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Chapter 10: Problem 50

A 70.0 -kg circus performer is fired from a cannon that is elevated at an angle of \(40.0^{\circ}\) above the horizontal. The cannon uses strong elastic bands to propel the performer, much in the same way that a sling-shot fires a stone. Setting up for this stunt involves stretching the bands by 3.00 \(\mathrm{m}\) from their unstrained length. At the point where the performer flies free of the bands, his height above the floor is the same as the height of the net into which he is shot. He takes 2.14 s to travel the horizontal distance of 26.8 \(\mathrm{m}\) between this point and the net. Ignore friction and air resistance and determine the effective spring constant of the firing mechanism.

Short Answer

Expert verified
The effective spring constant of the firing mechanism is approximately 2073.55 N/m.

Step by step solution

01

Calculate Horizontal Velocity

We start by determining the horizontal component of the velocity. The horizontal distance traveled by the performer is given as 26.8 meters, and the time taken is 2.14 seconds. Use the formula \( \text{velocity} = \frac{\text{distance}}{\text{time}} \) to find the horizontal velocity \( v_x \): \[ v_x = \frac{26.8 \text{ m}}{2.14 \text{ s}} = 12.52 \text{ m/s}\]
02

Calculate Initial Velocity

The horizontal component of the velocity is related to the initial velocity and the angle of projection by \( v_x = v_0 \cos(\theta) \), where \( \theta = 40.0^{\circ} \). Rearrange to solve for \( v_0 \): \[ v_0 = \frac{v_x}{\cos(40.0^{\circ})} = \frac{12.52}{\cos(40.0^{\circ})} \approx 16.33 \text{ m/s}\]
03

Calculate Kinetic Energy

The kinetic energy \( KE \) when the performer leaves the cannon is given by the formula \( KE = \frac{1}{2} m v_0^2 \), where \( m = 70.0 \text{ kg} \). Substitute the known values: \[ KE = \frac{1}{2} \times 70.0 \times (16.33)^2 \approx 9330.96 \text{ J}\]
04

Use Hooke's Law to Find Spring Constant

The elastic potential energy stored when the bands are stretched by 3.0 meters should equal the kinetic energy when the performer is fired. According to Hooke's Law, \( E_{PE} = \frac{1}{2} k x^2 \), where \( x = 3.0 \text{ m} \). Equate the elastic potential energy to the kinetic energy and solve for \( k \): \[ \frac{1}{2} k (3.0)^2 = 9330.96 \quad \Rightarrow \quad k \times 4.5 = 9330.96 \quad \Rightarrow \quad k = \frac{9330.96}{4.5} \approx 2073.55 \text{ N/m}\]

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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 all about the energy of motion. Imagine you're watching a performer flying through the air after being launched from a cannon. At that moment, they're right in the middle of showcasing kinetic energy. The main formula to calculate kinetic energy
  • involves the mass of the object and its velocity: \( KE = \frac{1}{2} mv^2 \).
In this formula: - \( KE \) is the kinetic energy, - \( m \) is the mass of the object or performer, and - \( v \) is the velocity at which the object is moving.
For example, if your performer has a mass of 70 kg and is moving with a velocity of 16.33 m/s, you can substitute these values into the equation to find their kinetic energy, which turns out to be around 9330.96 joules.
Joules is a unit of energy that measures how much work is done when forces cause an object to move. Higher mass or speed results in more kinetic energy.
Hooke's Law
Hooke's Law is a principle that describes how elastic materials like springs work. When you stretch or compress a spring, it wants to return to its original shape. Hooke's Law tells us how much force the spring exerts during this process using the formula:
  • \( F = kx \),
where:
  • \( F \) is the force the spring exerts,
  • \( k \) is the spring constant (a measure of a spring's stiffness), and
  • \( x \) is the distance the spring is stretched or compressed from its resting position.
So when the performer in the circus stunt is fired from the cannon, the stretched elastic bands are behaving like stretched springs. Understanding Hooke's Law helps can help you determine the spring constant of these bands. With the performer, they need to store up enough potential energy through these stretched bands to successfully make the big jump. If you know how far they stretched, what's left is to find out what spring constant results can support their mass and kinetic feat.
Elastic Potential Energy
Think of elastic potential energy as the energy stored in objects like springs or elastic bands when they are stretched or compressed. This type of energy is ready to convert into kinetic energy, launching objects into motion. Just like a stretched elastic band stores energy - it's the same for the bands used in the cannon. The formula for elastic potential energy is:
  • \( E_{PE} = \frac{1}{2} k x^2 \),
where:
  • \( E_{PE} \) is the elastic potential energy,
  • \( k \) is the spring constant, and
  • \( x \) is the amount of stretch or compression.
In the case of our circus performer, the energy stored in the stretched bands is transferred to them as they fly off through the air. As the bands are stretched by 3 meters, the energy stored needs to match the kinetic energy required for the performer to soar across a 26.8-meter distance.
This conversion is crucial to ensure a smooth and correctly powered stunt.

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Most popular questions from this chapter

Astronauts on a distant planet set up a simple pendulum of length 1.2 \(\mathrm{m}\) . The pendulum executes simple harmonic motion and makes 100 complete vibrations in 280 s. What is the magnitude of the acceleration due to gravity on this planet?

A 0.60 -kg metal sphere oscillates at the end of a vertical spring.As the spring stretches from 0.12 to 0.23 \(\mathrm{m}\) (relative to its unstrained length), the speed of the sphere decreases from 5.70 to 4.80 \(\mathrm{m} / \mathrm{s}\) . What is the spring constant of the spring?

An 86.0 -kg climber is scaling the vertical wall of a mountain. His safety rope is made of nylon that, when stretched, behaves like a spring with a spring constant of \(1.20 \times 10^{3} \mathrm{N} / \mathrm{m}\) . He accidentally slips and falls freely for 0.750 \(\mathrm{m}\) before the rope runs out of slack. How much is the rope stretched when it breaks his fall and momentarily brings him to rest?

A 15.0 -kg block rests on a horizontal table and is attached to one end of a massless, horizontal spring. By pulling horizontally on the other end of the spring, someone causes the block to accelerate uniformly and reach a speed of 5.00 \(\mathrm{m} / \mathrm{s}\) in 0.500 \(\mathrm{s}\) . In the process, the spring is stretched by 0.200 \(\mathrm{m}\) . The block is then pulled at a constant speed of 5.00 \(\mathrm{m} / \mathrm{s}\) , during which time the spring is stretched by only 0.0500 \(\mathrm{m} .\) Find \((\mathrm{a})\) the spring constant of the spring and \((\mathrm{b})\) the coefficient of kinetic friction between the block and the table.

In 0.750 s, a \(7.00-\mathrm{kg}\) block is pulled through a distance of 4.00 \(\mathrm{m}\) on a frictionless horizontal surface, starting from rest. The block has a constant acceleration and is pulled by means of a horizontal spring that is attached to the block. The spring constant of the spring is 415 \(\mathrm{N} / \mathrm{m} .\) By how much does the spring stretch?
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