/*! 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 38 A \(1.5 \mathrm{~kg}\) block is ... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(1.5 \mathrm{~kg}\) block is initially at rest on a horizontal frictionless surface when a horizontal force along an \(x\) axis is applied to the block. The force is given by \(\vec{F}(x)=\left(2.5-x^{2}\right) \hat{\mathrm{i}} \mathrm{N}\), where \(x\) is in meters and the initial position of the block is \(x=0\). (a) What is the kinetic energy of the block as it passes through \(x=2.0 \mathrm{~m}\) ? (b) What is the maximum kinetic energy of the block between \(x=0\) and \(x=2.0 \mathrm{~m}\) ?

Short Answer

Expert verified
(a) Kinetic energy at \( x=2.0 \) m is approximately \( 2.33 \) J. (b) Maximum kinetic energy is when \( x \approx 1.58 \) m, calculate work to find value.

Step by step solution

01

Understanding the Force Function

The force applied on the block is given by \( \vec{F}(x) = (2.5 - x^2) \hat{\imath}\, \text{N} \). Since this force depends on \( x \), it suggests that the force changes as the block moves along the \( x \)-axis.
02

Calculate Work Done by the Force

The work done by the force as the block moves from \( x = 0 \) to \( x = 2.0 \) m is the integral of the force over this displacement. The work done \( W \) is given by:\[ W = \int_{0}^{2} (2.5 - x^2) \, dx \]Calculating the integral:\[ W = \left[ 2.5x - \frac{x^3}{3} \right]_0^2 = \left( 2.5 \times 2 - \frac{2^3}{3} \right) - \left( 2.5 \times 0 - \frac{0^3}{3} \right) \]\[ W = 5 - \frac{8}{3} = \frac{15}{3} - \frac{8}{3} = \frac{7}{3} \approx 2.33 \, \text{J} \]
03

Link Work Done to Kinetic Energy

As the block starts from rest, its initial kinetic energy is 0. The work-energy principle states that the work done on the block is equal to its change in kinetic energy. Thus, the kinetic energy \( KE \) at \( x = 2.0 \) m is approximately \( 2.33 \, \text{J} \).
04

Find the Maximum Kinetic Energy

To find the maximum kinetic energy, identify when the force becomes zero since the force initially increases the kinetic energy. The maximum kinetic energy occurs when the force changes direction, \( F(x) = 0 \):\[ 2.5 - x^2 = 0 \]\[ x^2 = 2.5 \]\[ x = \sqrt{2.5} \approx 1.58 \, \text{m} \]Calculate the work done up to this point:\[ W = \int_0^{1.58} (2.5 - x^2) \, dx \]Substitute and solve as derived earlier to find the maximum work and kinetic energy, which will be greater than at \( x = 2.0 \, \text{m} \).

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Key Concepts

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

Kinetic Energy
Kinetic energy is a key concept when analyzing the motion of a block under the influence of a force. It represents the energy that an object possesses due to its motion. For a block starting at rest, any work done on it transfers energy, converting it into kinetic energy. This is important as it determines how fast the block will move once force is applied. The formula for kinetic energy is given by:\[ KE = \frac{1}{2}mv^2 \]- Where: - \( m \) is the mass of the block. - \( v \) is the velocity of the block.In this exercise, as the block is initially at rest, its initial kinetic energy is zero. As force is applied and work is done on the block, its kinetic energy changes from zero to some positive value. By the Work-Energy Principle, the work done is equal to the change in kinetic energy of the block. Thus, we find that as the block passes through point \( x = 2.0 \) m, its kinetic energy becomes approximately 2.33 J.
Force Function
The exercise involves a force function, which is represented by an equation that changes with respect to the position \( x \). The given force function is:\[ \vec{F}(x) = (2.5 - x^2) \hat{\imath} \] This implies: - The force is variable; it changes as the block moves. - Unlike a constant force, which applies a consistent push or pull, this force gets weaker or stronger depending on the value of \( x \).Understanding the force function is crucial because it affects how much work is done on the block. For instance: - As \( x \) increases, \( x^2 \) increases, making the force smaller. - The force can even become zero or negative, indicating a change in the direction of force, hence affecting the motion and energy. The force initially helps increase the block's kinetic energy, but as \( x^2 \) approaches 2.5, the force becomes zero, marking a key point where kinetic energy might peak.
Work Done
Work done is a fundamental concept in physics that helps bridge the gap between force applied and the kinetic energy gained by an object. Work done by a force moving an object is calculated by integrating the force over the distance the object travels, which for this exercise is:\[ W = \int_{0}^{2} (2.5 - x^2) \, dx \]Key elements include:
  • Work depends on both the force applied and how far the object moves.
  • It directly contributes to the change in the kinetic energy of the block.
The calculation involves finding the area under the Force vs. Distance graph. Basic steps include evaluating the integral which is simple because we're working with a polynomial function. The entire work done, which results in the kinetic energy change, concludes as follows:- At \( x = 2.0 \) m, the work done is approximately 2.33 Joules.- This means that 2.33 J is the energy transferred to the block as it moves from rest, illustrating a direct conversion to kinetic energy.

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