/*! 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 71 A magician pulls a tablecloth fr... [FREE SOLUTION] | 91Ó°ÊÓ

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A magician pulls a tablecloth from under a \(200-\mathrm{g}\) mug located \(30.0 \mathrm{cm}\) from the edge of the cloth. The cloth exerts a friction force of \(0.100 \mathrm{N}\) on the mug, and the cloth is pulled with a constant acceleration of \(3.00 \mathrm{m} / \mathrm{s}^{2} .\) How far does the mug move relative to the horizontal tabletop before the cloth is completely out from under it? Note that the cloth must move more than \(30 \mathrm{cm}\) relative to the tabletop during the process.

Short Answer

Expert verified
The mug moves 2.5 cm relative to the tabletop before the cloth is completely pulled out from under it.

Step by step solution

01

Calculate mug's acceleration

The net force on the mug is the frictional force, which equals the mass times the mug's acceleration. We can use the formula from Newton's second law \(F = ma\), where \(m\) is the mass of the mug and \(a\) is its acceleration. Solving for acceleration \(a\), we find \(a = F/m = 0.100 \mathrm{N} / 0.2 \mathrm{kg} = 0.5 \mathrm{m/s^{2}}\). Note that the frictional force acts opposite to the motion of the mug so the acceleration is negative.
02

Calculate the time

The time \(t\) required to pull the cloth out from under the mug can be calculated by dividing the distance that the edge of the cloth travels by the speed of the cloth. Hence, \(t = 30 \mathrm{cm} / 3.00 \mathrm{m/s} = 0.1 \mathrm{s}\).
03

Calculate the distance the mug moves

Next, we determine the distance \(d\) the mug moves while the cloth is still under it by using the formula for the distance traveled under constant acceleration: \(d = v_{i}t + 0.5at^{2}\). Here, \(v_{i}\) is the initial velocity, which is zero, \(a\) is the acceleration derived from step 1, and \(t\) is the time calculated in step 2. So, \(d = 0.5*(-0.5 \mathrm{m/s^{2}})*(0.1 \mathrm{s})^{2} = -0.0025 \mathrm{m}\) or -2.5 cm. The negative sign indicates that the mug moves in the opposite direction to the motion of the cloth.

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Key Concepts

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

Newton's second law
Understanding Newton's second law is crucial when examining forces and motion. This law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Mathematically, this relationship is represented as \( F = ma \), where \( F \) is the net force, \( m \) is the object's mass, and \( a \) is the acceleration. In our magician's example, the mug's acceleration is caused by the frictional force of the tablecloth. With a known force and mass, we can calculate that the mug experiences an acceleration of \( 0.5 \(\mathrm{m/s^2}\) \). It's interesting to note that the acceleration of the mug is lesser than the cloth, due to the friction not being the only force at play—gravity and the object's inertia are also crucial factors.

When applying this law, we assume that the forces are constant and that they result in a uniform acceleration. This is key in predicting the object's motion over time, which leads us to the concept of constant acceleration.
Constant acceleration
In physics, constant acceleration is a condition where the rate of change of velocity of an object is consistent over time. This means that the object speeds up or slows down at the same rate during the entire duration of its motion. In the context of our problem, the cloth is pulled with a constant acceleration, while the mug also encounters a steady frictional force, leading to its constant negative acceleration.

Constant acceleration allows us to use kinematic equations to predict an object's future position and velocity at any time. The basic equation for calculating distance under constant acceleration is \( d = v_i t + \frac{1}{2}at^2 \), where \( v_i \) is the initial velocity, \( t \) is time, and \( a \) is the constant acceleration. These equations are particularly helpful when forces are steady, as is the case in this exercise, where the magician pulls the cloth steadily enough to maintain a constant acceleration.
Friction force calculation
The friction force is a resistive force that acts opposite to the direction of motion and is crucial in problems involving motion and acceleration. To calculate this force, we need to understand the two surfaces in contact—in our case, the mug and the tablecloth. The frictional force can be found using the equation \( f = \mu N \) where \( \mu \) is the coefficient of friction and \( N \) is the normal force, which is equal to the object's weight when on a horizontal surface. However, in this problem, the frictional force has been given directly.

Using the given frictional force \( 0.100 \mathrm{N} \) and the mug's mass \( 0.200 \mathrm{kg} \) (or \( 200 \mathrm{g} \) as in the exercise), we applied Newton's second law to determine the acceleration of the mug. This accelerative force is the result of the friction between the mug and the cloth. As the cloth is pulled out with a known acceleration, the friction force acting on the mug brings about a corresponding motion. The mug's negative acceleration (\( -0.5 \mathrm{m/s^2} \) in our solution) reflects that the motion induced by the frictional force is opposite to that of the tablecloth's pull.

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Most popular questions from this chapter

A 15.0 -Ib block rests on the floor. (a) What force does the floor exert on the block? (b) If a rope is tied to the block and run vertically over a pulley, and the other end is attached to a free-hanging 10.0 -lb weight, what is the force exerted by the floor on the 15.0 -lb block? (c) If we replace the 10.0 -lb weight in part (b) with a 20.0 -lb weight, what is the force exerted by the floor on the 15.0 -lb block?

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A force \(\mathbf{F}\) applied to an object of mass \(m_{1}\) produces an acceleration of \(3.00 \mathrm{m} / \mathrm{s}^{2} .\) The same force applied to a second object of mass \(m_{2}\) produces an acceleration of \(1.00 \mathrm{m} / \mathrm{s}^{2} .\) (a) What is the value of the ratio \(m_{1} / m_{2} ?\) (b) If \(m_{1}\) and \(m_{2}\) are combined, find their acceleration under the action of the force \(\mathbf{F}\).

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