/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 31 A block with mass \(m=2.00 \math... [FREE SOLUTION] | 91Ó°ÊÓ

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A block with mass \(m=2.00 \mathrm{~kg}\) is placed against a spring on a frictionless incline with angle \(\theta=30.0^{\circ}\) (Fig. 8-44). (The block is not attached to the spring.) The spring, with spring constant \(k=19.6 \mathrm{~N} / \mathrm{cm},\) is compressed \(20.0 \mathrm{~cm}\) and then released. (a) What is the elastic potential energy of the compressed spring? (b) What is the change in the gravitational potential energy of the block- Earth system as the block moves from the release point to its highest point on the incline? (c) How far along the incline is the highest point from the release point?

Short Answer

Expert verified
(a) 39.2 J; (b) 39.2 J; (c) 4.00 m

Step by step solution

01

Convert Units

First, convert the spring constant from N/cm to N/m. Since there are 100 cm in a meter, we have \( k = 19.6 \, \text{N/cm} = 1960 \, \text{N/m} \). Also convert the compression distance from cm to meters: \( x = 20.0 \, \text{cm} = 0.20 \, \text{m} \).
02

Calculate Elastic Potential Energy

The elastic potential energy stored in the spring is given by the formula \( U_{\text{elastic}} = \frac{1}{2} k x^2 \). Substituting the known values, \( U_{\text{elastic}} = \frac{1}{2} \times 1960 \, \text{N/m} \times (0.20 \, \text{m})^2 = 39.2 \, \text{J} \).
03

Calculate Change in Gravitational Potential Energy

At the highest point on the incline, the gravitational potential energy is equal to the initial elastic potential energy because the system is frictionless and energy is conserved. Thus, the change in gravitational potential energy \( \Delta U_{\text{gravity}} \) is the same as \( U_{\text{elastic}} = 39.2 \, \text{J} \).
04

Calculate the Height Change

The change in gravitational potential energy can also be expressed as \( \Delta U_{\text{gravity}} = mgh \), where \( h \) is the vertical height. Rearranging gives \( h = \frac{\Delta U_{\text{gravity}}}{mg} = \frac{39.2 \, \text{J}}{2.00 \, \text{kg} \times 9.81 \, \text{m/s}^2} \approx 2.00 \, \text{m} \).
05

Calculate the Distance Along the Incline

Using the relation \( h = d \sin\theta \), where \( d \) is the distance along the incline, we solve for \( d \): \( d = \frac{h}{\sin(30.0^\circ)} = \frac{2.00 \, \text{m}}{0.5} = 4.00 \, \text{m} \).

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Key Concepts

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

Gravitational Potential Energy
Gravitational potential energy is an intriguing aspect of physics. It represents the energy an object has due to its position in a gravitational field, typically above the ground. In simpler terms, it is the energy stored in an object because of its height. The formula for calculating gravitational potential energy is \( U = mgh \), where:
  • \( m \) is the mass of the object
  • \( g \) is the acceleration due to gravity, approximately \( 9.81 \, \text{m/s}^2 \) on Earth
  • \( h \) is the height above a reference point
In scenarios where a spring launches an object up an incline, like in our example, the energy from the spring is converted into gravitational potential energy as the object ascends. As the object rises, it loses kinetic energy but gains gravitational potential energy, reflecting the energy change associated with height increase. Understanding these transfers can simplify tackling problems in mechanics by giving a clear view of how energy forms transform into one another.
Inclined Plane
An inclined plane is a flat surface tilted at an angle, other than horizontal. It helps in understanding how forces and energy affect objects on slopes. When a block is on a frictionless inclined plane, it moves under the action of gravity. The slope's angle \( \theta \) plays a significant role in determining the component forms of forces acting on the block.To find the distance the block travels along the incline, one must consider how this angle affects the forces and energy conversions. In our exercise, the block reaches a height \( h \) by traveling a distance \( d \) along the plane. The height is calculated using \( h = d \sin \theta \). This relationship links the vertical rise to the slanted path the block covers. When working with inclined planes, keep in mind the effects of the incline angle on motion and energy conversion.
Conservation of Energy
The law of conservation of energy is a cornerstone concept in physics, declaring that energy in an isolated system remains constant—it can neither be created nor destroyed. However, energy forms can convert into each other.In a frictionless system on an inclined plane, as in the exercise, we see elastic potential energy from the spring converting entirely into gravitational potential energy as the block rises to its highest point. The initial elastic potential energy is calculated using \( U_{\text{elastic}} = \frac{1}{2} k x^2 \), where \( k \) is the spring constant and \( x \) is the spring compression. Once the spring releases, this energy transforms as the block progresses upward.Examining such transformations within the laws of energy conservation broadens one's grasp of energy's behavior in physical systems, enabling resolution of complex problems involving motion, forces, and energy conversions.

