/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 26 The catapulting mechanism is use... [FREE SOLUTION] | 91Ó°ÊÓ

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The catapulting mechanism is used to propel the 10 -kg slider \(A\) to the right along the smooth track. The propelling action is obtained by drawing the pulley attached to rod \(B C\) rapidly to the left by means of a piston \(P .\) If the piston applies a constant force \(F=20 \mathrm{kN}\) to rod \(B C\) such that it moves it \(0.2 \mathrm{m}\), determine the speed attained by the slider if it was originally at rest. Neglect the mass of the pulleys, cable, piston, and rod \(B C\)

Short Answer

Expert verified
The speed attained by the slider \( A \) is \( 28.28 \mathrm{m/s} \).

Step by step solution

01

Calculations of the work done by the piston

The work done by the piston on the rod is the force applied, \( F \), multiplied by the distance the piston moves, \( d \). This can be calculated using the formula \(W = F \cdot d\). Substitute \( F = 20 \mathrm{kN} = 20000 \mathrm{N} \) and \( d= 0.2 \mathrm{m} \) into the formula to get the work done: \( W = 20000 \cdot 0.2 = 4000 \mathrm{J} \).
02

Calculate kinetic energy and velocity of slider

Assuming no energy losses in the setup, the work done by the piston is converted into kinetic energy of the slider (\( A \)). Therefore, the kinetic energy of the slider (\( KE \)) is equal to the work done (\( W \)): \( KE = W = 4000 \mathrm{J} \). Knowing kinetic energy, the velocity of the slider can be calculated from \( KE = \frac{1}{2} m v^2 \), where \( m = 10 \mathrm{kg} \) is the mass of the slider and \( v \) is its velocity. Solving for \( v \) gives \( v = \sqrt{\frac{2 KE}{m}} \). Substituting \( KE = 4000 \mathrm{J} \) and \( m = 10 \mathrm{kg} \) gives \( v = \sqrt{\frac{2 \cdot 4000}{10}} = 28.28 \mathrm{m/s} \).

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Key Concepts

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

Work-Energy Principle
The work-energy principle is a fundamental concept in physics that relates the work done on an object to its change in kinetic energy. The basic idea is that when work is done on an object, it either increases or decreases the object's kinetic energy.

In mathematical terms, this is expressed as \( KE_{final} - KE_{initial} = W \), where \( KE_{final} \) is the final kinetic energy of the object, \( KE_{initial} \) is the initial kinetic energy, and \( W \) is the work done by the forces acting on the object.

For example, in the textbook exercise, the slider started from rest, meaning its initial kinetic energy was zero \( (KE_{initial} = 0) \). The work done by the piston was converted entirely into the slider’s kinetic energy \( (KE_{final} = W) \), which we computed using the work done by the constant force applied over a certain distance. This demonstrates the work-energy principle in action, bridging the gap between force, displacement, and kinetic energy.
Dynamics of a Slider
When considering the dynamics of a slider, it's essential to understand how the forces acting on the slider affect its motion over a smooth track. A slider mechanism, like the one in our exercise, usually involves a flat or curved surface on which it moves horizontally or vertically with negligible friction.

The dynamics of such a system are governed by Newton's second law, which states that the force applied is equal to the mass of the slider times its acceleration \( (F = ma) \). However, if we apply the work-energy principle, we focus on the work done through the distance the slider travels, not just the instantaneous force.

In the absence of friction and other resistive forces, all the work done by external forces (like the piston in our scenario) converts to the kinetic energy of the slider. Thus, understanding the dynamics involves recognizing the interplay between force, mass, acceleration, and the resulting displacement that leads to changes in kinetic energy.
Kinetic Energy and Velocity
Kinetic energy, which is the energy of motion, can often be a more practical quantity to consider than forces and acceleration when studying the motion of objects. This is because it quantifies the total energy associated with the object's motion and can be directly related to its velocity.

The mathematical relationship between kinetic energy (KE) and velocity (v) of an object is given by the formula \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass of the object. From this formula, we can see that the kinetic energy of an object increases with the square of its velocity, meaning that even small increases in velocity result in significant increases in kinetic energy.

In our textbook exercise, once we calculated the work done (which is equal to the kinetic energy for a frictionless system), we were able to determine the slider's velocity. By rearranging the kinetic energy formula to solve for velocity, we find \( v = \sqrt{\frac{2 KE}{m}} \). This equation gave us a clear connection between the work done by the piston and the resulting speed of the slider.

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Most popular questions from this chapter

A 60 -kg satellite travels in free flight along an elliptical orbit such that at \(A\), where \(r_{A}=20 \mathrm{Mm}\), it has a speed \(v_{A}=40 \mathrm{Mm} / \mathrm{h} .\) What is the speed of the satellite when it reaches point \(B,\) where \(r_{B}=80 \mathrm{Mm} ?\) Hint: See Prob. \(14-82, \quad\) where \(\quad M_{e}=5.976\left(10^{24}\right) \mathrm{kg} \quad\) and \(G=66.73\left(10^{-12}\right) \mathrm{m}^{3} /\left(\mathrm{kg} \cdot \mathrm{s}^{2}\right) \)

The "flying car" is a ride at an amusement park which consists of a car having wheels that roll along a track mounted inside a rotating drum. By design the car cannot fall off the track, however motion of the car is developed by applying the car's brake, thereby gripping the car to the track and allowing it to move with a constant speed of the track, \(v_{t}=3 \mathrm{m} / \mathrm{s} .\) If the rider applies the brake when going from \(B\) to \(A\) and then releases it at the top of the drum, \(A\), so that the car coasts freely down along the track to \(B(\theta=\pi \mathrm{rad})\) determine the speed of the car at \(B\) and the normal reaction which the drum exerts on the car at \(B\). Neglect friction during the motion from \(A\) to \(B\). The rider and car have a total mass of \(250 \mathrm{kg}\) and the center of mass of the car and rider moves along a circular path having a radius of \(8 \mathrm{m}\)

Determine the required height \(h\) of the roller coaster so that when it is essentially at rest at the crest of the hill \(A\) it will reach a speed of \(100 \mathrm{km} / \mathrm{h}\) when it comes to the bottom \(B\). Also, what should be the minimum radius of curvature \(\rho\) for the track at \(B\) so that the passengers do not expericnce a normal force greater than \(4 m g=(39.24 m)\) N? Neglect the size of the car and passenger.

The block has a mass of \(150 \mathrm{kg}\) and rests on a surface for which the coefficients of static and kinetic friction are \(\mu_{s}=0.5\) and \(\mu_{k}=0.4,\) respectively. If a force \(F=\left(60 t^{2}\right) \mathrm{N},\) where \(t\) is in seconds, is applied to the cable, determine the power developed by the force when \(t=5\) s. Hint: First determine the time needed for the force to cause motion.

A pan of negligible mass is attached to two identical springs of stiffness \(k=250 \mathrm{N} / \mathrm{m}\). If a 10 -kg box is dropped from a height of \(0.5 \mathrm{m}\) above the pan, determine the maximum vertical displacement \(d\). Initially each spring has a tension of \(50 \mathrm{N}\)

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