/*! 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 81 A wooden block with mass 1.50 \(... [FREE SOLUTION] | 91Ó°ÊÓ

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A wooden block with mass 1.50 \(\mathrm{kg}\) is placed against a compressed spring at the bottom of an incline of slope \(30.0^{\circ}\) (point \(A )\) . When the spring is released, it projects the block up the incline. At point \(B\) , a distance of 6.00 \(\mathrm{m}\) up the incline from \(A\) . the block is moving up the incline at 7.00 \(\mathrm{m} / \mathrm{s}\) and is no longer in contact with the spring. The coefficient of kinetic friction between the block and the incline is \(\mu_{\mathrm{k}}=0.50 .\) The mass of the spring is negligible. Calculate the amount of potential energy that was initially stored in the spring.

Short Answer

Expert verified
The potential energy initially stored in the spring is approximately 119.07 J.

Step by step solution

01

Identify Known Values

We are given a block with mass \( m = 1.50 \, \text{kg} \), slope of the incline \( \theta = 30.0^{\circ} \), distance moved on the slope \( d = 6.00 \, \text{m} \), initial velocity at point B \( v_B = 7.00 \, \text{m/s} \), and the coefficient of kinetic friction \( \mu_k = 0.50 \). The goal is to calculate the initial potential energy stored in the spring, \( U_i \).
02

Calculate Kinetic Energy at Point B

The kinetic energy at point B is calculated using the formula \( KE_B = \frac{1}{2}mv_B^2 \). Substituting the values, we have \( KE_B = \frac{1}{2} \times 1.50 \, \text{kg} \times (7.00 \, \text{m/s})^2 = 36.75 \, \text{J} \).
03

Calculate Work Done Against Friction

The work done against friction is calculated by \( W_f = \mu_k \cdot mg \cdot d \cdot \cos(\theta) \). Substituting the given values, \( W_f = 0.50 \times 1.50 \, \text{kg} \times 9.81 \, \text{m/s^2} \times 6.00 \, \text{m} \times \cos(30.0^{\circ}) \approx 38.17 \, \text{J} \).
04

Calculate Gravitational Potential Energy at Point B

The gravitational potential energy at point B is given by \( U_{gB} = mgh \), where \( h = d \cdot \sin(\theta) \). Substituting values, \( h = 6.00 \, \text{m} \cdot \sin(30.0^{\circ}) = 3.00 \, \text{m} \). Thus, \( U_{gB} = 1.50 \, \text{kg} \times 9.81 \, \text{m/s^2} \times 3.00 \, \text{m} = 44.145 \, \text{J} \).
05

Apply Energy Conservation Principles

The initial potential energy stored in the spring, \( U_i \), is equal to the sum of the kinetic energy, work done against friction, and gravitational potential energy at point B. Using the equation, \( U_i = KE_B + W_f + U_{gB} \), we find \( U_i = 36.75 \, \text{J} + 38.17 \, \text{J} + 44.145 \, \text{J} = 119.065 \, \text{J} \).
06

Final Calculation and Result

Having summed all the energy components, the initial potential energy stored in the spring is approximately \( 119.07 \, \text{J} \).

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Key Concepts

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

Kinetic Friction
Kinetic friction occurs when two surfaces slide past each other. It acts as a force that opposes motion. The kinetic friction force can be calculated using the formula:
  • \( F_k = \mu_k \cdot N \)
in which \( \mu_k \) is the coefficient of kinetic friction and \( N \) is the normal force. Typically, the normal force is the perpendicular force exerted by a surface to support the weight of an object resting on it. In this exercise, the block faces kinetic friction from sliding up the slope, resisting its motion.

On an incline, the normal force \( N \) is given by:
  • \( N = mg \cdot \cos(\theta) \)
Here, \( m \) is the mass of the block, \( g \) is the acceleration due to gravity, and \( \theta \) is the angle of the incline. To find the work done against friction, it is then given by:
  • \( W_f = \mu_k \cdot mg \cdot d \cdot \cos(\theta) \)
Where \( d \) is the distance moved along the incline. For this problem, the kinetic friction force worked along the 6.00-meter slope to resist the movement of the block upward.
Gravitational Potential Energy
Gravitational potential energy is the energy stored in an object due to its position above the ground. It can be calculated using:
  • \( U_g = mgh \)
where:
  • \( m \) is the mass of the object,
  • \( g \) is the acceleration due to gravity (approximately \( 9.81 \ \text{m/s}^2 \)), and
  • \( h \) is the height above the reference point (usually the base level or ground).


In this exercise, the block moves up an incline and achieves additional height when it reaches point B. The calculation involves finding the vertical displacement using the sine of the incline angle. So, \( h = d \, \sin(\theta) \), where \( d \) is the inclined distance, gives the height \( h \) through trigonometric relationships. For the wooden block moving up the slope, gravitational potential energy is consequential because it contributes to the total energy needed from the initially compressed spring.
Energy Conservation
The principle of energy conservation states that energy cannot be created or destroyed. Instead, energy can transform from one type to another. Throughout the scenario presented, several forms of energy transitions occur, starting with spring potential energy being converted to kinetic energy, work done against friction, and gravitational potential energy as the block moves along the incline.

In using the energy conservation principle, all these energy forms must account for the initial energy stored in the spring when released. The initial spring potential energy \( U_i \) can be found using the equation:
  • \( U_i = KE_B + W_f + U_{gB} \)
Where:
  • \( KE_B \) is the kinetic energy of the block at point B,
  • \( W_f \) is the work done against friction, and
  • \( U_{gB} \) is the gravitational potential energy at point B.

The initial energy stored in the spring plays a vital role in determining the energies experienced by the block along its path. It equates to the interplay of kinetic actions and resistances as the block finally exits contact with the spring, moving up the incline.

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Most popular questions from this chapter

cp A small block with mass 0.0400 kg slides in a vertical circle of radius \(R=0.500 \mathrm{m}\) on the inside of a circular track. During one of the revolutions of the block, when the block is at the bottom of its path, point \(A\) , the magnitude of the normal force exerted on the block by the track has magnitude 3.95 \(\mathrm{N}\) . In this same revolution, when the block reaches the top of its path, point \(B,\) the magnitude of the normal force exerted on the block has magnitude 0.680 \(\mathrm{N} .\) How much work was done on the block by friction during the motion of the block from point \(A\) to point \(B ?\)

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