/*! 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 27 A small block is at rest on the ... [FREE SOLUTION] | 91Ó°ÊÓ

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A small block is at rest on the floor at the front of a rail. road boxcar that has length \(\ell .\) The coefficient of kinetic friction between the floor of the car and the block is \(\mu_{k}\) The car, originally at rest, begins to move with acceleration a. The block slides back horizontally until it hits the back wall of the car. At that moment, what is its speed (a) relative to the car? (b) relative to Earth?

Short Answer

Expert verified
The speed of the block when it hits the back of the car, relative to the car is \(v_{b} = a_{b} * t\). The speed of the block relative to the earth at the moment it hits the back of the car is given by \(v = v_{b} + a * t\).

Step by step solution

01

Identify the forces acting on the block

In this step, identify the forces acting on the block. The block is subjected to kinetic friction (\(f_{k}\)) which acts in the opposite direction to the car's acceleration. The friction is given by \(f_{k} = \mu_{k} * m * g\), where \(m\) is the mass of the block and \(g\) is the acceleration due to gravity.
02

Apply Newton's second law

In this step, apply Newton's second law in the horizotal direction to the block as it slides along the floor of the car. The horizontal acceleration of the block is caused by kinetic friction and can be found from \(m*a_{b} = f_{k}\), where \(a_{b}\) is the acceleration of the block.
03

Calculate the time for the block to hit the car's back

With the known acceleration of the block relative to the car, we can now calculate the time \(t\) it takes for the block to slide across the length of the car, which is given by \(t = \sqrt{\frac{2*l}{a_{b}}}\). Here, we are using the kinematic equation \(d = \frac{1}{2} a t^2\) for distance traveled in time \(t\) under acceleration \(a\), with starting velocity 0.
04

Calculate the speed of the block relative to the car

In this step, calculate the speed of the block relative to the car as it hits the back using the formula \(v_{b} = a_{b} * t\). Here, \(v_{b}\) is the speed of the block relative to the car, \(a_{b}\) is the acceleration of the block and \(t\) is the time calculated in the previous step.
05

Calculate the speed of the block relative to the earth

Finally, calculate the speed of the block relative to the earth when it hits the car's back. This is given by \(v = v_{b} + a * t\), where \(v_{b}\) is the speed of the block relative to the car, \(a\) is the acceleration of the car and \(t\) is the time calculated previously. This equation takes into account both the motion of the block and the car. The block's speed relative to the ground when it hits the back wall is its speed relative to the car plus the speed of the car itself.

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Key Concepts

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

Newton's Second Law Applications
One of the fundamental principles used in physics problems involving motion and forces is Newton's Second Law of Motion. This law states that the force exerted on an object is equal to the mass of the object multiplied by its acceleration, commonly written as \( F = m * a \). In practice, this law allows us to analyze situations where an object is accelerating due to various forces, such as friction, gravity, or applied forces.

When a block slides on an accelerating surface, the friction force acting on it depends on the kinetic friction coefficient and is exerted in the opposite direction of the acceleration of the surface. By applying Newton's Second Law, we can determine the block's acceleration relative to the surface. This is crucial to solving problems like the one in our exercise, as it gives us a starting point to find the acceleration and eventually the velocity of the block.
Kinematic Equations
Kinematic equations are the mathematical formulations that describe the motion of objects under constant acceleration. They connect the dots between an object's initial conditions (such as initial position and velocity), its acceleration, and its position and velocity at any later point in time. With these equations, we can calculate missing variables like time, distance, initial or final velocities without needing to consider the forces involved.

One of these equations, \( d = \frac{1}{2} a t^2 \), where \(d\) is distance, \(a\) is acceleration, and \(t\) is time, is often used when an object starts from rest, as is the case with the block in our example. This equation is particularly relevant for determining how long it takes for the block to reach the end of the car's floor.
Coefficient of Kinetic Friction
The coefficient of kinetic friction, denoted as \( \mu_{k} \), is a dimensionless number that characterizes the friction between two surfaces in relative motion. It is determined experimentally and varies based on the materials' properties and surface conditions. The kinetic frictional force itself is calculated by multiplying this coefficient by the normal force acting on the moving object, given as \( f_{k} = \mu_{k} * m * g \) where \(m\) is mass and \(g\) is the gravitational acceleration.

