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Chapter 8: Problem 8

A system consists of two identical cubes, each of mass \(3 \mathrm{~kg}\), linked together by a compressed weightless spring of force constant \(1000 \mathrm{~N} \mathrm{~m}^{-1}\). The cubes are also connected by a thread which is burnt at a certain moment. At what minimum value of initial compression \(x_{0}\) (in \(\mathrm{cm}\) ) of the spring will the lower cube bounce up after the thread is burnt through?

Short Answer

Expert verified
The minimum value of initial compression is 2.94 cm.

Step by step solution

01

Identify the forces involved

In this problem, we have two forces to consider: the gravitational force acting on the lower cube and the restoring force from the spring. We will need to calculate when the restoring force from the spring becomes strong enough to overcome gravity.
02

Calculate the gravitational force

Each cube has a mass of 3 kg. Thus, the gravitational force (F_g) acting on each cube is given by F_g = mg, where \( m = 3 \text{ kg} \) and \( g = 9.8 \text{ m/s}^2 \) (acceleration due to gravity). Therefore, \( F_g = 3 \times 9.8 = 29.4 \text{ N} \).
03

Equate forces for minimum compression

The minimum compression of the spring must provide an upward force equal to at least the gravitational force on the lower cube to make it bounce. For the spring, the force is given by F_s = k x_0 , where \( k = 1000 \text{ N/m} \) is the spring constant, and \( x_0 \) is the initial compression. Equate F_s to F_g to find x_0: \[ 1000 imes x_0 = 29.4 \]
04

Solve for initial compression

Solve the equation from Step 3 for x_0: \[ x_0 = \frac{29.4}{1000} = 0.0294 \text{ m} \]Convert this into centimeters: \[ 0.0294 \text{ m} = 2.94 \text{ cm} \].

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Key Concepts

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

Gravitational Force
Gravitational force is the attraction exerted by the Earth on any object towards its center. It's a fundamental force that acts on every mass. In our exercise, each cube experiences gravitational force that pulls it downwards. This force is calculated using the formula:
  • Formula: \( F_g = mg \)
  • \( m \) is the mass of the object, \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \) on Earth's surface).
For the cubes in this problem, each has a mass of 3 kg, resulting in a gravitational force of 29.4 N. Understanding this force is crucial because the spring force needs to counteract this, making the gravitational force a key consideration in our calculations. Gravitational force sets the initial condition that the spring must overcome to move the lower cube upwards.
Spring Force
Spring force is the restoring force exerted by a spring when it is compressed or stretched. It acts in the opposite direction to the displacement and tries to bring the system back to its equilibrium state. This force is given by Hooke's Law:
  • Hooke's Law: \( F_s = kx \)
  • \( k \) is the spring constant, indicating the stiffness of the spring.
  • \( x \) is the displacement from the equilibrium position, which is the spring's compression or extension.
In our scenario, the spring is compressed, and the goal is to find this initial compression, \( x_0 \), that produces a force equal to or greater than the gravitational force acting downward on the lower cube. The spring constant \( k \) here is 1000 N/m, which signifies that this spring is quite stiff, allowing for significant force generation even with small compressions.
Compression
Compression in a spring refers to the reduction in length when the spring is pressed together. For springs, the compression must be calculated in such a way that it can exert a sufficient force to counter any opposing forces鈥攊n this case, the gravitational force on the lower cube. This exercise requires calculating the minimum compression, \( x_0 \), necessary to ensure that the spring's restoring force is at least equal to the gravitational force. Using the formula for spring force, \( F_s = k x_0 \), and equating it to the gravitational force, we solve for \( x_0 \):
  • \( x_0 = \frac{F_g}{k} \)
  • Using \( F_g = 29.4 \, \text{N} \) and \( k = 1000 \, \text{N/m} \), \( x_0 \) is calculated to be 0.0294 m or 2.94 cm.
Understanding the compression specifics helps us ensure that the system can work effectively as desired, allowing the lower cube to bounce upward once the thread is burnt and the spring fully extends.

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Most popular questions from this chapter

Which of the following statements is correct? (1) Kinetic energy of a system can be changed without changing its momentum. (2) Kinetic energy of a system cannot be changed without changing its momentum. (3) Momentum of a system cannot be changed without changing its kinetic energy. (4) A system cannot have energy without having momentum.

One end of an unstretched vertical spring is attached to the ceiling and an object attached to the other end is slowly lowered to its equilibrium position. If \(S\) is the gain in spring energy and \(G\) is the loss in gravitational potential energy in the process, then (1) \(S=G\) (2) \(S=2 G\) (3) \(G=2 S\) (4) None of these

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A block of mass \(M\) rests on a table. It is fastened to the lower end of a light, vertical spring. The upper end of the spring is fastened to a block of mass \(m\). The upper block is pushed down by an additional force \(3 \mathrm{mg}\), so the spring compression is \(4 \mathrm{mg} / \mathrm{k}\). In this configuration, the upper block is released from rest. The spring lifts the lower block off the table. What is the greatest possible value for \(\frac{M}{m}\) ?

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