/*! 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 35 A toy cannon uses a spring to pr... [FREE SOLUTION] | 91Ó°ÊÓ

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A toy cannon uses a spring to project a \(5.30-\mathrm{g}\) soft rubber ball. The spring is originally compressed by \(5.00 \mathrm{cm}\) and has a force constant of \(8.00 \mathrm{N} / \mathrm{m}\). When the cannon is fired, the ball moves \(15.0 \mathrm{cm}\) through the horizontal barrel of the cannon, and there is a constant friction force of \(0.0320 \mathrm{N}\) between the barrel and the ball. (a) With what speed does the projectile leave the barrel of the cannon? (b) At what point does the ball have maximum speed? (c) What is this maximum speed?

Short Answer

Expert verified
(a) The projectile leaves the barrel of the cannon with a speed of \(20.1 \mathrm{m/s}\). (b) The ball has maximum speed at the point where the spring is fully extended before the beginning of the barrel. (c) This maximum speed is \(21.1 \mathrm{m/s}\).

Step by step solution

01

Calculate the potential energy of the compressed spring

The energy stored in a compressed spring is given by the formula \(U_{\mathrm{spring}}=1/2 k x^2\), where \(k\) is the spring constant and \(x\) is the distance the spring is stretched or compressed. Given \(k = 8.00 \mathrm{N/m}\) and \(x = 5.00 \mathrm{cm} = 0.05 \mathrm{m}\), we find: \(U_{\mathrm{spring}} = 1/2\ * 8.00\ \mathrm{N/m} * (0.05 \mathrm{m})^2 = 1.0 \mathrm{J}\).
02

Compute the work done against friction

The work done against friction can be calculated using the formula \(W=F*d\), where \(F\) refers to the force of friction and \(d\) refers to the distance over which the force is applied. Hence, \(W = 0.0320\ \mathrm{N} * 0.15\ \mathrm{m} = 0.0048\ \mathrm{J}\).
03

Determine the projectile's final speed

According to the conservation of mechanical energy, the initial potential energy of the spring minus the work done against friction should be equal to the projectile's kinetic energy when it leaves the barrel, which can be expressed as \( \frac{1}{2} m v_f^2\), where\( m = 5.30\ \mathrm{g} = 0.00530\ \mathrm{kg}\) is the mass of the ball and \( v_f\) is its speed. Solving for \( v_f\) we get: \( v_f = \sqrt{\frac{2 * (U_{\mathrm{spring}} - W)}{m}} = \sqrt{\frac{2 * (1.0\ \mathrm{J} - 0.0048\ \mathrm{J})}{0.00530\ \mathrm{kg}}} = 20.1\ \mathrm{m/s}\).
04

Identify the point of maximum speed

The maximum speed of the ball occurs at the point where the spring is fully extended, just prior to reaching the end of the barrel where external friction starts to encounter. This is due to the fact that, at this point, the ball has transformed all of its potential energy into kinetic energy while no energy has been wasted on work done against friction yet.
05

Compute the maximum speed

Just like in Step 3, but now consider the absence of any work done due to friction: \( v_{max} = \sqrt{\frac{2 * (U_{\mathrm{spring}} - 0)}{m}} = \sqrt{\frac{2 * 1.0\ \mathrm{J}}{0.00530\ \mathrm{kg}}} = 21.1\ \mathrm{m/s}\).

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Key Concepts

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

Spring Potential Energy
When you compress or stretch a spring, you're doing work on it, and this work is stored as potential energy within the spring. This energy is called spring potential energy. The amount of energy stored is directly related to how much the spring is compressed or stretched from its original position.

The mathematical formula for calculating spring potential energy is given by \( U_{\mathrm{spring}} = \frac{1}{2} k x^2 \), where \( k \) is the spring constant which tells us how stiff the spring is, and \( x \) is the amount of compression or stretch. In our exercise, with \( k = 8.00 \mathrm{N/m} \) and \( x = 0.05 \mathrm{m} \), the spring stores an energy of \( 1.0 \mathrm{J} \) when it's compressed. Spring potential energy is a form of mechanical energy, one of the core concepts of conservation of mechanical energy.

Connection with Conservation of Energy

Under ideal conditions (no non-conservative forces like friction), this stored energy will be entirely converted into kinetic energy when the spring is released. In real-world applications, though, some energy is always lost, often to friction, which leads to the important concept of work done against friction.
Work Done Against Friction
Friction is a force that opposes motion, and when a moving object experiences friction, some of its energy is converted into heat and sound, which is generally considered as 'lost' from the system we're interested in. The work done against friction can be calculated as the product of the frictional force and the distance over which it acts. In mathematical terms, it's expressed as \( W = F * d \).