Understanding how to use this coefficient is critical for solving problems involving a moving object on a surface. It allows us to find the magnitude of the friction force, which is necessary for applying Newton's Second Law, as described in the exercise.
Block Sliding on Accelerating Surface
When we have a block on an accelerating surface, such as the floor of a boxcar, it's essential to understand the interaction between the block's motion and the car's motion. As the car accelerates, the block moves relative to the car due to the friction between them, which acts as the unbalanced force causing the block to accelerate in the opposite direction of the car's acceleration. This relative motion is essentially an application of Newton's Second Law, with the force being the frictional force calculated by the coefficient of kinetic friction.

The real-world outcome of this interaction can be visualized as a block that begins at rest (relative to the car and Earth) and ends up sliding to the back of the car until it hits the rear wall; its motion has direct implications for the relative velocity which we are attempting to find in the exercise.
Relative Velocity
Relative velocity is a measure of how fast one object is moving with respect to another. In many physics problems, like the sliding block in our exercise, we are interested in comparing the velocities of two objects in different reference frames. In the exercise, we determine the block's speed relative to both the car and the Earth. The relative velocity between two objects can be found by vector addition if they are in linear motion.

For the block, its relative velocity with respect to the car is simply the velocity it gains due to friction. However, when considering its velocity relative to the Earth, we must combine this with the boxcar's velocity since the car is the surface upon which our block slides. This additive property of velocities is fundamental in problems dealing with relative motion and is showcased in the final step of the exercise solution.

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Most popular questions from this chapter

A crate of eggs is located in the middle of the flat bed of a pickup truck as the truck negotiates an unbanked curve in the road. The curve may be regarded as an arc of a circle of radius \(35.0 \mathrm{m}\). If the coefficient of static friction between crate and truck is \(0.600,\) how fast can the truck be moving without the crate sliding?

A \(50.0-\mathrm{kg}\) parachutist jumps from an airplane and falls to Earth with a drag force proportional to the square of the speed, \(R=C v^{2} .\) Take \(C=0.200 \mathrm{kg} / \mathrm{m}\) (with the parachute closed) and \(C=20.0 \mathrm{kg} / \mathrm{m}\) (with the chute open). (a) Determine the terminal speed of the parachutist in both configurations, before and after the chute is opened. (b) Set up a numerical analysis of the motion and compute the speed and position as functions of time, assuming the jumper begins the descent at \(1000 \mathrm{m}\) above the ground and is in free fall for 10.0 s before opening the parachute. (Suggestion: When the parachute opens, a sudden large acceleration takes place; a smaller time step may be necessary in this region.)

Whenever two Apollo astronauts were on the surface of the Moon, a third astronaut orbited the Moon. Assume the orbit to be circular and \(100 \mathrm{km}\) above the surface of the Moon, where the acceleration due to gravity is \(1.52 \mathrm{m} / \mathrm{s}^{2}\) The radius of the Moon is \(1.70 \times 10^{6} \mathrm{m} .\) Determine (a) the astronaut's orbital speed, and (b) the period of the orbit.

A small piece of Styrofoam packing material is dropped from a height of \(2.00 \mathrm{m}\) above the ground. Until it reaches terminal speed, the magnitude of its acceleration is given by \(a=g-b v .\) After falling \(0.500 \mathrm{m},\) the Styrofoam effectively reaches terminal speed, and then takes \(5.00 \mathrm{s}\) more to reach the ground. (a) What is the value of the constant \(b ?\) (b) What is the acceleration at \(t=0 ?\) (c) What is the acceleration when the speed is \(0.150 \mathrm{m} / \mathrm{s} ?\)

The Earth rotates about its axis with a period of \(24.0 \mathrm{h}\) Imagine that the rotational speed can be increased. If an object at the equator is to have zero apparent weight, (a) what must the new period be? (b) By what factor would the speed of the object be increased when the planet is rotating at the higher speed? Note that the apparent weight of the object becomes zero when the normal force exerted on it is zero.

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