Problem 37 - Physics

Question

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 slingshot 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 that 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.

Step-by-Step Solution

Answer

The effective spring constant is approximately 622 N/m.

Step 1: Understanding the Situation

The circus performer is being launched from a cannon at an angle of \(40^{\circ}\). Given initial horizontal distance \(d = 26.8 \, \text{m}\) covered in \(2.14 \, \text{s}\), we will calculate the horizontal component of the velocity.

Step 2: Calculate Horizontal Velocity

Using the equation for horizontal distance, \(d = v_{x} \cdot t\), we solve for horizontal velocity: \(v_{x} = \frac{26.8}{2.14} = 12.52 \, \text{m/s}\). This is the horizontal component of the velocity with which he was launched.

Step 3: Calculate Initial Velocity

Using the horizontal velocity and launch angle: \(v_{x} = v_0 \cdot \cos(40^{\circ})\), we find \(v_0\), the magnitude of the initial velocity. \(v_0 = \frac{v_{x}}{\cos(40^{\circ})} = \frac{12.52}{0.766} \approx 16.34 \, \text{m/s}\).

Step 4: Energy Conservation Setup

The elastic potential energy from the bands is transferred into kinetic energy. Initially, potential energy \(E_p = \frac{1}{2} k x^2\) where \(x = 3.00 \, \text{m}\), and kinetic energy is \(E_k = \frac{1}{2} m v_0^2\) with \(v_0 \approx 16.34 \, \text{m/s}\).

Step 5: Solve for Spring Constant \(k\)

Equating elastic potential energy to kinetic energy, \(\frac{1}{2} k x^2 = \frac{1}{2} m v_0^2\), solve for \(k\). Rearrange to find \(k = \frac{m v_0^2}{x^2}\). Substitute \(m = 70.0 \, \text{kg}\), \(v_0 = 16.34 \, \text{m/s}\), and \(x = 3.00 \, \text{m}\) to get \(k \approx \frac{70 \times (16.34)^2}{3^2} \approx 622 \, \text{N/m}\).

Key Concepts

Elastic Potential Energy

Elastic potential energy is the energy stored in an object when it is stretched or compressed. In this exercise, the circus cannon operates like a giant slingshot, using elastic bands to store energy when they are stretched by 3 meters. This stored energy is what launches the circus performer into the air.
The energy stored in the elastic bands is given by the formula:

• \[E_p = \frac{1}{2} k x^2\]

where:

• \(E_p\) is the elastic potential energy,
• \(k\) is the spring constant (how stiff the band is),
• \(x\) is the stretch or compression distance (here, 3 meters).

At the launch, all this elastic potential energy is converted into the kinetic energy of the performer, propelling them towards the net. Understanding this concept is crucial as it highlights how energy transformations occur, key to solving the problem.

Initial Velocity

Initial velocity refers to the speed and direction at which an object begins its motion. In projectile motion, like the circus performer's flight, initial velocity has both horizontal and vertical components.
To find the total initial velocity (\(v_0\)), the horizontal component (\(v_x\)) and the launch angle are used. From the exercise:

• Horizontal distance \(d = 26.8 \, \text{m}\) and time \(t = 2.14 \, \text{s}\) determine \(v_x\):\[v_x = \frac{d}{t} = 12.52 \, \text{m/s}\]
• Use the launch angle \(\theta = 40^{\circ}\):\[v_x = v_0 \cdot \cos(\theta) \Rightarrow v_0 = \frac{v_x}{\cos(40^{\circ})} \approx 16.34 \, \text{m/s}\]

This initial velocity is essential for predicting the path and distance the performer will cover during the stunt.

Conservation of Energy

The principle of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. In our exercise, the elastic potential energy in the stretched bands is transformed into kinetic energy as the performer is launched.
Initially, the elastic bands store potential energy:

• \[E_p = \frac{1}{2} k x^2\]

As the performer launches, this is converted to kinetic energy (\(E_k\)):

• \[E_k = \frac{1}{2} m v_0^2\]

The equation \(\frac{1}{2} k x^2 = \frac{1}{2} m v_0^2\) signifies this transformation, enabling us to solve for the spring constant \(k\). This constant indicates the force needed to stretch the band to store enough energy for the performance, helping us understand the mechanics behind projectile launches.