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Ch. 08 - Conservation of Energy
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoGiancoli Douglas 5th editionCh. 08 - Conservation of EnergyProblema 38b
Capitolo 8, Problema 38b

A spring ( k = 75 N/m) has an equilibrium length of 1.00 m. The spring is compressed to a length of 0.50 m and a mass of 2.0 kg is placed at its free end on a frictionless slope which makes an angle of 41° with respect to the horizontal (Fig. 8–41). The spring is then released. If the mass is attached to the spring, how far up the slope will the mass move before coming to rest?
A compressed spring on a frictionless slope, with a mass at its end, angled at 41° and compressed to 0.50 m.

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Determine the potential energy stored in the compressed spring using the formula for elastic potential energy: \( U_s = \frac{1}{2} k x^2 \), where \( k \) is the spring constant and \( x \) is the compression distance. Here, \( x = 1.00 \ \text{m} - 0.50 \ \text{m} = 0.50 \ \text{m} \).
Recognize that the spring's potential energy will be converted into gravitational potential energy as the mass moves up the slope. The gravitational potential energy is given by \( U_g = m g h \), where \( m \) is the mass, \( g \) is the acceleration due to gravity, and \( h \) is the vertical height gained by the mass.
Relate the vertical height \( h \) to the distance \( d \) traveled along the slope using the angle \( \theta \): \( h = d \sin \theta \). Substitute this into the gravitational potential energy formula to express \( U_g \) in terms of \( d \): \( U_g = m g d \sin \theta \).
Set the elastic potential energy equal to the gravitational potential energy to find the distance \( d \): \( \frac{1}{2} k x^2 = m g d \sin \theta \). Rearrange this equation to solve for \( d \): \( d = \frac{\frac{1}{2} k x^2}{m g \sin \theta} \).
Substitute the known values into the equation: \( k = 75 \ \text{N/m} \), \( x = 0.50 \ \text{m} \), \( m = 2.0 \ \text{kg} \), \( g = 9.8 \ \text{m/s}^2 \), and \( \sin \theta = \sin 41^\circ \). This will give the distance \( d \) the mass moves up the slope before coming to rest.

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Concetti chiave

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Hooke's Law

Hooke's Law states that the force exerted by a spring is directly proportional to its displacement from the equilibrium position, expressed as F = -kx, where F is the force, k is the spring constant, and x is the displacement. In this scenario, the spring's compression will generate a force that acts on the mass when released, influencing its motion up the slope.
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Spring Force (Hooke's Law)

Conservation of Energy

The principle of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. In this problem, the potential energy stored in the compressed spring will convert into kinetic energy as the mass moves up the slope, and eventually into gravitational potential energy at the highest point of its ascent.
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Conservation Of Mechanical Energy

Inclined Plane Dynamics

When analyzing motion on an inclined plane, the forces acting on the mass include gravitational force, normal force, and the spring force. The component of gravitational force acting parallel to the slope affects how far the mass will move up the incline before coming to rest, requiring the use of trigonometric functions to resolve forces based on the angle of the slope.
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Intro to Inclined Planes
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