/*! 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 crate of mass M starts from re... [FREE SOLUTION] | 91Ó°ÊÓ

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A crate of mass M starts from rest at the top of a frictionless ramp inclined at an angle a above the horizontal. Find its speed at the bottom of the ramp, a distance d from where it started. Do this in two ways: Take the level at which the potential energy is zero to be (a) at the bottom of the ramp with y positive upward, and (b) at the top of the ramp with y pos-itive upward. (c) Why didn’t the normal force enter into your solution?

Short Answer

Expert verified
The speed of the crate at the bottom of the ramp is \( V = sqrt(2 * g * d * sin(a)) \). The normal force did not factor into the solution, as it acts perpendicular to the direction of motion, doing no work and not contributing to kinetic or potential energy.

Step by step solution

01

Setup the Problem

Firstly, recall the conservation of mechanical energy principle which states that if there are only conservative forces acting on an object or system, the total mechanical energy (kinetic energy \( KE = 0.5 * M * V^2 \) plus potential energy \( PE = M* g * h \) where g is acceleration due to gravity and h is the height) remains constant. In this case, the crate starts from rest so the initial kinetic energy is zero and it also ends up on a ramp where the potential energy is zero.
02

Solve for Part (a)

If we consider the bottom of the ramp as our reference \( y = 0 \), the crate initially has a potential energy given by \( PE_{initial}= M * g * h \). As it slides down, this potential energy gets converted into kinetic energy. Therefore, at the bottom of the ramp, \( KE_{final} = PE_{initial} \), or \( 0.5 * M * V^2 = M * g * h \). By simplifying this we get \( V = sqrt(2 * g * h) \) where h is the height of the ramp from the ground, which is \( d* sin(a) \). So, \( V = sqrt(2 * g * d * sin(a)) \)
03

Solve for Part (b)

Now, if we take the top of the ramp as our reference point \( y = 0 \). Initially, the potential energy is zero since this is our reference point. While sliding, the crate loses potential energy equal to \( M * g * h \) but gains an equal amount of kinetic energy. Using the same energy conservation principle, the final kinetic energy at the bottom of the ramp would still be \( 0.5 * M * V^2 = M * g * h \) with the same solution found in part (a).
04

Answer Part (c)

The normal force doesn't enter our calculations because it acts perpendicular to the direction of motion of the crate. In the context of work and energy, the work done by a force perpendicular to the direction of motion is always zero. Because work and energy are directly related, the normal force doesn't impact the kinetic or potential energy and hence doesn't affect our solution.

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Key Concepts

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

Kinetic Energy
Kinetic energy represents the energy that an object possesses due to its motion. It is given by the equation KE = 0.5 \( M \times V^2 \), where \(M\) is the mass of the object and \(V\) is its velocity. In the context of the crate on the ramp, as the crate slides down, its kinetic energy increases as its velocity increases. This energy transformation from potential to kinetic energy is crucial in understanding how the crate gains speed as it descends the incline.

It's important to note that kinetic energy is a form of mechanical energy, and it's scalar, meaning it doesn't have a direction but only a magnitude. This concept becomes especially relevant when analyzing the motion of objects on inclined planes, such as in the provided textbook exercise.
Potential Energy
Potential energy is the stored energy in an object due to its position or height. Specifically, gravitational potential energy on Earth is calculated using the equation PE = M \times g \times h, where \(g\) is the acceleration due to gravity, and \(h\) is the height above the reference point. In the given exercise, the crate has its maximum potential energy at the top of the ramp, which is then converted into kinetic energy as the crate moves downwards.

When solving physics problems involving potential energy, it's essential to clearly define the reference level where \(PE\) is zero, as this can affect the calculation of height and subsequently the potential and kinetic energy.
Inclined Plane Physics
The physics of an inclined plane simplifies the analysis of objects moving along slopes. When an object like a crate slides down a frictionless ramp, gravity does all the work, transforming potential energy into kinetic energy. The angle of the incline, denoted as \(\alpha\) in the exercise, significantly impacts the calculations - it determines the component of gravitational force acting parallel to the slope, which in turn affects the rate at which potential energy is converted to kinetic energy.

To determine the speed at the bottom, you would use the incline's height, which can be found using the sine component of the ramp's angle multiplied by the distance along the ramp, denoted as \(d\times \sin(a)\). Therefore, understanding the relationships between the angle, distance along the incline, and gravitational force is key when solving problems involving inclined planes.
Work-Energy Principle
The work-energy principle states that the work done on an object is equal to the change in its mechanical energy (the sum of its potential and kinetic energy). In the absence of friction and other non-conservative forces, the total mechanical energy of an object remains constant. This principle underlies the conservation of mechanical energy and is foundational in solving the textbook exercise about the crate on the inclined plane.

Applying the work-energy principle helps to understand why the normal force, which acts perpendicular to the motion and therefore does no work, doesn't affect the kinetic or potential energy of the crate. It also explains how, no matter the chosen reference point for potential energy (top or bottom of the ramp), the speed at the bottom of the ramp remains the same because the total mechanical energy is conserved.

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Most popular questions from this chapter

A 90.0 kg mail bag hangs by a vertical rope 3.5 m long. A postal worker then displaces the bag to a position 2.0 m sideways from its original position, always keeping the rope taut. (a) What horizontal force is necessary to hold the bag in the new position? (b) As the bag is moved to this position, how much work is done (i) by the rope and (ii) by the worker?

A 25.0 kg child plays on a swing having support ropes that are 2.20 m long. Her brother pulls her back until the ropes are 42.0° from the vertical and releases her from rest. (a) What is her potential energy just as she is released, compared with the potential energy at the bottom of the swing’s motion? (b) How fast will she be moving at the bottom? (c) How much work does the tension in the ropes do as she swings from the initial position to the bottom of the motion?

A small box with mass \(0.600 \mathrm{~kg}\) is placed against a compressed spring at the bottom of an incline that slopes upward at \(37.0^{\circ}\) above the horizontal. The other end of the spring is attached to a wall. The coefficient of kinetic friction between the box and the surface of the incline is \(\mu_{k}=0.400\) The spring is released and the box travels up the incline, leaving the spring behind. What minimum elastic potential energy must be stored initially in the spring if the box is to travel \(2.00 \mathrm{~m}\) from its initial position to the top of the incline?

\(\mathrm{A} 0.150 \mathrm{~kg}\) block of ice is placed against a horizontal, compressed spring mounted on a horizontal tabletop that is \(1.20 \mathrm{~m}\) above the floor. The spring has force constant \(1900 \mathrm{~N} / \mathrm{m}\) and is initially compressed \(0.045 \mathrm{~m}\). The mass of the spring is negligible. The spring is released, and the block slides along the table, goes off the edge, and travels to the floor. If there is negligible friction between the block of ice and the tabletop, what is the speed of the block of ice when it reaches the floor?

A block with mass \(m=\) \(\begin{array}{lll}0.200 \mathrm{~kg} & \text { is placed against a com- }\end{array}\) pressed spring at the bottom of a ramp that is at an angle of \(53.0^{\circ}\) above the horizontal. The spring has \(8.00 \mathrm{~J}\) of elastic potential energy stored in it. The spring is released, and the block moves up the incline. After the block has traveled a distance of \(3.00 \mathrm{~m},\) its speed is \(4.00 \mathrm{~m} / \mathrm{s} .\) What is the magnitude of the friction force that the ramp exerts on the block while the block is moving?

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