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Chapter 3: Problem 60

An elevator starts from rest with a constant upward acceleration. It moves \(2.0 \mathrm{~m}\) in the first \(0.60 \mathrm{~s}\). A passenger in the elevator is holding a \(3.0-\mathrm{kg}\) package by a vertical string. What is the tension in the string during the accelerating process?

Short Answer

Expert verified
The tension in the string is approximately 62.73 N.

Step by step solution

01

Determine the Acceleration of the Elevator

First, use the equation of motion: \( s = ut + \frac{1}{2}at^2 \). Since the initial velocity \( u = 0 \), this simplifies to \( s = \frac{1}{2}at^2 \). Given \( s = 2.0 \, \text{m} \) and \( t = 0.60 \, \text{s} \), substitute these values: \( 2.0 = \frac{1}{2}a (0.60)^2 \). Solving for \( a \), we have \( a = \frac{2.0}{0.18} \approx 11.11 \, \text{m/s}^2 \).
02

Calculate the Net Force on the Package

The net force experienced by the package is due to the tension \( T \) in the string and the gravitational force. The equation is \( T - mg = ma \), where \( m = 3.0 \, \text{kg} \), \( g = 9.8 \, \text{m/s}^2 \), and \( a = 11.11 \, \text{m/s}^2 \).
03

Solve for the Tension in the String

Rearrange the force equation to solve for tension: \( T = mg + ma \). Substitute the values: \( T = 3.0 \times 9.8 + 3.0 \times 11.11 \). This gives \( T = 29.4 + 33.33 \approx 62.73 \, \text{N} \).

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Key Concepts

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

Newton's Laws of Motion
Newton's Laws of Motion are fundamental to understanding physics. They describe the relationship between a body and the forces acting upon it. There are three laws:
  • First Law (Law of Inertia): A body will remain at rest or move in a straight line with constant speed unless acted upon by a force.
  • Second Law (Law of Acceleration): The acceleration of an object depends on the mass of the object and the amount of force applied. Expressed as the formula: \( F = ma \), where \( F \) is force, \( m \) is mass, and \( a \) is acceleration.
  • Third Law (Action and Reaction): For every action, there is an equal and opposite reaction.
These laws help us understand why the elevator in the problem can accelerate upwards and how the package behaves inside. Applying the second law explains how force and mass relate to acceleration here, with tension in the string being crucial during the motion process.
Tension in Physics
Tension is the force exerted by a string, rope, or cable when it is pulled tight by forces acting from opposite ends. In physics, tension ensures that forces are transferred efficiently between objects. When considering the package in the elevator, tension is responsible for counteracting gravitational pull as well as contributing to the upward movement initiated by the elevator's acceleration.
  • The formula used to find tension when both gravitational and additional forces like acceleration are at play is: \( T = mg + ma \).
  • Here, \( T \) represents tension, \( m \) is the object's mass, \( g \) is acceleration due to gravity, and \( a \) is any additional acceleration, as seen when the elevator moves upwards.
When calculating tension, understanding the forces at play is vital, particularly in dynamic scenarios where acceleration alters tension levels beyond static situations.
Equations of Motion
Equations of motion are a set of equations used to describe the motion of a body under constant acceleration. These equations allow us to determine variables such as position, velocity, and time when details of the motion are available.
  • Key Equation: \( s = ut + \frac{1}{2}at^2 \)
  • Where \( s \) is the distance traveled, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time elapsed.
  • This calculation is crucial for finding the elevator's acceleration in the problem, where the initial velocity \( u = 0 \).
By implementing these equations, we analyze motions, predict future positions, and uncover initial conditions or forces, just as we solved for the elevator's dynamic behavior. Understanding these equations is essential for physics problem-solving, especially in cases involving variable acceleration, as calculations hinge upon accurately interpreting each element's role.