For our toy cannon, with a friction force of \( 0.0320 \mathrm{N} \) and a barrel length of \( 0.15 \mathrm{m} \), we compute the work against friction to be \( 0.0048 \mathrm{J} \). This small value is subtracted from the spring potential energy to find the final kinetic energy of the ball as it leaves the cannon.

Energy Conversion and Loss

It’s important to understand that in the presence of friction, not all the spring potential energy gets transferred to kinetic energy. Work done against friction essentially 'eats away' at the total mechanical energy available for conversion into kinetic energy, making it critical in calculations involving real-life applications where friction cannot be ignored.
Kinetic Energy
Kinetic energy is the energy of motion. Any object that is moving has kinetic energy, and the faster it moves, the more kinetic energy it has. This form of energy can be quantified by the equation \( KE = \frac{1}{2} m v^2 \), where \( m \) is the object's mass and \( v \) is its velocity.

In the context of our problem, when the soft rubber ball leaves the toy cannon's barrel, it is moving, and therefore, has kinetic energy. By applying the conservation of mechanical energy principle, we know that the energy isn't lost but transformed from spring potential energy into kinetic energy (minus any losses due to friction). The ball leaves the barrel with a kinetic energy equal to the initial spring potential energy of \( 1.0 \mathrm{J} \), minus the work done against friction, resulting in a final velocity of \( 20.1 \mathrm{m/s} \).

Maximum Speed and Energy Conversion

The ball reaches its maximum speed when all the spring potential energy is converted into kinetic energy, right before significant frictional forces start to act on it. That's where our maximum speed of \( 21.1 \mathrm{m/s} \) comes from. It’s important to recognize this point because it represents the maximum efficiency of energy transfer from potential to kinetic in the system.

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Most popular questions from this chapter

At time \(t_{i},\) the kinetic energy of a particle is \(30.0 \mathrm{J}\) and the potential energy of the system to which it belongs is \(10.0 \mathrm{J}\) At some later time \(t\), the kinetic energy of the particle is 18.0 J. (a) If only conservative forces act on the particle, what are the potential energy and the total energy at time \(t_{f} ?\) (b) If the potential energy of the system at time \(t_{f}\) is \(5.00 \mathrm{J},\) are there any nonconservative forces acting on the particle? Explain.

Dave Johnson, the bronze medalist at the 1992 Olympic decathlon in Barcelona, leaves the ground at the high jump with vertical velocity component \(6.00 \mathrm{m} / \mathrm{s}\). How far does his center of mass move up as he makes the jump?

The world's biggest locomotive is the MK5000C, a behemoth of mass 160 metric tons driven by the most powerful engine ever used for rail transportation, a Caterpillar diesel capable of 5000 hp. Such a huge machine can provide a gain in efficiency, but its large mass presents challenges as well. The engineer finds that the locomotive handles differently from conventional units, notably in braking and climbing hills. Consider the locomotive pulling no train, but traveling at \(27.0 \mathrm{m} / \mathrm{s}\) on a level track while operating with output power \(1000 \mathrm{hp} .\) It comes to a \(5.00 \%\) grade (a slope that rises \(5.00 \mathrm{m}\) for every \(100 \mathrm{m}\) along the track). If the throttle is not advanced, so that the power level is held steady, to what value will the speed drop? Assume that friction forces do not depend on the speed.

Assume that you attend a state university that started out as an agricultural college. Close to the center of the campus is a tall silo topped with a hemispherical cap. The cap is frictionless when wet. Someone has somehow balanced a pumpkin at the highest point. The line from the center of curvature of the cap to the pumpkin makes an angle \(\theta_{i}=0^{\circ}\) with the vertical. While you happen to be standing nearby in the middle of a rainy night, a breath of wind makes the pumpkin start sliding downward from rest. It loses contact with the cap when the line from the center of the hemisphere to the pumpkin makes a certain angle with the vertical. What is this angle?

A loaded ore car has a mass of \(950 \mathrm{kg}\) and rolls on rails with negligible friction. It starts from rest and is pulled up a mine shaft by a cable connected to a winch. The shaft is inclined at \(30.0^{\circ}\) above the horizontal. The car accelerates uniformly to a speed of \(2.20 \mathrm{m} / \mathrm{s}\) in \(12.0 \mathrm{s}\) and then continues at constant speed. (a) What power must the winch motor provide when the car is moving at constant speed? (b) What maximum power must the winch motor provide? (c) What total energy transfers out of the motor by work by the time the car moves off the end of the track, which is of length \(1250 \mathrm{m} ?\)

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