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Chapter 6: Problem 11

A light spring of natural length \(a\) is placed on a horizontal floor in the upright position. When a block of mass \(M\) is resting in equilibrium on top of the spring, the compression of the spring is \(a / 15\). The block is now lifted to a height \(3 a / 2\) above the floor and released from rest. Find the compression of the spring when the block first comes to rest.

Short Answer

Expert verified
To find the new compression of the spring when the block first comes to rest, you need to set up an energy balance equation using the concept of conservation of energy. At the final state, all energy is potential energy, considering both gravitational potential energy and elastic (spring) potential energy. You then need to solve this equation for the additional compression \( x \).

Step by step solution

01

Initial state of the system

Initially, when the block is on top of the spring and the spring is compressed by \(a/15\), the system is in equilibrium. This implies that the force exerted by the block due to gravity is balanced by the force exerted by the spring. We can then write the equation as follows: \( Mg = k(a/15) \). Here, \(M\) is the mass of the block, \(g\) is the acceleration due to gravity, and \(k\) is the spring constant.
02

Final state of the system

When the block is lifted and then released, it again comes to rest after compressing the spring further. Let us denote this additional compression as \( x \). Then the block will have risen by \( (3a/2) - (a/15 + x) \) above its equilibrium position. Because the block's kinetic energy is zero when it first comes to rest, all its initial energy must have been converted into potential energy. The energy balance can be expressed as: \( Mg(3a/2 - a/15 - x) = (1/2)k(a/15 + x)^2 \). We need to solve this equation for \( x \).
03

Solving for the spring constant

From the equilibrium condition, we can find the spring constant in terms of \(M\), \(a\), and \(g\). Rearranging the equilibrium equation \( Mg = k(a/15) \), we obtain the spring constant as \( k = 15Mg/a \).
04

Substitute \( k \) into the energy balance equation

Substitute \( k = 15Mg/a \) into the energy balance equation: \( Mg(3a/2 - a/15 - x) = (1/2)(15Mg/a)(a/15 + x)^2 \), and then simplify and solve for \( x \).
05

Solve the equation for \( x \)

By simplifying the equation and solving for \( x \), you will find the new compression when the block first comes to rest.

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Key Concepts

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

Mechanical Equilibrium
Understanding the state of mechanical equilibrium is crucial when analyzing systems at rest or in constant motion. In physics, mechanical equilibrium refers to a scenario where all the forces acting on a system cancel out, resulting in no acceleration of the system. This can be depicted by the equation \(\sum\vec{F} = 0\), indicating the sum of all vectors of force equals zero.

In the given problem, when a block of mass \(M\) is resting on a spring, it compresses the spring by \(a/15\) because the gravitational force \(Mg\) (downward force) is exactly countered by the spring's force (upward force), which is characterized by the spring's constant, \(k\), and the compression distance. This condition of no net force signifies mechanical equilibrium, which is essential for determining the spring's constant and the initial setup before any movement occurs.
Energy Conservation
The principle of energy conservation plays a fundamental role in physics problems involving motion and forces. It states that the total energy within an isolated system remains constant, although energy can be transformed from one form to another, such as from kinetic to potential energy and vice versa.

In our spring compression scenario, energy conservation is exemplified when the block, initially at rest, is lifted to a height \(3a/2\) and then released. As the block falls, its potential energy, due to its elevation, is converted to kinetic energy. The moment it hits the spring and comes to rest for the first time, its kinetic energy is completely transformed into spring potential energy. Through energy conservation, we can relate the initial gravitational potential energy to the spring potential energy to find the compression of the spring at that moment, which is expressed mathematically and determined in the step-by-step solution.
Spring Constant Calculation
The spring constant, commonly denoted as \(k\), is a measure of the stiffness of a spring, quantifying the force required per unit extension or compression. Calculating the spring constant is essential to understand how much a spring will compress or extend under a given load.

