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Chapter 10: Problem 9

In 0.750 s, a \(7.00-\mathrm{kg}\) block is pulled through a distance of 4.00 \(\mathrm{m}\) on a frictionless horizontal surface, starting from rest. The block has a constant acceleration and is pulled by means of a horizontal spring that is attached to the block. The spring constant of the spring is 415 \(\mathrm{N} / \mathrm{m} .\) By how much does the spring stretch?

Short Answer

Expert verified
The spring stretches approximately 0.239 m.

Step by step solution

01

Determine the acceleration

Since the block starts from rest and reaches 4.00 m in 0.750 seconds, use the kinematic equation: \( s = ut + \frac{1}{2}at^2 \), where \( s = 4.00 \, \text{m} \), \( u = 0 \), and \( t = 0.750 \, \text{s} \). Plug the known values in to find \( a \): \[ 4.00 = 0 + \frac{1}{2}a (0.750^2) \] Solve for \( a \): \[ 4.00 = \frac{1}{2}a (0.5625) \] \[ a = \frac{4.00}{0.28125} \approx 14.2 \, \text{m/s}^2 \].
02

Calculate the force exerted by the spring

Using Newton's second law, \( F = ma \), to find the force exerted by the spring: \[ F = 7.00 \, \text{kg} \times 14.2 \, \text{m/s}^2 \] \[ F \approx 99.4 \, \text{N} \].
03

Determine the stretch of the spring

Using Hooke's Law \( F = kx \) to find the stretch \( x \) of the spring, where \( k = 415 \, \text{N/m} \): \[ 99.4 = 415x \] Solve for \( x \): \[ x = \frac{99.4}{415} \approx 0.239 \, \text{m} \].

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Key Concepts

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

Kinematics
When analyzing the motion of an object, like the block in the exercise, we often refer to kinematics to describe its path. Kinematics focuses on the geometry of motion and not the forces causing it. For our problem, we needed to determine how far the block traveled over a period, using the kinematic equation:
  • \( s = ut + \frac{1}{2}at^2 \)
This relates the displacement \( s \), initial velocity \( u \), time \( t \), and acceleration \( a \). Since the block starts from rest, \( u = 0 \), simplifying our calculations. Taking known values of 4.00 m displacement and 0.750 seconds time, we could determine the block's constant acceleration.
Newtons Second Law
Newton's second law of motion is a key principle in understanding forces and motion. It states:
  • \( F = ma \)
Where \( F \) is the force acting on the object, \( m \) is its mass, and \( a \) is its acceleration. In this exercise, once we identified the acceleration at approximately 14.2 m/s虏, the next step was to find the force the spring exerted on the block. Multiplying the block's mass, 7.00 kg, by its acceleration gave us the force of 99.4 N. This calculation confirms the direct relationship between mass, acceleration, and resultant force on the block as it is pulled.
Hookes Law
Hooke's Law helps us understand how springs operate. It is expressed as:
  • \( F = kx \)
Here, \( F \) represents the force applied, \( k \) is the spring constant, and \( x \) is the spring's stretch (or compression). In our problem, we already found the force exerted by the spring to be 99.4 N. Knowing the spring constant is 415 N/m, we could then solve for the stretch \( x \). The outcome shows a stretch of approximately 0.239 meters, revealing how far the spring extends due to the applied force.
Constant Acceleration
Constant acceleration is a scenario where an object's acceleration remains uniform over time. This condition is vital for using specific kinematic equations to predict motion accurately. In this exercise, we assumed constant acceleration while calculating the displacement of the 7.00 kg block. The advantage of this assumption is that it allows straightforward solutions using initial conditions and the equation \( s = ut + \frac{1}{2}at^2 \).When acceleration is constant, computations become easier, lending clarity to motion problems. The block's uniform acceleration enabled us to derive necessary mechanical insights such as net force and spring behavior with simplicity.

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Most popular questions from this chapter

A tray is moved horizontally back and forth in simple harmonic motion at a frequency of \(f=2.00 \mathrm{Hz}\) . On this tray is an empty cup. Obtain the coefficient of static friction between the tray and the cup, given that the cup begins slipping when the amplitude of the motion is \(5.00 \times 10^{-2} \mathrm{m} .\)

A uniform \(1.4-\mathrm{kg}\) rod that is 0.75 \(\mathrm{m}\) long is suspended at rest from the ceiling by two springs, one at each end of the rod. Both springs hang straight down from the ceiling. The springs have identical lengths when they are unstretched. Their spring constants are 59 \(\mathrm{N} / \mathrm{m}\) and 33 \(\mathrm{N} / \mathrm{m}\) . Find the angle that the rod makes with the horizontal.

A hand exerciser utilizes a coiled spring. A force of 89.0 \(\mathrm{N}\) is required to compress the spring by 0.0191 \(\mathrm{m} .\) Determine the force needed to compress the spring by 0.0508 \(\mathrm{m} .\)

A 1.1\(\cdot \mathrm{kg}\) object is suspended from a vertical spring whose spring constant is 120 \(\mathrm{N} / \mathrm{m}\) . (a) Find the amount by which the spring is stretched from its unstrained length. (b) The object is pulled straight down by an additional distance of 0.20 \(\mathrm{m}\) and released from rest. Find the speed with which the object passes through its original position on the way up.

Two stretched cables both experience the same stress. The first cable has a radius of \(3.5 \times 10^{-3} \mathrm{m}\) and is subject to a stretching force of 270 \(\mathrm{N}\) . The radius of the second cable is \(5.1 \times 10^{-3} \mathrm{m} .\) Determine the stretching force acting on the second cable.
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