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Chapter 6: Problem 78

One end of a horizontal spring with force constant \(76.0 \mathrm{~N} / \mathrm{m}\) is attached to a vertical post. A \(2.00 \mathrm{~kg}\) block of frictionless ice is attached to the other end and rests on the floor. The spring is initially neither stretched nor compressed. A constant horizontal force of 54.0 \(\mathrm{N}\) is then applied to the block, in the direction away from the post. (a) What is the speed of the block when the spring is stretched \(0.400 \mathrm{~m} ?\) (b) At that instant, what are the magnitude and direction of the acceleration of the block?

Short Answer

Expert verified
The speed of the block when the spring is stretched is calculated using kinetic energy equations. And, the acceleration of the block at that instant is found by using the net force exerted on the block and Newton's Second Law.

Step by step solution

01

Determine the Displacement of the Spring

Compute the displacement of the spring \(x\) exerted by the force using Hooke's Law for springs, \(F = kx\), where \(F\) is the applied force, \(k\) is the spring constant, and \(x\) is the displacement of the spring. Resolving for \(x = F/k\), substituting the given values, \(x = 54.0\,N / 76.0\,N/m = 0.7105\,m.\)
02

Computing the kinetic energy of the block when the spring is compressed

When a compressing or stretching force is removed from a spring, it oscillates and the potential energy is converted to kinetic energy of the block. The kinetic energy \(KE\) is given by \(KE = \frac{1}{2}kx^2\), where \(k = 76.0\,N/m\) is the spring constant and \(x = 0.400\,m\) is the spring stretch. The kinetic energy is converted to the speed of the block \(v\), using the relation \(KE = \frac{1}{2}mv^2\), where \(m = 2.00\,kg\) is the block mass. Resolving for \(v\), we get \(v = \sqrt{ (2*KE) / m }\). Calculate the kinetic energy and use it to find the speed.
03

Calculating the Magnitude and Direction of the Block's Acceleration

Now, determine the magnitude of the block's acceleration with the formula for Newton's Second Law, \(F = ma\), where \(F\) is the force exerted on the block, \(m\) is the mass of the block, and \(a\) is the acceleration. So, \(a = F/m\). At this instant the force will be equal to the restoring force of the spring \(Fs = kx\), where \(x = 0.400\,m\). Use the calculated force on the spring to determine acceleration. The direction of the acceleration will be in the opposite direction of the applied force, towards the post.

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Key Concepts

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

Spring Constant Calculation
The spring constant, symbolized by the letter k, is a measure of the stiffness of a spring. It denotes the force required to stretch or compress the spring by one unit of length (usually meters). In the given exercise, Hooke's Law, which is represented by the formula F = kx, plays a critical role in determining the spring constant.

Hooke's Law states that the force needed to extend or compress a spring is directly proportional to the distance the spring is stretched or compressed from its resting position. With the given force F and the displacement x, the spring constant can be calculated by rearranging the law to k = F/x. It's important for students to understand that the spring constant is unique to each spring and is a measure of its potential to store elastic potential energy.
Kinetic Energy Calculation
Kinetic energy, often abbreviated as KE, is the energy an object possesses due to its motion. The kinetic energy of an object can be calculated using the equation KE = 1/2 mv^2, where m represents the mass of the object and v is its velocity.

In the problem given, the kinetic energy of the block is found by equating it with the potential energy stored in the spring when it is compressed or stretched. The spring's potential energy at displacement x is given by KE = 1/2 kx^2. This formula shows that the kinetic energy of the block is directly proportional to the square of the displacement and to the spring constant. Knowing the kinetic energy allows us to backtrack to find the velocity, providing insights into the block's speed when the spring is stretched by a certain amount.
Newton's Second Law Application
Newton's Second Law of Motion is foundational to understanding dynamics in physics. The law is written as F = ma, where F is the net force acting on an object, m is the mass of the object, and a is its acceleration.

In the context of our problem, we apply this law to calculate the acceleration of the block when the spring is stretched. At the instant the spring is stretched, the force exerted is the restoring force of the spring, which can be found using Hooke's Law (Fs = kx). The direction of the acceleration is always in the opposite direction of the applied force. Therefore, knowing the force applied by the spring at a given extension allows us to find how much the block is accelerating towards the equilibrium position.

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Most popular questions from this chapter

When its \(75 \mathrm{~kW}\) (100 hp) engine is generating full power, a small single-engine airplane with mass \(700 \mathrm{~kg}\) gains altitude at a rate of \(2.5 \mathrm{~m} / \mathrm{s}(150 \mathrm{~m} / \mathrm{min},\) or \(500 \mathrm{ft} / \mathrm{min}) .\) What fraction of the engine power is being used to make the airplane climb? (The remainder is used to overcome the effects of air resistance and of inefficiencies in the propeller and engine.)

BIO Power of the Human Heart. The human heart is a powerful and extremely reliable pump. Each day it takes in and discharges about \(7500 \mathrm{~L}\) of blood. Assume that the work done by the heart is equal to the work required to lift this amount of blood a height equal to that of the average American woman ( \(1.63 \mathrm{~m}\) ). The density (mass per unit volume) of blood is \(1.05 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\). (a) How much work does the heart do in a day? (b) What is the heart's power output in watts?

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You throw a \(3.00 \mathrm{~N}\) rock vertically into the air from ground level. You observe that when it is \(15.0 \mathrm{~m}\) above the ground, it is traveling at \(25.0 \mathrm{~m} / \mathrm{s}\) upward. Use the work-energy theorem to find (a) the rock's speed just as it left the ground and (b) its maximum height.

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