/*! 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 18 A small block of mass \(m\) on a... [FREE SOLUTION] | 91Ó°ÊÓ

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A small block of mass \(m\) on a horizontal frictionless surface is attached to a horizontal spring that has force constant \(k .\) The block is pushed against the spring, compressing the spring a distance \(d\). The block is released, and it moves back and forth on the end of the spring. During its motion, what is the maximum speed of the block?

Short Answer

Expert verified
The maximum speed of the block is \( v = \sqrt{\frac{kd^2}{m}} \).

Step by step solution

01

Identify the energy transformation

The potential energy of the spring is being transformed into the kinetic energy of the block. Let the potential energy, \( PE_{spring} = \frac{1}{2}kd^2 \) and the kinetic energy, \( KE_{block} = \frac{1}{2}mv^2 \). The maximum speed of the block will be reached when the spring is at its equilibrium point and all potential energy released has been converted to kinetic energy.
02

Equate energies and solve for speed

At the point of maximum speed, the block will have expended all its potential energy from the spring. This means that \( PE_{spring} = KE_{block} \). Therefore, we set \( \frac{1}{2}kd^2 = \frac{1}{2}mv^2 \) and solve for \(v\). The \( \frac{1}{2} \) cancels out. Dividing both sides by \(m\) to isolate the variable, you get \( v = \sqrt{\frac{kd^2}{m}} \). Hence, the maximum speed of the block is \( v = \sqrt{\frac{kd^2}{m}} \).

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Key Concepts

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

Conservation of Energy
In the context of a block on a spring, the principle of conservation of energy plays a pivotal role in finding the maximum speed of the block. Conservation of energy states that energy cannot be created or destroyed; it can only be transformed from one form to another or transferred from one object to another. When the block is attached to the spring and the spring is compressed, potential energy is stored in the spring. Once the block is released, this stored potential energy begins to transform into kinetic energy, which is the energy of motion.

At the point where the spring returns to its natural length—referred to as the equilibrium position—there's an important transition occurring; the maximum potential energy stored has been completely converted to kinetic energy. This is when the block reaches its maximum speed because all the energy is now driving the movement of the block with none remaining in the spring.
Spring Potential Energy
Spring potential energy is the energy stored in a spring when it is compressed or stretched. It's dependent upon two key factors: the spring constant and the distance compressed or stretched. The spring constant, denoted by the variable 'k', is a measure of the stiffness of the spring. A higher 'k' means the spring is stiffer, and more force is required to compress it. On the other hand, 'd' is the measure of the spring's deformation.

The formula to calculate spring potential energy is given by \( PE_{spring} = \frac{1}{2}kd^2 \). Now, when the spring in our exercise is compressed by the distance 'd', the same formula applies. The potential energy stored is maximum at this point, and as the block is released and the spring extends, this energy is converted to kinetic energy, eventually causing the block to move.
Kinetic Energy of a Block
Now, let's delve into the kinetic energy of the block, which is directly tied to the block's speed. Kinetic energy is the energy that an object possesses due to its motion, calculated by the formula \( KE_{block} = \frac{1}{2}mv^2 \), where 'm' represents the mass of the block and 'v' is its velocity. When the block reaches the equilibrium point of the spring, all the spring's potential energy has been transferred to the block as kinetic energy. At this point, the velocity of the block is at its peak, indicating the maximum kinetic energy.

The textbook exercise walks us through equating the spring's potential energy to the block's kinetic energy because, at maximum speed, the energy in the system is purely kinetic. Solving the equation for 'v' gives us the maximum velocity as \( v = \sqrt{\frac{kd^2}{m}} \). This shows that the maximum speed depends on the spring's stiffness, the distance initially compressed, and the mass of the block. Remarkably, the block's maximum speed doesn't depend on external factors like gravity or surface friction because the surface is assumed frictionless, emphasizing the simplicity and beauty of the conservation of energy in isolated systems.

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Most popular questions from this chapter

Tarzan and Jane. Tarzan, in one tree, sights Jane in another tree. He grabs the end of a vine with length \(20 \mathrm{~m}\) that makes an angle of \(45^{\circ}\) with the vertical, steps off his tree limb, and swings down and then up to Jane's open arms. When he arrives, his vine makes an angle of \(30^{\circ}\) with the vertical. Determine whether he gives her a tender embrace or knocks her off her limb by calculating Tarzan's speed just before he reaches Jane. Ignore air resistance and the mass of the vine.

The food calorie, equal to \(4186 \mathrm{~J},\) is a measure of how much energy is released when the body metabolizes food. A certain fruit-and-cereal bar contains 140 food calories. (a) If a 65 kg hiker eats one bar, how high a mountain must he climb to "work off" the calories, assuming that all the food energy goes into increasing gravitational potential energy? (b) If, as is typical, only \(20 \%\) of the food calories go into mechanical energy, what would be the answer to part (a)? (Note: In this and all other problems, we are assuming that \(100 \%\) of the food calories that are eaten are absorbed and used by the body. This is not true. A person's "metabolic efficiency" is the percentage of calories eaten that are actually used; the body eliminates the rest. Metabolic efficiency varies considerably from person to person.

An ideal spring stores potential energy \(U_{0}\) when it is compressed a distance \(x_{0}\) from its uncompressed length. (a) In terms of \(U_{0}\). how much energy does the spring store when it is compressed (i) twice as much and (ii) half as much? (b) In terms of \(x_{0}\). how much must the spring be compressed from its uncompressed length to store (i) twice as much energy and (ii) half as much energy?

You are testing a new amusement park roller coaster with an empty car of mass 120 kg. One part of the track is a vertical loop with radius 12.0 m. At the bottom of the loop (point A) the car has speed 25.0 m>s, and at the top of the loop (point B) it has speed 8.0 m>s. As the car rolls from point A to point B, how much work is done by friction?

CALC The potential energy of two atoms in a diatomic molecule is approximated by \(U(r)=\left(a / r^{12}\right)-\left(b / r^{6}\right),\) where \(r\) is the spacing between atoms and \(a\) and \(b\) are positive constants. (a) Find the force \(F(r)\) on one atom as a function of \(r .\) Draw two graphs: one of \(U(r)\) versus \(r\) and one of \(F(r)\) versus \(r\). (b) Find the equilibrium distance between the two atoms. Is this equilibrium stable? (c) Suppose the distance between the two atoms is equal to the equilibrium distance found in part (b). What minimum energy must be added to the molecule to dissociate it - that is, to separate the two atoms to an infinite distance apart? This is called the dissociation energy of the molecule. (d) For the molecule CO, the cquilibrium distance between the carbon and oxygen atoms is \(1.13 \times 10^{-10} \mathrm{~m}\) and the dissociation energy is \(1.54 \times 10^{-18} \mathrm{~J}\) per molecule. Find the values of the constants \(a\) and \(b\).

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