/*! 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 6 A \(1.5-\mathrm{kg}\) block is i... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 6

A \(1.5-\mathrm{kg}\) block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of \(x\)-axis is applied to the block. The force is given by \(\vec{F}=\left(4-x^{2}\right) \vec{i} \mathrm{~N}\), where \(x\) is in meter and the initial position of the block is \(x=0\). The maximum positive displacement \(x\) is (1) \(2 \sqrt{3} \mathrm{~m}\) (2) \(2 \mathrm{~m}\) (3) \(4 \mathrm{~m}\) (4) \(\sqrt{2} \mathrm{~m}\)

Short Answer

Expert verified
The maximum positive displacement is 2 meters (option 2).

Step by step solution

01

Understand the Force Equation

The force acting on the block is given by \( \vec{F} = (4 - x^2) \vec{i} \) N. Here, \( x \) is the displacement in meters. The force is a function of \( x \), and as \( x \) increases, \( F \) decreases.
02

Set Force Equal to Zero to Find Maximum Displacement

To find the maximum displacement, we need to find when the force becomes zero since after this point, the block will stop advancing in the positive \( x \)-direction. Set \( F = 0 \): \[ 4 - x^2 = 0 \] Solve for \( x \).
03

Solve the Quadratic Equation

Rearrange the equation to find \( x \): \[ x^2 = 4 \] Take the square root of both sides to solve for \( x \): \[ x = \sqrt{4} = 2 \] Since \( x \) is positive, the solution is \( x = 2 \) meters.
04

Verify the Solution

Substitute \( x = 2 \) meters into the original force equation: \[ F = 4 - (2)^2 = 4 - 4 = 0 \] The force becomes zero, confirming that 2 meters is indeed the maximum positive displacement.

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Key Concepts

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

Understanding the Force Equation
In this mechanics problem, the force applied to a block on a frictionless surface is expressed as a function of displacement. The force equation given is \( \vec{F} = (4 - x^2) \vec{i} \ \mathrm{N} \), where \( x \) represents the displacement in meters from the initial position. This equation tells us how the force changes as the block moves.

- **Understanding Variables**: - \( \vec{F} \) is the force exerted on the block. - \( x^2 \) is the square of the displacement, affecting the force inversely. - **Force Decreases with Displacement**: As \( x \) increases, the value of \( x^2 \) increases, reducing \( F \). This means the block experiences less force as it moves further.

Knowing how to interpret this force equation allows us to predict the block's behavior as it moves on the surface without friction. The force stops when the force equation equals zero, which helps us identify how far the block will go before it stops.
Finding Maximum Displacement
To determine the block's maximum displacement, we find when the force acting on it becomes zero. This is crucial because a force of zero means the block's acceleration ceases, causing it to stop moving forward.

To find this point, we set the force equation to zero: \[ 4 - x^2 = 0 \]- **Rearranging the Equation**: By rearranging, we find \( x^2 = 4 \).- **Solving for \( x \)**: Taking the square root gives \( x = \sqrt{4} = 2 \). Since displacement is positive, we identify the maximum displacement as \( x = 2 \) meters.

Once \( x \) reaches 2 meters, the force pushing the block forward drops to zero, meaning the block will not continue to displace further, thus confirming it's the maximum displacement. Always check solutions by substituting back into the force equation to verify calculations.
The Importance of a Frictionless Surface
This problem is simplified by the fact that the block moves on a frictionless surface. Understanding the impact of this condition is key in mechanics problems.

- **No Friction Means No Opposing Force**: - The absence of friction implies there's no opposing force to the block's movement. - All the applied force contributes solely to the block's displacement.

- **Purely Driven by Applied Force**: The only factor limiting displacement in this case is the force \( \vec{F} = (4 - x^2) \vec{i} \). As the force diminishes to zero, no other forces affect the block's motion.

- **Simplification for Learning**: In educational contexts, frictionless surface problems allow students to focus purely on the force's effects on displacement without worrying about friction, which simplifies calculations and understanding of core physics principles.

This assumption helps illustrate key mechanics concepts like Newton鈥檚 first law of motion, making it easier to learn and apply these principles effectively.

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Most popular questions from this chapter

A particle is taken from point \(A\) to point \(B\) under the influence of a force field. Now it is taken back from \(B\) to \(A\) and it is observed that the work done in taking the particle from \(A\) to \(B\) is not equal to the work done in taking it from \(B\) to \(A\). If \(W_{n c}\) and \(W_{c}\) are the work done by non- conservative and conservative forces present in the system, respectively \(\Delta U\) is the change in potential energy and \(\Delta k\) is the chan in kinetic energy, then (1) \(W_{n c}-\Delta U=\Delta k\) (2) \(W_{e}=-\Delta U\) (3) \(W_{n c}^{n}+W_{c}=\Delta k\) (4) \(W_{n c}-\Delta U=-\Delta k\)

Blocks \(A\) and \(B\) of mass \(m\) each are connected with spring of constant \(k\). Both blocks lie on frictionless ground and are imparted horizontal velocity \(v\) as shown when spring is unstretched. Find the maximum stretch of spring. (1) \(v \sqrt{\frac{m}{k}}\) (2) \(v \sqrt{\frac{m}{2 k}}\) (3) \(v \sqrt{\frac{2 m}{k}}\) (4) None of the above

A block of \(4 \mathrm{~kg}\) mass starts at rest and slides a distance \(d\) down a friction less incline (angle \(30^{\circ}\) ) where it runs into a spring of negligible mass. The block slides an additional \(25 \mathrm{~cm}\) before it is brought to rest momentarily by compressing the spring. The force constant of the spring is \(400 \mathrm{~N} \mathrm{~m}^{-1}\), The value of \(d\) is (take \(g=10 \mathrm{~m} \mathrm{~s}^{-2}\) ) (1) \(25 \mathrm{~cm}\) (2) \(37.5 \mathrm{~cm}\) (3) \(62.5 \mathrm{~cm}\) (4) None of the above

Choose the correct statement(s) from the following. (1) Force acting on a particle for equal time intervals can produce the same change in momentum but different change in kinetio energy. (2) Force acting on a particle for equal displacements can produce same change in kinetic energy but different change in momentum. (3) Force acting on a particle for equal time intervals can produce different change in momentum but same change in kinetic energy. (4) Force acting on a particle for equal displacements can produce different change in kinetic energy but same change in momentum.

A man slowly pulls a bucket of water from a well of depth \(h=20 \mathrm{~m}\). The mass of the uniform rope and bucket full of water are \(m=200 \mathrm{~g}\) and \(M 19.9 \mathrm{~kg}\), respectively. Find the work done (in kJ) by the man.
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