91Ó°ÊÓ

Ice flake is released from the edge of a hemispherical bowl whose radius \(r\) is \(22.0\) \(\mathrm{cm}\). The flake-bowl contact is frictionless. (a) How much work is done on the flake by the gravitational force during the flake's descent to the bottom of the bowl? (b) What is the change in the potential energy of the flake-Earth system during that descent? (c) If that potential energy is taken to be zero at the bottom of the bowl, what is its value when the flake is released? (d) If, instead, the potential energy is taken to be zero at the release point, what is its value when the flake reaches the bottom of the bowl? (e) If the mass of the flake were doubled, would the magnitudes of the answers to (a) through (d) increase, decrease, or remain the same?

Step by step solution

01

Understanding the Problem

We are analyzing an ice flake sliding down a frictionless hemispherical bowl. The primary forces at play are gravitational, and we need to calculate work done and potential energy changes based on various reference points for potential energy.

02

Calculate Work Done by Gravitational Force (Part a)

The work done on the flake by the gravitational force is equal to the change in gravitational potential energy, which is calculated as \( \Delta U = mgh \), where \( h \) is the change in vertical height. Since the flake moves from the rim to the bottom, \( h \) is equal to the radius \( r = 22.0 \) cm. Thus, the work done is \( W = mgh = mg(22 \text{ cm}) = mg(0.22 \text{ m}) \).

03

Calculate Change in Potential Energy (Part b)

The change in the potential energy \( \Delta U \) of the flake-Earth system is the same as the work done when the flake descends to the bottom, given by \( \Delta U = mgh = mg(0.22 \text{ m}) \).

04

Potential Energy at Release Point (Part c)

If the potential energy at the bottom is taken to be zero, then the potential energy at the release point is \( U = mgh = mg(0.22 \text{ m}) \) since it's at the height of the radius.

05

Potential Energy at the Bottom (Different Reference, Part d)

If the potential energy at the release point is zero, then at the bottom, the potential energy would be \( U = -mgh = -mg(0.22 \text{ m}) \), implying a decrease by \( mg(0.22 \text{ m}) \).

06

Effect of Doubling Mass (Part e)

Doubling the mass of the flake would increase the magnitude of work done and changes in potential energy, since both are directly proportional to mass \( m \). Hence, the answers to (a) through (d) would increase in magnitude.

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Key Concepts

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

Work Done

When we talk about the work done on an object, we refer to the amount of energy transferred to that object by a force causing it to move. For gravitational force in this context, the formula for work done is \( W = mgh \), where:

• \( m \) is the mass of the object,
• \( g \) is the acceleration due to gravity (approximately \(9.8 \, \text{m/s}^2\) on Earth),
• \( h \) is the change in height.

In our exercise, an ice flake slides down from the top of a frictionless bowl to the bottom, resulting in a change in height equal to the bowl's radius, \( r = 22.0 \) cm or \( 0.22 \) meters. Since the bowl is frictionless, only gravity does work on the flake. Hence, the work done equates to the change in gravitational potential energy as it moves from the rim to the bottom.

Frictionless Motion

Frictionless motion is when an object moves without any resistance from the surface it is on. This ideal condition allows for the simplification of many physics problems, as it means that all the energy in the system comes solely from gravitational forces. In the case of the ice flake, the absence of friction means:

• The potential energy of the flake is fully converted into kinetic energy as it slides down.
• There are no energy losses to heat or other forms of energy that usually result from friction.

Thus, the only factor affecting the flake's motion is the gravitational force, leading to straightforward calculations for work done and energy changes.

Reference Points in Potential Energy

Potential energy can have different values depending on where you set your reference point. In this exercise, different scenarios illustrate how the reference point affects the potential energy values:

• If potential energy is zero at the bottom of the bowl, then the potential energy at the top (release point) is \( U = mgh \).
• Alternatively, if zero is set at the release point, the potential energy at the bottom becomes \( U = -mgh \), indicating a potential energy loss.

Choosing a reference point helps interpret energy changes in physical terms, as energy must be conserved. This exercise clearly demonstrates the importance of defining where zero potential energy is for ease of solving problems.

Effect of Mass on Energy Change

Mass plays a key role in determining the magnitudes of gravitational potential energy and work done. Both quantities are directly proportional to mass, as expressed in the equation \( W = mgh \).

• If the mass doubles, both potential energy and work done amounts will double as well.
• For example, changing from a single flake to a heavier double-massed flake doesn't alter the height or gravity, but the force due to the flake's weight and corresponding energy changes increase.

Understanding this concept helps in predicting energy dynamics in physical systems, especially in situations involving differences in mass.