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Chapter 13: Problem 13

A box with a mass of 5 \(\mathrm{kg}\) is sliding across a table at a speed of 2 \(\mathrm{m} / \mathrm{s}\) . The coefficient of kinetic friction between the book and table is \(\mu=\) \(0.25 .\) What is the minimum force that has to be applied on the box to maintain this speed? (A) 0 N (B) 10 N (C) 12.3 N (D) 24.6 N

Short Answer

Expert verified
The minimum force that has to be applied on the box to maintain its speed is \( 12.25 \, N \), so the closest answer choice is (C) 12.3 N

Step by step solution

01

Calculate the force of friction

First, calculate the force of kinetic friction (\( F_k \)) using the formula \( F_k = \mu \cdot F_n \), where \( \mu \) is the coefficient of kinetic friction and \( F_n \) is the normal force. The normal force for an object lying on a horizontal surface is equal to its weight, which is its mass times the acceleration due to gravity (\( g \)). So, \( F_n = m \cdot g \) where \( m = 5 \, kg \) and \( g = 9.8 \, m/s^2 \). Substituting these values and \( \mu = 0.25 \) into the equations gives: \( F_k = 0.25 \cdot 5 \cdot 9.8 = 12.25 \, N \)
02

Determine the minimum force to maintain speed

Since the box is not accelerating or decelerating, and only sliding across the table with a constant speed, the sum of the forces acting on it in the horizontal direction should be zero according to Newton's first law. Therefore, the minimum force that needs to be applied on the box to balance the friction and hence maintaining its speed is equal to the force of kinetic friction. Thus the minimum force required is \( F = F_k = 12.25 \, N \)

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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 against each other, and it plays a crucial role in the scenario with the sliding box on the table. This frictional force acts in the direction opposite to the motion, making it necessary to apply an external force to keep the object moving.
The coefficient of kinetic friction (\( \mu \)) is a dimensionless value that represents how much two surfaces resist sliding motion against each other. The formula used to calculate the kinetic frictional force is:
  • \( F_k = \mu \cdot F_n \)
where \( F_k \) is the kinetic frictional force, \( \mu \) is the coefficient of kinetic friction, and \( F_n \) is the normal force.
The normal force (\( F_n \)) is typically the weight of the object, which on a horizontal surface, equals the object's mass multiplied by gravitational acceleration (\( g \approx 9.8 \ m/s^2 \)). Thus, for our box:
  • \( F_n = m \cdot g = 5 \ kg \cdot 9.8 \ m/s^2 = 49 \ N \)
  • \( F_k = 0.25 \cdot 49 \ = 12.25 \ N \)
In this problem, the kinetic friction must be overcome to maintain the box's constant velocity.
Newton's First Law
Newton's First Law, often known as the law of inertia, states that a body at rest will stay at rest, and a body in motion will stay in motion at a constant velocity unless acted upon by an unbalanced force. This principle helps in understanding the forces at play when maintaining a box's speed on a table.
For the box sliding on the table, if no additional force is applied, the kinetic friction will cause the box to slow down and eventually stop.
  • To keep moving at a constant velocity, the net force acting on the box must be zero.
  • This means the applied force to maintain the speed should equal the force of kinetic friction.
In this situation, the exerted force works to counterbalance the frictional force, allowing the box to continue moving at 2 \( m/s \). This illustrates Newton's First Law in action by demonstrating that the box will only move at a steady pace if the applied and frictional forces are in equilibrium.
Forces in Equilibrium
The concept of forces in equilibrium relates to the conditions necessary for an object to maintain constant velocity or remain at rest. Equilibrium means all forces acting upon the object are balanced.
In the context of the box on the table, achieving equilibrium requires the applied force to exactly counterbalance the force of kinetic friction.
  • Since the friction force (\( F_k = 12.25 \ N \)) attempts to oppose the motion, the applied force should be equal and opposite, resulting in equilibrium.
  • When forces are in equilibrium, the net force is zero, aligning with the calculated minimum force to maintain speed being also 12.25 \( N \).
By understanding the interplay of forces, students can predict movement or stasis of objects in physics problems involving sliding friction and equilibrium conditions.

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