How Much Potential Energy Is Lost to Friction on an Incline?

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A 0.40 kg mass slides down an incline, reaching a maximum velocity of 1.00 m/s, resulting in a kinetic energy of 0.2 Joules. To determine the percentage of potential energy lost to friction, the potential energy at the top of the incline must be calculated, which is approximately 0.2744 Joules based on a height of 7 cm. The difference between the potential energy and the kinetic energy gives the energy lost to friction. This loss can then be expressed as a percentage of the initial potential energy. Understanding these calculations is essential for solving the problem effectively.
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Homework Statement



A mass 0.40 kg is released and slides down the incline. The maximum velocity ( taken the instant before the mass reaches the bottom of the incline ) is 1.00 m/s. What is the kinetic energy at that time? What percentage of the potential energy of the system is not converted into kinetic energy due to friction? (converted into heat instead)

I don't quite understand how to find out the percent of potential energy lost..thanks for any help!

Homework Equations



I solved for the kinetic energy correctly by using

W=KE=1/2mv^2


The Attempt at a Solution



=1/2(.4)(1)^2
=.2 Joules

i'm not sure about how to solve for the percent of potential energy lost..thanks
 
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You're going to need to know the height to figure the PE.
 
The height is 7 cm (.007 m)
 
Potential Energy!

Homework Statement



A mass 0.40 kg is released and slides down the incline. The maximum velocity ( taken the instant before the mass reaches the bottom of the incline ) is 1.00 m/s. What is the kinetic energy at that time? What percentage of the potential energy of the system is not converted into kinetic energy due to friction? (converted into heat instead)

The height of the incline is 7 cm..and the hypotenuse is 40 cm

I don't quite understand how to find out the percent of potential energy lost..thanks for any help!

Homework Equations



I solved for the kinetic energy correctly by using

W=KE=1/2mv^2


The Attempt at a Solution



=1/2(.4)(1)^2
=.2 Joules

i'm not sure about how to solve for the percent of potential energy lost..thanks!
 
OK. So what is the potential energy available to go to kinetic?

How much does it need to go 1 m/s? Well you found that.

So how much potential isn't accounted for in going to kinetic?
 


What is the potential energy at the top of the inclined plane?
 


Is it .2744 Joules?
 

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