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Chapter 7: Problem 37

CE A block of mass \(m\) and speed \(v\) collides with a spring. compressing it a distance \(\Delta x .\) What is the compression of the spring if the mass of the block is halved and its speed is doubled?

Short Answer

Expert verified
The compression of the spring remains unchanged at \( \Delta x \).

Step by step solution

01

Understand the Initial Energy Relationship

Initially, the kinetic energy of the block is converted into the elastic potential energy of the spring. This can be expressed as: \[ \frac{1}{2}mv^2 = \frac{1}{2}k(\Delta x)^2 \] where \( k \) is the spring constant and \( \Delta x \) is the compression of the spring.
02

Halve the Mass and Double the Speed

Consider the new scenario where the mass of the block is halved \( (m/2) \) and the speed is doubled \( (2v) \). The new kinetic energy is: \[ \frac{1}{2}(m/2)(2v)^2 \] Simplifying this, we find: \[ \frac{1}{2} \cdot \frac{m}{2} \cdot 4v^2 = \frac{1}{2} \cdot 2mv^2 = mv^2 \]
03

Determine New Spring Compression

The new kinetic energy is equal to the original kinetic energy of the block \( mv^2 \), which was equal to its elastic potential energy previously. Therefore, the compression \( \Delta x \) of the spring remains the same as before since both the energy and spring constant \( k \) are unchanged.

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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 that an object possesses due to its motion. It depends on two key factors: the mass of the object and its velocity. The formula used to calculate kinetic energy is:
  • \( KE = \frac{1}{2}mv^2 \)
Here, \( m \) is the mass of the object, and \( v \) is its velocity.
The kinetic energy increases with both mass and velocity. However, it is important to note that velocity has a direct squared relationship with kinetic energy.
This means that even a small increase in velocity can significantly increase kinetic energy.
Elastic Potential Energy
Elastic potential energy is the stored energy in elastic materials when they are either compressed or stretched. Springs, rubber bands, and similar objects exhibit elastic potential energy.
  • The formula for calculating elastic potential energy is \( EPE = \frac{1}{2}k(\Delta x)^2 \)
  • Here, \( k \) is the spring constant, and \( \Delta x \) is the displacement or compression of the spring from its equilibrium position.
This kind of energy is always present in systems where forces cause an elastic change in shape or configuration. In the context of springs, it's the compression (or stretching) that harbors potential energy, ready to convert back into kinetic energy when the force is released.
Spring Constant
The spring constant, denoted as \( k \), is a measure of a spring's stiffness. It defines how resistant a spring is to being compressed or stretched.
  • The larger the spring constant, the stiffer the spring, requiring more force to achieve the same compression as a spring with a lower constant.
  • The spring constant has the units of Newtons per meter (N/m).
Understanding the spring constant helps us predict how a spring will behave when forces are applied and how it stores elastic potential energy. This concept is critical when calculating how much compression or elongation occurs for a given force or energy input.
Speed and Mass Relationships
Speed and mass both play crucial roles in determining the kinetic energy of an object. Though they are often considered independently, they have interconnected relationships when energy conservation is involved.
  • When the mass is halved and the speed is doubled, the overall kinetic energy remains the same, as seen in the initial exercise.
  • This is because the increase in velocity has a squared effect in the kinetic energy equation, able to offset changes in mass.
Thus, in systems where speed and mass change proportionally, understanding these relationships helps in predicting how energy is conserved or transformed.

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Most popular questions from this chapter

IP A \(1300-\mathrm{kg}\) car delivers a constant 49 hp to the drive wheels. We assume the car is traveling on a level road and that all frictional forces may be ignored. (a) What is the acceleration of this car when its speed is \(14 \mathrm{m} / \mathrm{s} ?\) (b) If the speed of the car is doubled, does its acceleration increase, decrease, or stay the same? Explain. (c) Calculate the car's acceleration when its speed is \(28 \mathrm{m} / \mathrm{s}\).

A 65-kg bicyclist rides his 8.8-kg bicycle with a speed of \(14 \mathrm{m} / \mathrm{s} .\) (a) How much work must be done by the brakes to bring the bike and rider to a stop? (b) How far does the bicycle travel if it takes \(4.0 \mathrm{s}\) to come to rest? \((\mathrm{c})\) What is the magnitude of the braking force?

A juggling ball of mass \(m\) is thrown straight upward from an initial height \(h\) with an initial speed \(v_{0}\). How much work has gravity done on the ball (a) when it reaches its greatest height, \(h_{\text {max }}\), and \((b)\) when it reaches ground level? (c) Find an expression for the kinetic energy of the ball as it lands.

CE A block of mass \(m\) and speed \(v\) collides with a spring, compressing it a distance \(\Delta x\). What is the compression of the spring if the force constant of the spring is increased by a factor of four?

IP A sled with a mass of \(5.80 \mathrm{kg}\) is pulled along the ground through a displacement given by \(\overrightarrow{\mathrm{d}}=(4.55 \mathrm{m}) \hat{\mathrm{x}}\). (Let the \(x\) axis be horizontal and the \(y\) axis be vertical.) (a) How much work is done on the sled when the force acting on it is \(\overrightarrow{\mathbf{F}}=(2.89 \mathrm{N}) \hat{\mathrm{x}}+\) \((0.131 \mathrm{N}) \hat{\mathrm{y}} ?\) (b) How much work is done on the sled when the force acting on it is \(\overrightarrow{\mathbf{F}}=(2.89 \mathrm{N}) \hat{\mathrm{x}}+(0.231 \mathrm{N}) \hat{\mathrm{y}} ?\) (c) If the mass of the sled is increased, does the work done by the forces in parts (a) and (b) increase, decrease, or stay the same? Explain.
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