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Most popular questions from this chapter

A \(5.0 \mathrm{~g}\) marble is fired vertically upward using a spring gun. The spring must be compressed \(8.0 \mathrm{~cm}\) if the marble is to just reach a target \(20 \mathrm{~m}\) above the marble's position on the compressed spring. (a) What is the change \(\Delta U_{g}\) in the gravitational potential energy of the marble-Earth system during the \(20 \mathrm{~m}\) ascent? (b) What is the change \(\Delta U_{s}\) in the elastic potential energy of the spring during its launch of the marble? (c) What is the spring constant of the spring?

A 1500 kg car begins sliding down a \(5.0^{\circ}\) inclined road with a speed of \(30 \mathrm{~km} / \mathrm{h} .\) The engine is turned off, and the only forces acting on the car are a net frictional force from the road and the gravitational force. After the car has traveled \(50 \mathrm{~m}\) along the road, its speed is \(40 \mathrm{~km} / \mathrm{h} .\) (a) How much is the mechanical energy of the car reduced because of the net frictional force? (b) What is the magnitude of that net frictional force?

At a certain factory, \(300 \mathrm{~kg}\) crates are dropped vertically from a packing machine onto a conveyor belt moving at \(1.20 \mathrm{~m} / \mathrm{s}\) (Fig. 8 -64). (A motor maintains the belt's constant speed.) The coefficient of kinetic friction between the belt and each crate is \(0.400 .\) After a short time, slipping between the belt and the crate ceases, and the crate then moves along with the belt. For the period of time during which the crate is being brought to rest relative to the belt, calculate, for a coordinate system at rest in the factory, (a) the kinetic energy supplied to the crate, (b) the magnitude of the kinetic frictional force acting on the crate, and (c) the energy supplied by the motor. (d) Explain why answers (a) and (c) differ.

A \(20 \mathrm{~kg}\) object is acted on by a conservative force given by \(F=-3.0 x-5.0 x^{2},\) with \(F\) in newtons and \(x\) in meters. Take the potential energy associated with the force to be zero when the object is at \(x=0 .\) (a) What is the potential energy of the system associated with the force when the object is at \(x=2.0 \mathrm{~m} ?\) (b) If the object has a velocity of \(4.0 \mathrm{~m} / \mathrm{s}\) in the negative direction of the \(x\) axis when it is at \(x=5.0 \mathrm{~m},\) what is its speed when it passes through the origin? (c) What are the answers to (a) and (b) if the potential energy of the system is taken to be \(-8.0 \mathrm{~J}\) when the object is at \(x=0 ?\)

You drop a \(2.00 \mathrm{~kg}\) book to a friend who stands on the ground at distance \(D=10.0 \mathrm{~m}\) below. If your friend's outstretched hands are at distance \(d=1.50 \mathrm{~m}\) above the ground (Fig. 8-30), (a) how much work \(W_{g}\) does the gravitational force do on the book as it drops to her hands? (b) What is the change \(\Delta U\) in the gravitational potential energy of the book-Earth system during the drop? If the gravitational potential energy \(U\) of that system is taken to be zero at ground level, what is \(U\) (c) when the book is released and (d) when it reaches her hands? Now take \(U\) to be \(100 \mathrm{~J}\) at ground level and again find (e) \(W_{g},(\mathrm{f}) \Delta U,(\mathrm{~g}) U\) at the release point, and (h) \(U\) at her hands.

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