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Chapter 8: Problem 55

A block of mass \(m=2.5 \mathrm{~kg}\) slides head on into a spring of spring constant \(k=320 \mathrm{~N} / \mathrm{m}\). When the block stops, it has compressed the spring by \(7.5 \mathrm{~cm}\). The coefficient of kinetic friction between block and floor is \(0.25\). While the block is in contact with the spring and being brought to rest, what are (a) the work done by the spring force and (b) the increase in thermal energy of the block-floor system? (c) What is the block's speed just as it reaches the spring?

Short Answer

Expert verified
(a) 0.9 Joules, (b) 0.459375 Joules, (c) 1.045 m/s.

Step by step solution

01

Convert units and gather initial data

First, convert the compression of the spring from centimeters to meters since SI units are preferred. So, \(7.5 \text{ cm} = 0.075 \text{ m}\). Gather the other known quantities: mass \(m = 2.5 \mathrm{~kg}\), spring constant \(k = 320 \mathrm{~N/m}\), and the coefficient of kinetic friction \(\mu_k = 0.25\).
02

Calculate work done by the spring force

The work done by the spring force when compressed by distance \(x\) is given by \(W = \frac{1}{2} k x^2\). Substitute the known values: \[ W = \frac{1}{2} \times 320 \times (0.075)^2 = 0.9 \text{ Joules}.\]
03

Find the increase in thermal energy due to friction

The increase in thermal energy \( \Delta E_{th}\) is given by the work done against friction, which is the friction force times the distance. The friction force \(f_k = \mu_k m g\), where \(g = 9.8 \text{ m/s}^2\). Thus, \(f_k = 0.25 \times 2.5 \times 9.8 = 6.125 \text{ N}\). The work done by this force is \(W_{f_k} = f_k x = 6.125 \times 0.075 = 0.459375 \text{ Joules}.\)
04

Apply energy principles to find initial speed

The initial mechanical energy is all kinetic \(E_k = \frac{1}{2} m v^2\) and when the block stops, all energy is transferred to the spring and lost due to friction. According to energy conservation: \[ \frac{1}{2} m v^2 = \frac{1}{2} k x^2 + \Delta E_{th}.\] Plugging known values: \(\frac{1}{2} \times 2.5 \times v^2 = 0.9 + 0.459375\), compute: \(1.25v^2 = 1.359375\). Solve for \(v\), yielding: \(v = \sqrt{\frac{1.359375}{1.25}} = 1.045 \text{ m/s}.\)

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Key Concepts

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

Spring Compression
Spring compression is a fundamental concept in mechanics, especially in the context of energy storage and transfer. When a spring is compressed or stretched from its equilibrium position, it stores potential energy.
  • This potential energy is given by the formula: \( E_{spring} = \frac{1}{2} k x^2 \), where \( k \) is the spring constant and \( x \) is the displacement from the natural length of the spring.
  • In our exercise, the spring constant \( k \) is 320 N/m and the compression \( x \) is 0.075 m, translating to a work done of 0.9 Joules when the block compresses the spring.
  • This energy value signifies the potential energy stored in the spring due to compression and can be "released" when the spring is allowed to return to its natural length.
By understanding spring compression, one can appreciate how energy is transformed and conserved in mechanical systems with springs, laying the groundwork for more complex topics.
Kinetic Friction
Kinetic friction arises when two surfaces slide against each other, converting kinetic energy into thermal energy. This force opposes the motion of the sliding object and is proportional to the normal force exerted between the surfaces.
  • The force of kinetic friction \( f_k \) is calculated using the equation: \( f_k = \mu_k m g \), where \( \mu_k \) is the coefficient of kinetic friction, \( m \) is the mass, and \( g \) is the acceleration due to gravity.
  • In the given problem, \( \mu_k = 0.25 \), and with the block's mass at 2.5 kg, the kinetic friction force is approximately 6.125 N.
  • The work done by this friction, which leads to an increase in thermal energy, is the product of the friction force and the distance moved, calculated here as 0.459375 Joules.
This concept is crucial in understanding how energy dissipation occurs in real-world systems and how mechanical energy is not always conserved, especially in systems where friction is present.
Energy Conservation
Energy conservation is one of the cornerstones of physics, dictating that energy cannot be created or destroyed, only transformed. This principle helps solve many mechanics problems by equating initial energy states to final energy states.
  • In our scenario, we apply energy conservation to find the block's speed as it hits the spring. Initially, the block鈥檚 energy is purely kinetic, given by \( E_k = \frac{1}{2} m v^2 \).
  • As the block compresses the spring and comes to rest, its kinetic energy is transformed into spring potential energy and thermal energy due to friction. The total initial energy is thus distributed as: \( \frac{1}{2} m v^2 = \frac{1}{2} k x^2 + \Delta E_{th} \).
  • Using this equation and inserting the known values, we solve for the initial speed, revealing \( v = 1.045 \text{ m/s} \).
Energy conservation allows us to predict the behavior of physical systems over time, emphasizing the interconnectedness of kinetic, potential, and thermal energies.

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Most popular questions from this chapter

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The potential energy of a diatomic molecule (a two-atom system like \(\mathrm{H}_{2}\) or \(\mathrm{O}_{2}\) ) is given by $$U=\frac{A}{r^{12}}-\frac{B}{r^{6}}$$ where \(r\) is the separation of the two atoms of the molecule and \(A\) and \(B\) are positive constants. This potential energy is associated with the force that binds the two atoms together. (a) Find the equilibrium separation - that is, the distance between the atoms at which the force on each atom is zero. Is the force repulsive (the atoms are pushed apart) or attractive (they are pulled together) if their separation is (b) smaller and (c) larger than the equilibrium separation?
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