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A horizontal spring attached to a wall has a force constant of \(k=850 \mathrm{~N} / \mathrm{m}\). A block of mass \(m=1.00 \mathrm{~kg}\) is attached to the spring and rests on a frictionless, horizontal surface as in Figure P13.21. (a) The block is pulled to a position \(x_{i}=6.00 \mathrm{~cm}\) from equilibrium and released. Find the potential energy stored in the spring when the block is \(6.00 \mathrm{~cm}\) from equilibrium. (b) Find the speed of the block as it passes through the equilibrium position. (c) What is the speed of the block when it is at a position \(x_{i} / 2=3.00 \mathrm{~cm}\) ?

Short Answer

Expert verified
A) The potential energy stored in the spring when the block is \(6.00 \, cm\) from equilibrium is \(1.53 \, J\). B) The speed of the block as it passes through equilibrium is \(1.75 \, m/s\). C) The speed of the block when it is at a position \(3.00 \, cm\) from equilibrium is \(1.52 \, m/s\).

Step by step solution

01

Find the potential energy when the block is 6.00 cm from equilibrium

The potential energy (U) stored in a spring is given by the equation \(U = 0.5 k x^2\), where \(k\) is the spring constant and \(x\) is the distance from the equilibrium. Plug values to find: \(U = 0.5 \times 850 \times (0.06)^2 = 1.53 \, Joules \)
02

Find the speed of the block as it passes through the equilibrium position

At the equilibrium position, all the potential energy gets converted into kinetic energy due to conservation of mechanical energy. The kinetic energy (K) can be calculated using the equation \(K = 0.5 m v^2\), where \(m\) is the mass and \(v\) is the speed. We know the kinetic energy equals the potential energy from step 1. Isolating \(v\), we get: \(v = \sqrt{(2K)/m} = \sqrt{(2 \times 1.53)/1} = 1.75 \, m/s\)
03

Find the speed of the block when it is at a position 3.00 cm from equilibrium

At the position 3.00 cm from equilibrium, the total energy will still be the same (1.53 J) due to energy conservation. But, this energy is now split between potential and kinetic energy. First, we calculate the potential energy at this position: \(U' = 0.5 \times 850 \times (0.03)^2 = 0.383 J\). The kinetic energy (K') at this position is the total energy minus the potential energy: \(K' = 1.53 J - 0.383 J = 1.147 J\). The speed \(v'\) of the block can then be found by rearranging the kinetic energy formula: \(v' = \sqrt{(2K')/m} = \sqrt{(2 \times 1.147)/1} = 1.52 \, m/s\)

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Key Concepts

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

Potential Energy
Potential energy in spring mechanics is a fascinating concept related to the energy stored in the system when work is done to compress or stretch a spring. The potential energy (often symbolized as \( U \)) can be understood through the formula \( U = 0.5 k x^2 \). Here, \( k \) is the spring constant, indicating the stiffness of the spring, while \( x \) denotes the displacement from the equilibrium position. In our example, when the block is pulled 6.00 cm from equilibrium, the potential energy becomes significant and is calculated as 1.53 Joules.
This energy represents the work done on the spring, storing this energy until it is released, allowing the spring to return to its original shape. Potential energy can manifest as either compression or extension, depending on the nature of the force applied to the spring.
  • The more the spring is displaced, the higher the potential energy, as it depends on the square of the displacement.
  • The stiffer the spring (higher \( k \)), the more potential energy it can store with the same displacement.
Kinetic Energy
Kinetic energy is all about movement. It's the energy an object possesses due to its motion. The formula used to calculate kinetic energy \( K \) is \( K = 0.5 m v^2 \), where \( m \) is the mass of the object and \( v \) is its velocity. In our spring scenario, the block gains kinetic energy when it moves. As it reaches the equilibrium position, its speed is at its maximum because all potential energy from the spring has been converted into kinetic energy—as illustrated in our example with a speed of 1.75 m/s at the equilibrium.
This transition from a state of potential energy to kinetic energy highlights the continuous exchange in the system's energy.
  • Kinetic energy depends on the square of the speed, meaning if speed doubles, kinetic energy increases fourfold.
  • As potential energy decreases, kinetic energy increases, keeping the total mechanical energy constant.
Mechanical Energy Conservation
The principle of mechanical energy conservation is a cornerstone in understanding spring mechanics. This principle states that in a closed system where only conservative forces like springs are acting, the total mechanical energy remains constant. Mechanical energy is the sum of both potential energy and kinetic energy in the system.
As the block oscillates on the spring, mechanical energy conservation ensures that while the energy shifts between being stored in the spring and being possessed by the block's movement, the total energy remains unchanged, barring any external forces. For instance, between two points where potential energy and kinetic energy are exchanged, the total always equates to 1.53 Joules, as calculated.
  • This law allows prediction of velocities and positions of objects in the absence of non-conservative forces like friction.
  • Helps to analyze systems efficiently and predict outcomes using energy rather than forces.

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Most popular questions from this chapter

A \(50.0-\mathrm{g}\) object is attached to a horizontal spring with a force constant of \(10.0 \mathrm{~N} / \mathrm{m}\) and released from rest with an amplitude of \(25.0 \mathrm{~cm}\). What is the velocity of the object when it is halfway to the equilibrium position if the surface is frictionless?

A simple pendulum has a length of \(52.0 \mathrm{~cm}\) and makes \(82.0\) complete oscillations in \(2.00 \mathrm{~min}\). Find (a) the period of the pendulum and (b) the value of \(g\) at the location of the pendulum.

A horizontal block-spring system with the block on a frictionless surface has total mechanical energy \(E=\) \(47.0 \mathrm{~J}\) and a maximum displacement from equilibrium of \(0.240 \mathrm{~m}\). (a) What is the spring constant? (b) What is the kinetic energy of the system at the equilibrium point? (c) If the maximum speed of the block is \(3.45 \mathrm{~m} / \mathrm{s}\), what is its mass? (d) What is the speed of the block when its displacement is \(0.160 \mathrm{~m}\) ? (e) Find the kinetic energy of the block at \(x=0.160 \mathrm{~m}\). (f) Find the potential energy stored in the spring when \(x=0.160 \mathrm{~m}\). (g) Suppose the same system is released from rest at \(x=0.240 \mathrm{~m}\) on a rough surface so that it loses \(14.0 \mathrm{~J}\) by the time it reaches its first turning point (after passing equilibrium at \(x=0\) ). What is its position at that instant?

The position of an object connected to a spring varies with time according to the expression \(x=\) \((5.2 \mathrm{~cm}) \sin (8.0 \pi t)\). Find (a) the period of this motion, (b) the frequency of the motion, (c) the amplitude of the motion, and (d) the first time after \(t=0\) that the object reaches the position \(x=2.6 \mathrm{~cm}\).

A taut clothesline has length \(L\) and a mass \(M\). A transverse pulse is produced by plucking one end of the clothesline. If the pulse makes \(n\) round trips along the clothesline in \(t\) seconds, find expressions for (a) the speed of the pulse in terms of \(n, L\), and \(t\) and (b) the tension \(F\) in the clothesline in terms of the same variables and mass \(M\).

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