} ?\) (c) If you pull a pebble that is twice as heavy back \(20.0 \mathrm{~cm},\) how high will it \(\mathrm{go} ?\)

Short Answer

Expert verified
(a) 24 m (b) 28.3 cm (c) 3 m

Step by step solution

01

Understand Hooke's Law and Potential Energy

Hooke's Law states that the force exerted by a spring is proportional to its extension or compression, mathematically represented as \( F = -kx \). For potential energy stored in a spring, use the formula \( E_p = \frac{1}{2}kx^2 \), where \( x \) is the extension. When the pebble is launched, this energy is converted to gravitational potential energy \( E_g = mgh \). We're given conditions without knowing \( k \) or \( m \), but they remain constants. Thus, compare the ratios of heights to extensions to solve the problem.
02

Calculate Height for 40 cm Pull-back (a)

Using the relation between stored elastic potential energy and the height achieved: if a 20 cm pull-back results in a 6 m height, then doubling the extension to 40 cm results in energy \( E_{p1} = 4 \times E_{p2} \), since energy is proportional to the square of extension \((40/20)^2 = 4\). Thus, this results in a 4 times higher elevation: \( h = 4 \times 6 \text{ m} = 24 \text{ m} \).
03

Calculate Pull-back Distance for 12 m Height (b)

To find the extension needed for a 12 m height, calculate energy ratios. If a 20 cm pull-back results in a 6 m height, double that height requires a higher extension. Find \( x \), using the formula \((x/20)^2 = (12/6)\), leading to \( x = 20 \sqrt{2} \approx 28.3 \text{ cm} \).
04

Height for a Pebble Twice as Heavy (c)

Using energy conservation, if the mass is doubled, the height achieved will be halved because energy ratio remains constant and energy is converted to twice the gravitational potential energy. Thus, a pull-back of 20 cm for a twice-heavier pebble will reach \( 6/2 = 3 \text{ m} \).

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Key Concepts

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

Elastic Potential Energy
Elastic potential energy is the energy stored in objects like springs or elastic bands when they are stretched or compressed. This concept is essential to understanding phenomena like spring-loaded mechanisms or slingshots. Using Hooke's Law, the force exerted by a spring is directly proportional to how far it's stretched or compressed. Mathematically, we can express this as:\[ F = -kx \]where:
  • \( F \) is the force exerted by the spring,
  • \( k \) is the spring constant, which measures the stiffness of the spring,
  • \( x \) is the distance the spring is stretched or compressed.
The minus sign indicates that the force exerted by the spring is in the opposite direction of the displacement...The elastic potential energy (\( E_p \)) stored in a spring when it is stretched or compressed can be calculated using:\[ E_p = \frac{1}{2}kx^2 \]As you pull the slingshot back, you increase the elastic potential energy, which is later converted to other forms, like gravitational potential energy, when released.
Gravitational Potential Energy
Gravitational potential energy is the energy an object possesses because of its position in a gravitational field. When we lift an object, we work against gravitational forces, thus giving the object potential energy.We calculate gravitational potential energy (\( E_g \)) using the formula:\[ E_g = mgh \]where:
  • \( m \) is the mass of the object,
  • \( g \) is the acceleration due to gravity (approximately \( 9.81 \text{ m/s}^2 \) on Earth),
  • \( h \) is the height above the reference point.
As an object rises to a greater height, its gravitational potential energy increases. When the pebble from the slingshot ascends, this stored gravitational potential energy determines how high it will travel. The conversion of elastic potential energy to gravitational potential energy explains how high the pebble in the exercise can reach.
Spring Force
The spring force comes into play when a spring is either compressed or stretched. It's this force that propels the pebble in the slingshot exercise.Springs obey Hooke's Law, which means the behavior of the spring force is linear until the elastic limit is reached. Understanding this helps us predict how much energy a spring can store and subsequently release:\[ F = -kx \]This equation tells us that:
  • The more you stretch or compress the spring (increasing \( x \)),
  • The greater the force exerted by the spring.
Knowing the force and the spring constant \( k \) assists in understanding how springs work in various mechanisms, such as launching the pebble in this exercise. The spring force is what initiates the motion of the pebble upward, converting stored energy into kinetic energy, which we can further convert into gravitational potential energy.
Energy Conservation
Energy conservation is a fundamental principle in physics stating that energy cannot be created or destroyed, only transformed from one form into another. In the context of a slingshot, stored elastic potential energy is transferred to kinetic energy as the pebble is released, and eventually to gravitational potential energy as it ascends. In every transition:
  • The total energy before and after must be equal assuming no energy is lost due to external forces like friction or air drag.
  • This is why the initial potential energy stored in the spring when you stretch it is what translates to the energy needed to reach a certain height.
By understanding energy conservation, we can predict the pebble's behavior under different conditions, such as varying extension lengths or masses, just as illustrated in the original exercise. Each situation conserves energy, leading to predictable and calculable outcomes.

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Most popular questions from this chapter

A tandem (two-person) bicycle team must overcome a force of \(165 \mathrm{~N}\) to maintain a speed of \(9.00 \mathrm{~m} / \mathrm{s}\). Find the power required per rider, assuming that each contributes equally.

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An elevator has mass \(600 \mathrm{~kg}\), not including passengers. The elevator is designed to ascend, at constant speed, a vertical distance of \(20.0 \mathrm{~m}\) (five floors) in \(16.0 \mathrm{~s}\), and it is driven by a motor that can provide up to 40 hp to the elevator. What is the maximum number of passengers that can ride in the elevator? Assume that an average passenger has mass \(65.0 \mathrm{~kg}\).

Ski jump ramp. You are designing a ski jump ramp for the next Winter Olympics. You need to calculate the vertical height \(h\) from the starting gate to the bottom of the ramp. The skiers push off hard with their ski poles at the start, just above the starting gate, so they typically have a speed of \(2.0 \mathrm{~m} / \mathrm{s}\) as they reach the gate. For safety, the skiers should have a speed of no more than \(30.0 \mathrm{~m} / \mathrm{s}\) when they reach the bottom of the ramp. You determine that for an \(85.0 \mathrm{~kg}\) skier with good form, friction and air resistance will do total work of magnitude \(4000 \mathrm{~J}\) on him during his run down the slope. What is the maximum height \(h\) for which the maximum safe speed will not be exceeded?

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