/*! 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 10 A student presses a 0.5 \(\mathr... [FREE SOLUTION] | 91影视

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Chapter 14: Problem 10

A student presses a 0.5 \(\mathrm{kg}\) book against the wall. If the \(\mu_{\mathrm{s}}\) between the book and the wall is \(0.2,\) what force must the student apply to hold the book in place? (A) 0 \(\mathrm{N}\) (B) 15 \(\mathrm{N}\) (C) 25 \(\mathrm{N}\) (D) 35 \(\mathrm{N}\)

Short Answer

Expert verified
The student must apply a force of nearly 1N (option not given) to keep the book in place. The provided options do not include the calculated force value. There might be an error in the question or the given options.

Step by step solution

01

Calculate weight force

Firstly, we calculate the weight force of the book using the formula of \(F_{g}=m*g\), where \(m\) is the mass of the book and \(g\) is the acceleration due to gravity. Given that \(m=0.5 kg\) and \(g=9.8 m/s^2\), we have \(F_{g}=0.5kg*9.8m/s^2=4.9N\). Therefore,the weight force of the book is \(4.9N\).
02

Calculate the frictional force

Next, we calculate the frictional force using the formula \(F_{f}=\mu_{s}*F_{n}\), where \(\mu_{s}\) is the static friction coefficient and \(F_{n}\) is the normal force. In this case, the book is not moving vertically, so the normal force is equal to the weight force. Therefore, we have \(F_{f}=0.2*4.9N=0.98N\). So, the frictional force is \(0.98N\).
03

Determine the horizontal force applied by the student

Finally, put into mind that the student's force is the frictional force because it's applied in the opposite direction to prevent the book from falling. Therefore, the force applied by the student equals to the static frictional force and that is \(0.98N\).

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Key Concepts

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

Understanding Physics Problem Solving
Physics problem solving often requires a systematic approach. Start by identifying the forces at play and what they mean for the scenario.
  • Break the problem down into manageable steps.
  • Use known formulas and constants, like gravity which is typically 9.8 m/s虏 on Earth.
  • Check which forces balance each other, especially in static situations.
In this problem, we are dealing with a book being pressed against a wall. The main forces we need to consider are the weight force (acting downwards due to gravity) and the frictional force (acting horizontally, preventing the book from falling). By identifying these forces, we simplify the problem to effectively determine the necessary force to keep the book in place.
The Role of Normal Force
Normal force is the support force exerted by a surface perpendicular to the object in contact with it. When you push an object against a surface, the normal force usually balances out the weight force acting on the object.
  • In our scenario, the normal force prevents the book from falling through the wall.
  • It acts perpendicular to the surface (the wall, in this case).
  • It is equal in magnitude but opposite to the weight force when the object isn't moving vertically.
Therefore, understanding the normal force's role helps us figure out how much force is needed to maintain the book鈥檚 position. In the exercise, this involves calculating the weight of the book, which equals the normal force.
What is Weight Force?
Weight force is simply the force of gravity acting on an object. It is calculated using the formula:\[ F_g = m \times g \]
  • Where \( m \) is the mass of the object. For the book, \( m = 0.5 \) kg.
  • \( g \) is the acceleration due to gravity, usually \( 9.8 \ m/s^2 \).
  • Thus, the weight force \( F_g = 0.5 \times 9.8 = 4.9 \ N \).
Understanding weight force is crucial in this problem because it determines how much force the book exerts on the wall vertically. This helps us calculate the frictional and applied forces needed to keep it stationary.
Static Friction Coefficient Explained
The static friction coefficient \( \mu_s \) represents the ratio of the maximum static friction force to the normal force. It indicates how much two surfaces resist movement relative to each other without any motion.
  • A higher \( \mu_s \) means more resistance to start sliding, requiring stronger force to overcome.
  • In this exercise, \( \mu_s = 0.2 \) reflects the interaction between the book and the wall.
  • Static friction force is found with \[ F_f = \mu_s \times F_n \]. Given \( F_n = 4.9 \ N \), then the static friction \( F_f = 0.2 \times 4.9 = 0.98 \ N \).
Understanding this concept allows students to comprehend how forces balance and result in the necessity of any applied force to keep the book stationary against the wall.

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Most popular questions from this chapter

A 4 \(\mathrm{kg}\) box is accelerated upwards by a string with a breaking strength of 80 \(\mathrm{N}\) . What is the maximum upward acceleration that can be applied on the box without breaking the string? (A) 2.5 \(\mathrm{m} / \mathrm{s}^{2}\) (B) 5.0 \(\mathrm{m} / \mathrm{s}^{2}\) (C) 7.5 \(\mathrm{m} / \mathrm{s}^{2}\) (D) 10 \(\mathrm{m} / \mathrm{s}^{2}\)

A worker moves a 30 \(\mathrm{kg}\) box by pulling on it with a rope that makes a \(60^{\circ}\) angle with the horizontal. If the worker applies a force of 40 \(\mathrm{N}\) and pulls the box over a distance of \(20 \mathrm{m},\) how much work did the worker do? (A) 100 \(\mathrm{J}\) (B) 200 \(\mathrm{J}\) (C) 400 \(\mathrm{J}\) (D) 800 \(\mathrm{J}\)

A student pushes a 6 kg box up an inclined plane with a height of 10 m. How much work does gravity do on the box during this process? \((\mathrm{A})-1,200 \mathrm{J}\) (B) \(-600 \mathrm{J}\) (C) 600 \(\mathrm{J}\) (D) \(1,200 \mathrm{J}\)

Which of the following is a true statement about standing waves? (A) Standing waves have zero amplitude at antinodes. (B) Standing waves have maximum amplitude at nodes. (C) Complete constructive interference occurs at nodes. (D) Complete destructive interference occurs at nodes.

If the speed of a satellite orbiting the Earth at a distance r from the center of the Earth is v, what is the speed of a second satellite orbiting the Earth at a distance 2r from the center of the Earth? (A) 2 \(\mathrm{V}\) (B) \(\sqrt{2} \mathrm{v}\) (C) \(\frac{1}{\sqrt{2}} v\) (D) \(\frac{1}{2} v\)
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