In our exercise, the spring constant is deduced from the equilibrium condition where the spring compression equals \(a/15\) due to the block's weight. By using Hooke's Law, which states \(F = kx\), we establish a relationship between the force \(F\) (the weight of the block in this case), the spring constant \(k\), and the displacement \(x\). From the solution, rearranging the equilibrium equation gives us \(k = 15Mg/a\), which allows us to proceed with further calculations in the exercise.
Kinetic and Potential Energy
Kinetic energy refers to the energy an object possesses due to its motion, while potential energy represents the stored energy of an object given its position or arrangement. These roles of energy are pivotal in analyzing various physics problems.

Considering the block-spring system, when the block is lifted, it gains gravitational potential energy corresponding to its height above the floor. As the block is released and falls towards the spring, this potential energy transitions to kinetic energy. Finally, when the block compresses the spring and momentarily stops, its kinetic energy is now stored as the spring's potential energy. The textbook exercise employs these concepts to trace the energy transformations that occur from the starting lift of the block to the point where it briefly comes to rest after engaging the spring.

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Most popular questions from this chapter

A particle \(P\) of mass \(m\) moves on the \(x\)-axis under the combined gravitational attraction of two particles, each of mass \(M\), fixed at the points \((0, \pm a, 0)\) respectively (see Figure 3.3). Example \(3.4\) shows that the force field \(F(x)\) acting on \(P\) is given by $$ F=-\frac{2 m M G x}{\left(a^{2}+x^{2}\right)^{3 / 2}} $$ Find the corresponding potential energy \(V(x)\) Initially \(P\) is released from rest at the point \(x=3 a / 4\). Find the maximum speed achieved by \(P\) in the subsequent motion.

A particle \(P\) carries a charge \(e\) and moves under the influence of the static magnetic field \(\boldsymbol{B}(\boldsymbol{r})\) which exerts the force \(\boldsymbol{F}=e \boldsymbol{v} \times \boldsymbol{B}\) on \(P\), where \(\boldsymbol{v}\) is the velocity of \(P\). Show that \(P\) travels with constant speed.

A particle \(P\) of mass \(m\) moves on the axis \(O z\) under the gravitational attraction of a uniform circular disk of mass \(M\) and radius \(a\) as shown in Figure 3.6. Example \(3.6\) shows that the force field \(F(z)\) acting on \(P\) is given by $$ F=-\frac{2 m M G}{a^{2}}\left[1-\frac{z}{\left(a^{2}+z^{2}\right)^{1 / 2}}\right] \quad(z>0) $$ Find the corresponding potential energy \(V(z)\) for \(z>0\) Initially \(P\) is released from rest at the point \(z=4 a / 3\). Find the speed of \(P\) when it hits the disk.

A smooth wire has the form of the parabola \(z=x^{2} / 2 b, y=0\), where \(b\) is a positive constant. The wire is fixed with the axis \(O z\) pointing vertically upwards. A particle \(P\), which can slide freely on the wire, is performing oscillations with \(x\) in the range \(-a \leq x \leq a\). Show that the period \(\tau\) of these oscillations is given by $$ \tau=\frac{4}{(g b)^{1 / 2}} \int_{0}^{a}\left(\frac{b^{2}+x^{2}}{a^{2}-x^{2}}\right)^{1 / 2} d x $$ By making the substitution \(x=a \sin \psi\) in the above integral, obtain a new formula for \(\tau\). Use this formula to find a two-term approximation to \(\tau\), valid when the ratio \(a / b\) is small.

A particle \(P\) of mass \(4 \mathrm{~kg}\) moves under the action of the force \(\boldsymbol{F}=4 \boldsymbol{i}+12 t^{2} \boldsymbol{j} \mathrm{N}\), where \(t\) is the time in seconds. The initial velocity of the particle is \(2 \boldsymbol{i}+\boldsymbol{j}+2 \boldsymbol{k} \mathrm{m} \mathrm{s}^{-1}\). Find the work done by \(\boldsymbol{F}\), and the increase in kinetic energy of \(P\), during the time interval \(0 \leq t \leq 1\) What principle does this illustrate?
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