/*! 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 small block of mass \(m=0.032 ... [FREE SOLUTION] | 91Ó°ÊÓ

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A small block of mass \(m=0.032 \mathrm{~kg}\) can slide along the frictionless loop-the-loop, with loop radius \(R=12 \mathrm{~cm}\). The block is released from rest at point \(P\), at height \(h=5.0 R\) above the bottom of the loop. How much work does the gravitational force do on the block as the block travels from point \(P\) to (a) point \(Q\) and (b) the top of the loop? If the gravitational potential energy of the block-Earth system is taken to be zero at the bottom of the loop, what is that potential energy when the block is (c) at point \(P\), (d) at point \(Q\), and (e) at the top of the loop? (f) If, instead of merely being released, the block is given some initial speed downward along the track, do the answers to (a) through (e) increase, decrease, or remain the same?

Short Answer

Expert verified
(a) 0.19 J, (b) 0.11 J, (c) 0.19 J, (d) 0 J, (e) 0.075 J, (f) remain the same.

Step by step solution

01

Understanding the Problem

The problem involves a block moving around a loop-the-loop. We are tasked with finding out the work done by gravity and potential energy changes at different points (P, Q, and the top of the loop). The loop radius is \(R = 12\, \text{cm} = 0.12\, \text{m}\), and the block is released from a height \(h = 5R\). There is no friction, so we care only about gravitational effects.
02

Calculate Work Done by Gravity (Point P to Q)

Point Q is at the base of the loop, where the gravitational potential energy is set to zero. The work done by gravity, \(W\), as the block moves from point P to Q is equal to the change in gravitational potential energy. At point P, the height is \(h = 5R = 0.6\, \text{m}\). Thus, \(W = mgh = 0.032\, \text{kg} \times 9.81\, \text{m/s}^2 \times 0.6\, \text{m} = 0.188352\, \text{J}\).
03

Calculate Work Done by Gravity (Point P to Top of Loop)

The height of the loop is \(2R = 0.24\, \text{m}\). The work done by gravity from point P to the top is the change in potential energy from height \(h = 5R\) to \(h = 2R\). Thus, \(h' = 5R - 2R = 3R = 0.36\, \text{m}\). Therefore, \(W = mgh' = 0.032\, \text{kg} \times 9.81\, \text{m/s}^2 \times 0.36\, \text{m} = 0.1135296\, \text{J}\).
04

Calculate Potential Energy at Point P

At Point P, the height is \(h = 5R\). Therefore, potential energy \(U = mgh = 0.032\, \text{kg} \times 9.81\, \text{m/s}^2 \times 0.6\, \text{m} = 0.188352\, \text{J}\).
05

Calculate Potential Energy at Point Q

At Point Q, the block is at the base, where the potential energy is zero by definition. Therefore, \(U = 0\, \text{J}\).
06

Calculate Potential Energy at the Top of the Loop

The top of the loop is at height \(2R\). So, \(U = mgh = 0.032\, \text{kg} \times 9.81\, \text{m/s}^2 \times 0.24\, \text{m} = 0.0753408\, \text{J}\).
07

Effect of Initial Speed on Answers

If the block is given initial speed downward, only kinetic energy changes at the start. Work done by gravity and potential energy differences between specific points remain the same, as they depend solely on height differences.