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Most popular questions from this chapter

A tow rope will break if the tension in it exceeds \(1500 \mathrm{~N}\). It is used to tow a \(700-\mathrm{kg}\) car along level ground. What is the largest acceleration the rope can give to the car? (Remember that 1500 has four significant figures; see Appendix A.) The forces acting on the car are shown in Fig. \(3-12\). Only the \(x\) -directed force is of importance, because the \(y\) -directed forces balance each other. Indicating the positive direction with a \(+\) sign and a little arrow, we write, $$ \pm \sum F_{x}=m a_{x} \quad \text { becomes } \quad 1500 \mathrm{~N}=(700 \mathrm{~kg})(a) $$ from which \(a=2.14 \mathrm{~m} / \mathrm{s}^{2}\).

A horizontal cable pulls a \(200-\mathrm{kg}\) cart along a horizontal track. The tension in the cable is \(500 \mathrm{~N}\). Starting from rest, (a) How long will it take the cart to reach a speed of \(8.0 \mathrm{~m} / \mathrm{s}\) ? (b) How far will it have gone?

A man who weighs \(1000 \mathrm{~N}\) on Earth stands on a scale on the surface of the mythical nonspinning planet Mongo. That body has a mass which is \(4.80\) times Earth's mass and a diameter which is \(0.500\) times Earth's diameter. Neglecting the effect of the Earth's spin, how much does the scale read?

Consider an essentially spherical homogeneous celestial body of mass \(M\). The acceleration due to gravity in its vicinity beyond its surface at a distance \(R\) from its center is \(g_{R}\). Show that $$ g_{R}=\frac{G M}{R^{2}} $$ Notice that the acceleration drops off as \(1 / R^{2}\). Imagine an object of mass \(m\) at a distance \(R\) from the center of our celestial body. Its weight is \(F_{W}=m g_{R}\), but that's also the gravitation force on it due to the mass \(M\), that is, \(F_{W}=F_{G}\). Hence, $$ m g_{R}=G \frac{m M}{R^{2}} $$

A \(600-\mathrm{kg}\) car is coasting along a level road at \(30 \mathrm{~m} / \mathrm{s}\). (a) How large a retarding force (assumed constant) is required to stop it in a distance of \(70 \mathrm{~m} ?(b)\) What is the minimum coefficient of friction between tires and roadway if this is to be possible? Assume the wheels are not locked, in which case we are dealing with static friction- there's no sliding. (a) First find the car's acceleration from a constant- \(a\) equation. It is known that \(v_{i x}=30 \mathrm{~m} / \mathrm{s}, v_{f x}=0\), and \(x=70 \mathrm{~m}\). Use \(v_{f x}^{2}=v_{i x}^{2}+2 a x\) to find $$ a=\frac{v_{f x}^{2}-v_{i x}^{2}}{2 x}=\frac{0-900 \mathrm{~m}^{2} / \mathrm{s}^{2}}{140 \mathrm{~m}}=-6.43 \mathrm{~m} / \mathrm{s}^{2} $$ Now write $$ F=m a=(600 \mathrm{~kg})\left(-6.43 \mathrm{~m} / \mathrm{s}^{2}\right)-3858 \mathrm{~N} \text { or }-3.9 \mathrm{kN} $$ (b) Assume the force found in ( \(a\) ) is supplied as the friction force between the tires and roadway. Therefore, the magnitude of the friction force on the tires is \(F_{\mathrm{f}}=3858 \mathrm{~N}\). The coefficient of friction is given by \(\mu_{s}=F_{\mathrm{f}} / F_{N}\), where \(F_{N}\) is the normal force. In the present case, the roadway pushes up on the car with a force equal to the car's weight. Therefore, $$ F_{N}=F_{W}=m g=(600 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=5886 \mathrm{~N} $$ so that $$ \mu_{s}=\frac{F_{\mathrm{f}}}{F_{N}}=\frac{3858}{5886}=0.66 $$ The coefficient of friction must be at least \(0.66\) if the car is to stop within \(70 \mathrm{~m}\).
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