/*! 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 65 Two boxes connected by a light h... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 65

Two boxes connected by a light horizontal rope are on a horizontal surface, as shown in Fig. P5.35. The coefficient of kinetic friction between each box and the surface is \(\mu_{\mathrm{k}}=0.30\) . One box (box \(B\) ) has mass \(5.00 \mathrm{kg},\) and the other box (box \(A )\) has mass \(m .\) A force \(F\) with magnitude 40.0 \(\mathrm{N}\) and direction \(53.1^{\circ}\) above the horizontal is applied to the \(5.00-\mathrm{kg}\) box, and both boxes move to the right with \(a=1.50 \mathrm{m} / \mathrm{s}^{2}\) . (a) What is the tension \(T\) in the rope that connects the boxes? (b) What is the mass \(m\) of the second box?

Short Answer

Expert verified
(a) The tension is 3.74 N. (b) The mass of the second box is 0.84 kg.

Step by step solution

01

Analyze Forces on Box B

First, consider the forces acting on box B. The horizontal component of the force, \( F_x \), is \( F\cos(53.1掳) \) and the vertical component \( F_y \) is \( F\sin(53.1掳) \). Thus, \( F_x = 40\cos(53.1掳) \approx 24.04 \text{ N} \). The normal force \( N_B \) on box B is \( N_B = mg - F_y \), where \( mg = 5.00kg \times 9.81m/s^2 \) and \( F_y = 40\sin(53.1掳) \).
02

Compute Friction on Box B

The kinetic friction force \( f_k \) on box B is \( \mu_k \times N_B \). Substituting from step 1, \( N_B = 5.00kg \times 9.81m/s^2 - 40\times\sin(53.1掳) \approx 37.67 \text{ N} \). Thus, \( f_k = 0.30 \times 37.67 \approx 11.30 \text{ N} \).
03

Write Equation for Box B's Acceleration

Using Newton's second law, \( F_x - f_k - T = ma_B \). Substituting the values, \( 24.04 - 11.30 - T = 5.00\times 1.50 \), solve for \( T \). This yields \( T = 11.24 - 7.50 = 3.74 \text{ N} \).
04

Analyze Forces on Box A

Consider the forces on box A. The tension \( T \) pulls it forward, and friction opposes the motion. According to Newton's second law, \( T - f_{kA} = ma_A \), where the friction on box A, \( f_{kA} = \mu_k \times N_A \) with \( N_A = mg \).
05

Compute Mass of Box A

Substitute the known values from previous steps into the equation for box A. Since \( T = 3.74 \text{ N} \), and \( f_{kA} = 0.30 \times 9.81m \), we have \( 3.74 - 0.30 \times 9.81m = ma \). Solve for \( m \), \( 3.74 = 1.50m + 2.943m \). Reorganize to find \( m = \frac{3.74}{4.443}\approx 0.84 \text{ kg} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Kinetic Friction
Kinetic friction is the resistive force that occurs when two surfaces slide against each other. For this exercise, we deal with kinetic friction between the boxes and the horizontal surface. The friction force is calculated using the equation:
  • \( f_k = \mu_k \times N \)
where \( \mu_k \) is the coefficient of kinetic friction and \( N \) is the normal force. Given that the coefficient of kinetic friction \( \mu_k \) is 0.30, we first determine the normal force. For box B, the normal force \( N_B \) is calculated by considering weight and vertical forces applied. The normal force is balanced by gravitational force corrected for any vertical component of an external force. This makes sure friction depends not only on surface type but also on how these forces interact.
Applying the equation allows us to determine that the friction opposing box B鈥檚 motion is around 11.30 N, exemplifying how effectively kinetic friction resists motion even without visible changes.
Tension in a Rope
Tension is the force transmitted through a string, rope, cable, or similar when it is pulled tight by forces acting from opposite ends. In this problem, tension acts as the connecting force between boxes A and B, influencing their movement across the surface. To find the tension, we need to apply Newton's Second Law which states that the sum of forces equals the mass times acceleration of the body, or motion segment:
  • \( F_x - f_k - T = ma_B \)
The tension \( T \) in the rope can then be calculated by isolating \( T \) in this equation. Here, box B鈥檚 horizontal movement is primarily aided by the horizontal force component of an applied force and resisted by kinetic friction. Thus, solving the equation, we find the tension in the rope is approximately 3.74 N.
This calculation reflects how tension ensures connectivity and equilibrium between moving parts in a system.
Acceleration Calculation
Calculating acceleration, particularly influenced by various forces, is crucial for understanding motion dynamics. Newton's Second Law relates force, mass, and acceleration in the form:
  • \( F = ma \)
For box A, it's essential to consider the net force acting upon it. The acceleration \( a \) is given as 1.50 m/s虏. The net force comprises the tension pulling forward and kinetic friction hindering the movement:
  • \( T - f_{kA} = ma_A \)
Using given data, recognizing that friction acts as an opposing force calculated with the mass-dependent normal force, and solving for \( m \), demonstrates the connection between these elements. This enables determining the mass of box A to be approximately 0.84 kg.
Thus, acceleration not only determines how fast objects speed up but also plays a pivotal role in calculating other related properties, such as unknown masses within dynamic systems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

You tie a cord to a pail of water, and you swing the pail in a vertical circle of radius 0.600 \(\mathrm{m} .\) What minimum speed must you give the pail at the highest point of the circle if no water is to spill from it?

The Cosmoclock 21 Ferris wheel in Yokohama City, Japan, has a diameter of 100 \(\mathrm{m}\) . Its name comes from its 60 arms, each of which can function as a second hand (so that it makes one revolution every 60.0 s). (a) Find the speed of the passengers when the Ferris wheel is rotating at this rate. (b) A passenger weighs 882 \(\mathrm{N}\) at the weight-guessing booth on the ground. What is his apparent weight at the highest and at the lowest point on the Ferris wheel? (c) What would be the time for one revolution if thepassenger's apparent weight at the highest point were zero? (d) What then would be the passenger's apparent weight at the lowest point?

A hammer is hanging by a light rope from the ceiling of a bus. The ceiling of the bus is parallel to the roadway. The bus is traveling in a straight line on a horizontal street. You observe that the hammer hangs at rest with respect to the bus when the angle between the rope and the ceiling of the bus is \(67^{\circ} .\) What is the acceleration of the bus?

Block \(A\) in Fig. P5.72 weighs 60.0 \(\mathrm{N} .\) The coefficient of static friction between the block and the surface on which it rests is \(0.25 .\) The weight \(w\) is 12.0 \(\mathrm{N}\) and the system is in equilibrium. (a) Find the friction force exerted on block \(A .\) (b) Find the maximum weight \(w\) for which the system will remain in equilibrium.

A 1125 -kg car and a 2250 -kg pickup truck approach a curve on the expressway that has a radius of 225 \(\mathrm{m}\) . (a) At what angle should the highway engineer bank this curve so that vehicles traveling at 65.0 \(\mathrm{mi} / \mathrm{h}\) can safely round it regardless of the condition of their tires? Should the heavy truck go slower than the lighter car? (b) As the car and truck round the curve at find the normal force on each one due to the highway surface.
See all solutions

Recommended explanations on Physics Textbooks

Torque and Rotational Motion

Read Explanation

Fundamentals of Physics

Read Explanation

Work Energy and Power

Read Explanation

Scientific Method Physics

Read Explanation

Linear Momentum

Read Explanation

Astrophysics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.