/*! 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 \(1-\mathrm{kg}\) block ' \(B\... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 6

A \(1-\mathrm{kg}\) block ' \(B\) ' rests as shown on a frictionless bracket \({ }^{\prime} A\) ' of same mass. Constant force \(F=3 \mathrm{~N}\) starts to act at time \(t=0\), when the distance of block \(B\) from the end of bracket is \(13.5 \mathrm{~m}\). Find time (in sec), when block \(B\) falls off the bracket.

Short Answer

Expert verified
The block falls off the bracket at approximately 4.24 seconds.

Step by step solution

01

Understanding Force Action on System

The force of 3 N acts on the 1-kg mass block B. As block B is on a frictionless bracket A, both would accelerate equally when force is applied to B. This results in a combined acceleration of the system.
02

Calculating Acceleration

The total mass of the system (block B and bracket A) is 2 kg. The acceleration can be calculated using Newton's second law: \( F = ma \), where \( F = 3 \, \mathrm{N} \) and \( m = 2 \), hence \( a = \frac{3}{2} = 1.5 \, \mathrm{m/s^2} \).
03

Relative Motion Analysis

Consider the motion of block B relative to bracket A. Since only block B is experiencing the force directly, it should be analyzed separately to find its displacement relative to A.
04

Finding Time for Displacement

The relative acceleration of block B with respect to bracket A is the same (1.5 m/s虏) because both start from rest. Using the equation \( s = ut + \frac{1}{2} a t^2 \), and with initial velocity \( u = 0 \), we can set \( s = 13.5 \, \mathrm{m} \) for when block B falls off: \( 13.5 = \frac{1}{2} \cdot 1.5 \cdot t^2 \).
05

Solving for Time

Simplifying the equation from the last step gives \( 13.5 = 0.75 t^2 \). Solving for \( t \) gives \( t^2 = \frac{13.5}{0.75} = 18 \). Thus, \( t = \sqrt{18} = 3 \sqrt{2} \approx 4.24 \) seconds.

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Key Concepts

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

Newton's Second Law
Newton's Second Law is a fundamental principle in physics that explains the relationship between force, mass, and acceleration.
This law is mathematically represented by the equation: \( F = ma \), where \( F \) is the force applied, \( m \) is the mass of the object, and \( a \) is the acceleration produced.
In the provided exercise, the second law is used to determine how a constant force of 3 N affects a system consisting of a block and a bracket, both with mass 1 kg each.
  • The force applied is 3 N.
  • The total mass of the system is 2 kg (since both the block and bracket are 1 kg).
  • The resulting acceleration is calculated as \( a = \frac{F}{m} = \frac{3}{2} = 1.5 \, \mathrm{m/s^2} \).
Understanding this relationship helps us predict how the velocity of an object will change over time when subjected to a specific force.
By applying this principle, we can solve for the system's acceleration and subsequently analyze the motion of each component.
Acceleration
Acceleration is a measure of how quickly the velocity of an object changes over time.
It is a vector quantity, meaning that it has both magnitude and direction. In the context of the exercise, the system of the block and bracket undergoes uniform acceleration.
This uniformity is vital because even though the force directly acts on block B, both the block and the bracket are frictionlessly linked, so they accelerate as one unit.
  • Calculated acceleration is \( 1.5 \, \mathrm{m/s^2} \) for both block B and bracket A.
  • This reflects how quickly their velocities are increasing due to the applied force.
When solving problems involving acceleration, knowing that force leads directly to acceleration as outlined in Newton's Second Law can simplify calculations.
Thus, even with multiple bodies involved, such as in this exercise, acceleration remains constant across all bodies when they share the same conditions.
Relative Motion
Relative motion refers to the movement of one object in relation to another.
This concept is crucial when analyzing how objects interact within a system.In the given exercise, although both the block and the bracket begin from rest, the problem requires evaluating the motion of block B relative to bracket A.
This means even while the entire system accelerates externally, block B's movement is studied concerning bracket A, to determine the moment when block B "falls off" the bracket.
  • This is achieved by analyzing the displacement needed for block B to move with respect to bracket A.
  • The relative acceleration simplifies to the same rate as the overall acceleration calculated (\( 1.5 \, \mathrm{m/s^2} \)) because of the initial conditions.
  • The displacement equation \( s = \frac{1}{2} a t^2 \) was used to find the time it took for block B to travel 13.5 m.
Understanding relative motion allows for breaking down complex systems into simpler parts, where movement, such as the block falling off, can be studied in more manageable terms.
Relative motion analysis is essential as it distinguishes actual perceived movement from that of an observational frame, herein bracket A.

