91Ó°ÊÓ

A small block of mass \(m\) can slide along the frictionless loop-the-loop. The block is released from rest at point \(P\), at height \(h=5 R\) 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 \((\mathrm{b})\) 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 \(P,(\mathrm{~d})\) at point \(Q\), and \((\mathrm{e})\) at the top of the loop? (f) If, instead of 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?

Step by step solution

01

- Identify Given Data

Given the mass of the block is \(m\), the height of point \(P\) is \(h = 5R\), and the loop-the-loop radius is \(R\). Since the track is frictionless, only gravitational force is doing work on the block.

02

- Work Done by Gravitational Force (Part a)

The work done by the gravitational force as the block travels from point \(P\) to point \(Q\) is equal to the change in gravitational potential energy. At point \(P\), the height \(h_P = 5R\). At point \(Q\), the height \(h_Q = R\):\[ W_{PQ} = mgh_P - mgh_Q = mg(5R - R) = 4mgR \].

03

- Work Done by Gravitational Force (Part b)

The work done by gravitational force as the block travels from point \(P\) to the top of the loop \(h_T\) is 2R above the bottom of the loop: \[ W_{PT} = mgh_P - mgh_T = mg(5R - 2R) = 3mgR \].

04

- Potential Energy at Point P (Part c)

The gravitational potential energy at point \(P\) is given by the height above the bottom of the loop: \[ U_P = mgh = mg(5R) = 5mgR \].

05

- Potential Energy at Point Q (Part d)

The gravitational potential energy at point \(Q\), which is at height \(R\): \[ U_Q = mgh_Q = mgR \].

06

- Potential Energy at the Top of the Loop (Part e)

The gravitational potential energy at the top of the loop, height \(2R\): \[ U_T = mgh_T = mg(2R) = 2mgR \].

07

- Effect of Initial Speed (Part f)

If the block is given some initial speed downward along the track, the work done by gravity (parts a and b) and the potential energy (parts c, d, and e) will remain unaffected because work done by gravity and gravitational potential energy depend only on the change in height and not on the initial speed.

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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 the energy an object possesses due to its position in a gravitational field. It depends on the object's height above a reference point and its mass. In the loop-the-loop problem, the gravitational potential energy of the block at any given point is given by the formula: \( U = mgh \). Here, \( m \) is the mass of the block, \( g \) is the acceleration due to gravity, and \( h \) is the height above the reference point, which is the bottom of the loop.
At point \( P \), the height is \( 5R \), leading to \( U_P = 5mgR \). Similarly, at points \( Q \) and the top of the loop, the gravitational potential energy can be found by substituting their respective heights into the formula.

work done by gravity

The work done by gravity when an object moves from one point to another is equal to the change in its gravitational potential energy. For a block sliding down a frictionless track, this work is simply the difference in potential energy at the starting and ending points.
From point \( P \) to \( Q \), the work done by gravity is the difference in potential energy between these points: \( W_{PQ} = U_P - U_Q = 4mgR \).
Similarly, from point \( P \) to the top of the loop, the work done by gravity can be calculated as \( W_{PT} = U_P - U_T = 3mgR \). These calculations show how energy is conserved and converted between potential and kinetic forms as the block moves.

conservation of energy

The principle of conservation of energy states that the total energy of an isolated system remains constant. In the case of the frictionless loop-the-loop, mechanical energy is conserved. This means that the sum of the block's kinetic and potential energy at any point along the track is equal to its total energy at the starting point.
As the block is released from rest, its initial energy is entirely gravitational potential energy at point \( P \), which is \( 5mgR \). As it moves down, this energy is converted into kinetic energy. At point \( Q \) and the top of the loop, the sum of kinetic and potential energy equals the initial potential energy, showing the conservation of energy in the system.

kinematics

Kinematics deals with the motion of objects without considering the forces causing the motion. In the context of the loop-the-loop problem, kinematics helps describe the block's motion in terms of its position, velocity, and acceleration over time.
For example, as the block moves from point \( P \) to the bottom of the loop, it accelerates due to gravity, increasing its velocity. This motion can be analyzed using kinematic equations: \( v = u + at \) and \( s = ut + \frac{1}{2}at^2 \), where \( u \) is the initial velocity, \( v \) is the final velocity, \( a \) is the acceleration, and \( s \) is the distance traveled. In this problem, however, the primary focus is on energy considerations rather than detailed kinematic analysis.

frictionless motion

Frictionless motion means that no energy is lost to friction, and the only force doing work on the block is gravity. This simplification allows us to analyze the problem using conservation of energy principles without needing to account for energy losses.
In a frictionless loop-the-loop, the block’s mechanical energy (the sum of potential and kinetic energy) remains constant throughout its motion. This idealization helps in understanding fundamental physics concepts like gravitational potential energy and work done by gravity, making it easier to apply these concepts to more complex real-world scenarios.