/*! 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 23 \(\mathrm{A} 2.50 \mathrm{~kg}\)... [FREE SOLUTION] | 91Ó°ÊÓ

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\(\mathrm{A} 2.50 \mathrm{~kg}\) mass is pushed against a horizontal spring of force constant \(25.0 \mathrm{~N} / \mathrm{cm}\) on a frictionless air table. The spring is attached to the tabletop, and the mass is not attached to the spring in any way. When the spring has been compressed enough to store \(11.5 \mathrm{~J}\) of potential energy in it, the mass is suddenly released from rest. (a) Find the greatest speed the mass reaches. When does this occur? (b) What is the greatest acceleration of the mass, and when does it occur?

Short Answer

Expert verified
The greatest speed the mass reaches is found by using the conservation of energy principle, setting the kinetic energy equal to the potential energy of the spring, and solving for the speed. The greatest acceleration of the mass occurs when the spring is fully extended or fully compressed and is found by applying Hooke's Law to find the force exerted by the spring, and then using the equation of motion to find the acceleration.

Step by step solution

01

Identify the Potential Energy

Identify the potential energy stored in the compressed spring. In this exercise, it's given as 11.5 J.
02

Apply Conservation of Energy

Using the conservation of energy, the kinetic energy (KE) of the mass when it's released is equal to the potential energy (PE) stored in the spring. Since half of the kinetic energy is \(KE = \frac{1}{2} m v^{2}\), where \(m\) is the mass and \(v\) is the velocity (or speed), you can set this equation equal to the potential energy and solve for \(v\). So you have \(11.5J = \frac{1}{2} (2.5kg) v^{2}\). Solving this equation will give the greatest speed the mass reaches.
03

Find the Maximum Acceleration

Acceleration is the rate of change of velocity over time. It can be found by using the equation of motion \(a = \frac{F}{m}\), where \(F\) is the force exerted by the spring when it's released and \(m\) is the mass of the object. The force \(F\) is obtained from Hooke's Law, which states that the force exerted by a spring is proportional to the displacement of the spring. In this case, the force is maximum when the spring is fully compressed or fully extended, which gives the greatest acceleration. Knowing the force constant \(k\) of the spring and the displacement \(x\) (which can be obtained from the potential energy - \(PE = \frac{1}{2} k x^{2}\)), you can plug these values into Hooke's Law which gives \(F = -kx\), and then plug \(F\) into the equation of motion to find the maximum acceleration.

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. In the context of the exercise, the mass starts from rest and then gains speed as the spring pushes it. This gain in speed translates to an increase in kinetic energy.

The kinetic energy of an object is given by the formula:
  • \( KE = \frac{1}{2} m v^2 \)
where:
  • \( KE \) is the kinetic energy
  • \( m \) is the mass of the object
  • \( v \) is the velocity (or speed) of the object
In our problem, the point at which the spring has completely transferred its stored potential energy to the mass is when the mass achieves its maximum speed, thus converting all the potential energy to kinetic energy.

By setting the stored potential energy equal to the kinetic energy formula, we can solve for the mass's greatest speed. This transformation exemplifies the conservation of energy principle, where no energy is lost, just transformed from one type to another. The understanding of kinetic energy and its conversion from potential energy is crucial in explaining motion dynamics.
Hooke's Law
Hooke's Law describes the behavior of springs and how they exert force when compressed or stretched. It is a fundamental principle explaining the linear relationship between the displacement of a spring and the force it exerts. This principle is very helpful for calculations involving springs, such as the one described in the exercise.

The law is expressed as:
  • \( F = -kx \)
where:
  • \( F \) represents the restoring force exerted by the spring
  • \( k \) is the spring constant, a measure of the spring's stiffness
  • \( x \) is the displacement from the spring's equilibrium position
In this exercise, we find the maximum force exerted by using this law at the point of maximum compression or extension. This is useful to determine the maximum acceleration of the mass once the spring is released. By knowing the spring constant and displacement, we can precisely calculate this force, which plays a critical role in understanding how the spring affects the object's motion.
Potential Energy
Potential energy in this scenario is stored energy in the compressed spring. When you compress a spring, you are doing work on it, which is stored as elastic potential energy.

The formula for the potential energy stored in a spring is:
  • \( PE = \frac{1}{2} k x^2 \)
where:
  • \( PE \) is the potential energy
  • \( k \) is the spring constant, indicating how stiff the spring is
  • \( x \) is the displacement from its rest position
The problem specifies that there is 11.5 Joules of potential energy when the spring is compressed. This energy becomes crucial as it provides the necessary force to propel the mass once released. It showcases how potential energy can be converted into kinetic energy during its motion, reinforcing the conservation of energy principle. Understanding potential energy is vital to predicting and explaining the behavior of the spring system, especially when calculating maximum speeds and accelerations involved.

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Most popular questions from this chapter

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?

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.

You are designing a delivery ramp for crates containing exercise equipment. The \(1470 \mathrm{~N}\) crates will move at \(1.8 \mathrm{~m} / \mathrm{s}\) at the top of a ramp that slopes downward at \(22.0^{\circ} .\) The ramp exerts a \(515 \mathrm{~N}\) kinctic friction force on cach crate, and the maximum static friction force also has this value. Each crate will compress a spring at the bottom of the ramp and will come to rest after traveling a total distance of \(5.0 \mathrm{~m}\) along the ramp. Once stopped, a crate must not rebound back up the ramp. Calculate the largest force constant of the spring that will be needed to meet the design criteria.

A small block with mass \(m\) slides without friction on the inside of a vertical circular track that has radius \(R .\) What minimum speed must the block have at the bottom of its path if it is not to fall off the track at the top of its path?

\(A 2.50 \mathrm{~kg}\) block on a horizontal floor is attached to a horizontal spring that is initially compressed \(0.0300 \mathrm{~m}\). The spring has force constant \(840 \mathrm{~N} / \mathrm{m}\). The coefficient of kinetic friction between the floor and the block is \(\mu_{k}=0.40 .\) The block and spring are released from rest, and the block slides along the floor. What is the speed of the block when it has moved a distance of \(0.0200 \mathrm{~m}\) from its initial position? (At this point the spring is compressed \(0.0100 \mathrm{~m}\).)

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