Calculate energy lost to a rough surface

AI Thread Summary
A block with a mass of 2kg and an initial velocity of 2.68 m/s slides over a rough surface, and the energy lost due to friction is calculated using the coefficient of kinetic friction of 0.4. The kinetic energy of the block is determined to be 7.18 Joules, while the energy lost to friction is calculated as 24 Joules. This raises questions about the discrepancy, as the energy lost exceeds the initial kinetic energy. The discussion concludes that the block will stop when the energy lost equals the initial kinetic energy. Understanding these calculations is crucial for accurately analyzing energy loss in friction scenarios.
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.
 
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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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