Potential Energy/ The Isolated system

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Homework Help Overview

The problem involves a ball of mass 8 kg dropped from a height of 13.9 m, with an initial downward speed of 4.9 m/s. The question seeks to determine the final speed of the ball when it reaches a height 0.7 m below the release point, while ignoring air resistance.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants express confusion regarding the wording of the question, particularly what is meant by "final speed when it hits the table 0.7 m below the release point." Some clarify that it refers to the speed just before impact with the table, while others question if the initial height of 13.9 m is relevant to the final speed calculation.

Discussion Status

The discussion is ongoing, with participants exploring interpretations of the question and clarifying the specific height from which the speed is to be calculated. There is no explicit consensus yet, but some guidance has been provided regarding the focus on the speed after falling 0.7 m.

Contextual Notes

Participants are navigating the implications of the initial conditions and the specific height mentioned in the problem. The potential energy and kinetic energy relationships are being referenced, but no definitive resolution has been reached regarding the calculations involved.

aaronb
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Homework Statement



A ball of mass m = 8 kg is dropped from rest at a height h = 13.9 m above the ground, as in Figure 7.4. Ignore air resistance.

If the ball is being released with a downward speed 4.9 m/s initially, what will be its final speed when it hits the table 0.7 m below the release point?

Homework Equations



Kf+Ugf=Ki+Ugi which is 1/2mvf2+mgy=1/2mvi2 +mgh

The Attempt at a Solution


I am confused in the wording of the question. What do they mean by "final speed when it hits the table 0.7m below the release point?"
 
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aaronb said:

Homework Statement



A ball of mass m = 8 kg is dropped from rest at a height h = 13.9 m above the ground, as in Figure 7.4. Ignore air resistance.

If the ball is being released with a downward speed 4.9 m/s initially, what will be its final speed when it hits the table 0.7 m below the release point?

Homework Equations



Kf+Ugf=Ki+Ugi which is 1/2mvf2+mgy=1/2mvi2 +mgh

The Attempt at a Solution


I am confused in the wording of the question. What do they mean by "final speed when it hits the table 0.7m below the release point?"

They mean the full speed it was going right when it hits the table. Of course the table stopping it will quickly accelerate it down to a velocity of 0, but we want to know the velocity before the table actually stops it.

The important information is .7m. How much kinetic energy will it have when it's fallen .7 meters from where it was dropped?
 
So it's just asking for the speed of the ball after it has fallen 0.7m?
 
aaronb said:
So it's just asking for the speed of the ball after it has fallen 0.7m?

Apparently so.

I will presume that the first question was the speed or kinetic energy when it hit 13.9m below?
 
Nope. The first question asked for the speed at an arbitrary height before the ball hit the ground. In that case it was 5.2m
 

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