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Key Concepts

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

Gravitational Potential Energy
Gravitational potential energy is a form of energy associated with the position of an object within a gravitational field. Imagine holding a book above the floor. The energy it has due to its height is gravitational potential energy.
In physics, gravitational potential energy (U) is calculated using the formula: \[ U = mgh \] where \( m\) is the mass of the object in kilograms, \( g\) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2\) on Earth), and \( h\) is the height of the object above a reference point.
  • Height is crucial in determining gravitational potential energy. The higher the object, the greater the energy.
  • This energy depends only on the vertical position, not the path taken to reach there.
In the context of our loop-the-loop, the block starts with maximum potential energy at point P because it's at the highest point. As it descends, this energy decreases and gets converted to kinetic energy.
Work Done by Gravity
Work done by gravity is the energy transferred when an object moves in relation to a gravitational force. This work can influence how energy changes from potential to kinetic form and vice versa.
The work done by gravity (W) can be expressed by the change in gravitational potential energy as the object moves: \[ W = mgh \] When an object moves higher against gravity, it loses kinetic energy, akin to gaining potential energy, as the gravitational force does negative work. Conversely, as the block travels down, gravity does positive work, transferring potential energy into kinetic energy.
  • From point P to point Q at the loop's base, the work done by gravity equals the total loss in potential energy from height \( h = 5R \) to \( 0 \).
  • Positive work signifies energy is added to the system, enhancing speed.
This understanding helps in calculating the energy changes happening as the block moves along the loop-the-loop.
Loop-the-Loop Physics
Consider a rollercoaster loop. Loop-the-loop physics explores the dynamics of objects in circular paths. Here, it's about how gravitational and kinetic energies interact during motion through the loop.
For successful looping without falling, the block needs a specific amount of kinetic energy when reaching the top. This ensures it has sufficient speed to remain in circular motion.
  • At the top of the loop, despite maximum height, some gravitational potential energy is converted back to kinetic energy to maintain speed.
  • The circular path involves centripetal force that keeps the block on track. This force is more critical at the loop’s top to overcome gravity and maintain motion.
Understanding these mechanics ensures safe and thrilling experiences in amusement park rides, illustrating energy transformations compellingly.
Conservation of Mechanical Energy
In physics, one important principle is the conservation of mechanical energy. It states that in the absence of external forces like friction, the total mechanical energy of a system remains constant.
Mechanical energy is the sum of potential and kinetic energies. In our loop-the-loop scenario, as the block slides down, potential energy is converted to kinetic energy, but their total stays unchanged.
  • This principle simplifies calculations by allowing direct energy exchanges without tracking every step.
  • It underlines orthogonal dynamics, where potential energies transform directly into kinetic forms while looping.
As the block moves under gravity, its energy exchanges between forms, demonstrating how energy persists, merely switching from type to type within the physical constraints of the path.

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Most popular questions from this chapter

A spring with \(k=170 \mathrm{~N} / \mathrm{m}\) is at the top of a frictionless incline of angle \(\theta=37.0^{\circ}\). The lower end of the incline is distance \(D=1.00 \mathrm{~m}\) from the end of the spring, which is at its relaxed length. A \(2.00 \mathrm{~kg}\) canister is pushed against the spring until the spring is compressed \(0.200 \mathrm{~m}\) and released from rest. (a) What is the speed of the canister at the instant the spring returns to its relaxed length (which is when the canister loses contact with the spring)? (b) What is the speed of the canister when it reaches the lower end of the incline?

A metal tool is sharpened by being held against the rim of a wheel on a grinding machine by a force of \(180 \mathrm{~N}\). The frictional forces between the rim and the tool grind off small pieces of the tool. The wheel has a radius of \(20.0 \mathrm{~cm}\) and rotates at \(2.50\) rev/s. The coefficient of kinetic friction between the wheel and the tool is \(0.320 .\) At what rate is energy being transferred from the motor driving the wheel to the thermal energy of the wheel and tool and to the kinetic energy of the material thrown from the tool?

A conservative force \(\vec{F}=(6.0 x-12) \hat{\mathrm{i}} \mathrm{N}\) where \(x\) is in meters, acts on a particle moving along an \(x\) axis. The potential energy \(U\) associated with this force is assigned a value of \(27 \mathrm{~J}\) at \(x=0\). (a) Write an expression for \(U\) as a function of \(x\), with \(U\) in joules and \(x\) in meters. (b) What is the maximum positive potential energy? At what (c) negative value and (d) positive value of \(x\) is the potential energy equal to zero?

A block of mass \(m=3.20 \mathrm{~kg}\) slides from rest a distance \(d\) down a frictionless incline at angle \(\theta=30.0^{\circ}\) where it runs into a spring of spring constant \(431 \mathrm{~N} / \mathrm{m}\). When the block momentarily stops, it has compressed the spring by \(21.0 \mathrm{~cm} .\) What are (a) distance \(d\) and (b) the distance between the point of the first block-spring contact and the point where the block's speed is greatest?

A \(25 \mathrm{~kg}\) bear slides, from rest, \(12 \mathrm{~m}\) down a lodgepole pine tree, moving with a speed of \(5.6 \mathrm{~m} / \mathrm{s}\) just before hitting the ground. (a) What change occurs in the gravitational potential energy of the bear-Earth system during the slide? (b) What is the kinetic energy of the bear just before hitting the ground? (c) What is the average frictional force that acts on the sliding bear?

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