Calculate energy lost to a rough surface

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Homework Help Overview

The discussion revolves around calculating the energy lost by a block sliding across a rough surface, specifically focusing on the effects of kinetic friction. The problem involves a block of mass 2kg moving at a velocity of 2.68 m/s over a distance of 3m with a coefficient of kinetic friction of 0.4.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the application of kinetic energy and friction equations, noting the calculated values of energy. Questions arise regarding the interpretation of the calculated energy lost and its relation to the initial kinetic energy of the block.

Discussion Status

Participants are exploring the implications of the energy calculations, questioning how the energy lost can exceed the initial kinetic energy and discussing the conditions under which the block would stop. There is a mix of interpretations regarding the meaning of the energy values derived.

Contextual Notes

There is uncertainty about the definitions and implications of the energy calculations, particularly regarding the relationship between energy lost to friction and the block's initial kinetic energy. Participants are also grappling with the physical meaning of the results in the context of the problem.

Toliver96
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Homework Statement


A block of mass 2kg, moving at velocity 2.68 m/s, slides across a rough 3m surface with a coefficient of kinetic friction of .4. Calculate the energy lost to the 3m rough surface.
M=2kg
uK=.4
V=2.68 m/s
D=3m

Homework Equations


1/2mv^2
uK*Fn*D

The Attempt at a Solution


I used the kinetic energy equation of 1/2mv^2 to get 7.18J. Then I did uK*Fn*D to get 24J, where Fn is normal force and Gravity = 10m/s^2. I have no idea where to go from here.
 
Last edited:
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Toliver96 said:

Homework Statement


A block of mass 2kg, moving at velocity 2.68 m/s, slides across a rough 3m surface with a coefficient of kinetic friction of .4. Calculate the energy lost to the 3m rough surface.
M=2kg
uK=.4
V=2.68 m/s
D=3m


Homework Equations


1/2mv^2
uK*Fn*D


The Attempt at a Solution


I used the kinetic energy equation of 1/2mv^2 to get 7.18J. Then I did uK*Fn*D to get 24J, where Fn is normal force and Gravity = 10m/s^2. I have no idea where to go from here.
What does that 24 J represent ?
 
It is supposed to be Jules, but I'm not sure if that's correct.
 
Toliver96 said:
It is supposed to be Jules, but I'm not sure if that's correct.

The units are Joules, but what does the 24 Joules represent?
 
The energy lost to the rough surface, but how can it lose more energy than what it has?
 
Toliver96 said:
The energy lost to the rough surface, but how can it lose more energy than what it has?
What happens to the block when the energy lost is equal to the initial Kinetic Energy ?
 
So the block will stop then, right?
 
Toliver96 said:
So the block will stop then, right?

Right.
 

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