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Most popular questions from this chapter

Two blocks of masses \(m_{1}\) and \(m_{2}\) are connected with a light spring of force constant \(k\) and the whole system is kept on a frictionless horizontal surface. The masses are applied forces \(F_{1}\) and \(F_{2}\) as shown in figure. At any time, the blocks have same acceleration \(a_{0}\) but in opposite directions. Now answer the following: The value of spring force is (1) \(\frac{m_{1} F_{2}+F_{1} m_{2}}{m_{1}-m_{2}}\) (2) \(\frac{m_{1} F_{2}-F_{1} m_{2}}{m_{1}+m_{2}}\) (3) \(\frac{m_{1} F_{2}+F_{1} m_{2}}{m_{1}+m_{2}}\) (4) \(\frac{m_{1} F_{2}-F_{1} m_{2}}{m_{1}-m_{2}}\)

A particle of mass \(2 \mathrm{~kg}\) moves with an initial velocity of \((4 \hat{i}+2 \hat{j}) \mathrm{ms}^{-1}\) on the \(x-y\) plane. A force \(\vec{F}=(2 \hat{i}-8 \hat{j}) \mathrm{N}\) acts on the particle. The initial position of the particle is \((2 \mathrm{~m}, 3 \mathrm{~m})\). Then for \(y=3 \mathrm{~m}\), (1) Possible value of \(x\) is only \(x=2 \mathrm{~m}\) (2) Possible value of \(x\) is not only \(x=2 \mathrm{~m}\), but there exists some other value of \(x\) also (3) Time taken is \(2 \mathrm{~s}\) (4) All of the above

A block of weight \(9.8 \mathrm{~N}\) is placed on a table. The table surface exerts an upward force of \(10 \mathrm{~N}\) on the block. Assume \(g=9.8 \mathrm{~m} / \mathrm{s}^{2} .\) Select the correct statement(s). (1) The block exerts a force of \(10 \mathrm{~N}\) on the table (2) The block exerts a force of \(19.8 \mathrm{~N}\) on the table (3) The block exerts a force of \(9.8 \mathrm{~N}\) on the table (4) The block has an upward acceleration

A pendulum of mass \(m\) hangs from a support fixed to a trolley. The direction of the string when the trolley rolls up a plane of inclination \(\alpha\) with acceleration \(a_{0}\) is (1) \(\theta=\tan ^{-1} \alpha\) (2) \(\theta=\tan ^{-1}\left(\frac{a_{0}}{g}\right)\) (3) \(\theta=\tan ^{-1}\left(\frac{g}{a_{0}}\right)\) (4) \(\theta=\tan ^{-1}\left(\frac{a_{0}+g \sin \alpha}{g \cos \alpha}\right)\)

A particle is moving in the \(x-y\) plane. At certain instant of time, the components of its velocity and acceleration are as follows: \(v_{x}=3 \mathrm{~ms}^{-1}, v_{y}=4 \mathrm{~m} \mathrm{~s}^{-1}, a_{x}=2 \mathrm{~m} \mathrm{~s}^{-2}\) and \(a_{y}=1 \mathrm{~ms}^{-2}\). The rate of change of speed at this moment is (1) \(\sqrt{10} \mathrm{~m} \mathrm{~s}^{-2}\) (2) \(4 \mathrm{~ms}^{-2}\) (3) \(\sqrt{5} \mathrm{~ms}^{-2}\) (4) \(2 \mathrm{~ms}^{-2}\)
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