91Ó°ÊÓ

Spring Gun. A hydraulic press is used to compress a spring that will then be used to project a \(25.0-\mathrm{kg}\) steel ball. The system is similar, but smaller in scale, to the car jack illustrated in Figure \(11.14 .\) In this case, the smaller cylinder has a diameter of \(d_{1}=0.75 \mathrm{cm}\) and has a manually operated plunger. The larger cylinder has a diameter of \(d_{2}=10.0 \mathrm{cm},\) and its piston compresses the spring. The idea is that the gun operator pulls a lever that pushes the plunger on the small cylinder, which transmits a pressure to the larger piston that, in turn, exerts a force on the spring and compresses it. Once the spring is compressed, the steel ball is loaded and the spring is released, ejecting the ball. (a) If a force of \(950 \mathrm{N}\) is exerted on the primary (smaller) piston to compress the spring \(1.25 \mathrm{m}\) from its equilibrium (uncompressed) position, what is the spring constant \(k\) of the spring? (b) What is the velocity of the steel ball just after it is ejected? (c) Neglecting air resistance, what is the maximum range of this "spring gun"?

Step by step solution

01

Calculate the Pressure Applied by the Smaller Cylinder

The pressure exerted in the hydraulic system is the same in both cylinders. Start by calculating the force applied per unit area (pressure) on the smaller cylinder with the given force of \(950 \text{ N}\). For a cylinder, the area \(A\) is given by \(A = \pi r^2\), where \(r\) is the radius.Given diameter \(d_1 = 0.75 \text{ cm}\), the radius \(r_1 = 0.375 \text{ cm} = 0.00375 \text{ m}\).The area of the smaller piston, \(A_1 = \pi (0.00375)^2 = 4.42 \times 10^{-5}\, \text{m}^2\).So the pressure \(P\) is:\[P = \frac{950}{4.42 \times 10^{-5}} = 2.15 \times 10^{7}\, \text{N/m}^2.\]

02

Determine the Force Exerted by the Larger Cylinder

Since the pressure is the same across both cylinders, calculate the force exerted by the larger piston using: \(F = PA\).The area of the larger cylinder \(A_2\) with diameter \(d_2 = 10.0 \text{ cm}\) gives radius \(r_2 = 5.0 \text{ cm} = 0.05 \text{ m}\).The area \(A_2 = \pi (0.05)^2 = 7.85 \times 10^{-3}\, \text{m}^2\).Thus, force \(F_2 = P \times A_2 = 2.15 \times 10^{7} \times 7.85 \times 10^{-3} = 1.69 \times 10^{5}\, \text{N} \).

03

Calculate the Spring Constant

Force exerted on the spring \(F = kx\) can be used to find the spring constant \(k\).Here, \(x = 1.25\, \text{m}\) is the compression distance.Rearrange to find \(k\):\[k = \frac{F}{x} = \frac{1.69 \times 10^{5}}{1.25} = 1.35 \times 10^{5}\, \text{N/m}.\]

04

Calculate the Velocity of the Steel Ball

The potential energy stored in the spring (elastic potential energy) is converted to the kinetic energy of the steel ball.Using the formula: \(\frac{1}{2} kx^2 = \frac{1}{2} mv^2\), solve for \(v\).\(\frac{1}{2} (1.35 \times 10^{5})(1.25)^2 = \frac{1}{2} (25.0) v^2\)Solving for \(v\):\[v^2 = \frac{1.35 \times 10^{5} \times 1.5625}{25.0} \v^2 = 8.4375 \times 10^3 \v = \sqrt{8.4375 \times 10^3} \approx 91.8 \, \text{m/s}\]

05

Calculate the Maximum Range of the Spring Gun

Use the formula for projectile range, where the range \(R\) is given by:\[R = \frac{v^2 \sin 2\theta}{g}\]Assume optimal angle \(\theta = 45^{\circ}\) for maximum range.Thus,\[R = \frac{(91.8)^2}{9.81} = \frac{8428.44}{9.81} \approx 859 \text{ m}.\]

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Hooke's Law

Imagine a spring that compresses when you press on it and returns to its original shape when released. This behavior can be explained by Hooke's Law. Hooke’s Law states that the force required to compress or extend a spring is proportional to the distance it is stretched or compressed. Mathematically, it's expressed as:

• \( F = kx \)

Here, \( F \) is the force applied on the spring, \( k \) is the spring constant, and \( x \) is the displacement from the spring’s rest position.
The spring constant, \( k \), tells us how stiff a spring is. A larger \( k \) value means a stiffer spring, needing more force to compress or stretch it a certain amount. In the problem with the hydraulic press, the spring is compressed by the force transferred from the larger piston, allowing us to calculate the spring constant using the known force and displacement.
Knowing Hooke's Law is fundamental for understanding how the spring helps store energy and how much force it can exert when returning to its original state.

Energy Conservation

Energy conservation is a key concept that explains how energy transforms from one form to another without being created or destroyed. In this scenario, the potential energy stored in the compressed spring converts to kinetic energy as the spring returns to its original form to project the steel ball.
The potential energy in the spring is given by:

• \( rac{1}{2} kx^2 \)

Once released, this energy transfers into kinetic energy of the projectile, described by:

• \( rac{1}{2} mv^2 \)

Here, \( m \) is mass, and \( v \) is velocity of the ball. By setting these energies equal

• \( rac{1}{2} kx^2 = rac{1}{2} mv^2 \)

we can solve for the velocity \( v \) of the ball.
This transfer validates the principle of energy conservation, showing energy neither appears nor disappears but only changes forms.

Projectile Motion

Projectile motion describes the trajectory of an object that once launched, travels freely under the influence of gravity. When the spring gun ejects the steel ball, it follows a curved path called a trajectory. The basic projectile motion equations can predict this path.
For maximum horizontal range, the ball should be launched at an angle of \(45^{\circ}\). The range \( R \) of the projectile can be calculated using:

• \( R = \frac{v^2 \sin 2\theta}{g} \)

where \( g \) is the acceleration due to gravity, and \( \theta \) is the launch angle.
This calculation uses the initial velocity achieved from the energy transformation of the compressed spring. By understanding projectile motion, we can determine not only the distance the ball travels but also how external forces like gravity affect its path.

Hydraulic Press

A hydraulic press is an engineering marvel that uses a small amount of force to produce a larger force, thus exemplifying Pascal's Principle. This system involves two cylinders of different sizes connected by a fluid.
The pressure created in the small cylinder by manually exerted force transfers to the larger cylinder. The pressure remains constant throughout the fluid, described by:

• \( P = \frac{F}{A} \)

where \( P \) is pressure, \( F \) is force, and \( A \) is the cross-sectional area. In this spring gun, using a hydraulic press provides the necessary large force to compress the spring significantly with less effort.
This principle of equal pressure enables the larger area of the second cylinder to exert a much greater force, showcasing how hydraulic systems efficiently multiply forces to accomplish work. Understanding hydraulic systems is crucial for manipulating mechanical advantage and influencing pressure distributions in various applications.