How Do You Calculate Mechanical Energy Lost Due to Friction on a Slide?

AI Thread Summary
To calculate the mechanical energy lost due to friction on a slide, the correct approach involves using the conservation of energy principle, where the work done by friction equals the change in kinetic and potential energy. The initial potential energy (U1) at height h1 and the kinetic energy (K2) at height h2 must be properly accounted for. The error in the initial calculation stemmed from incorrectly applying the energy equations, specifically in the order of subtraction for kinetic and potential energy. The final result should reflect the positive value of energy lost due to friction, correcting the sign of the work done by friction. Properly applying these principles will yield the accurate energy loss due to friction.
bbreezy
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Homework Statement


A 27.0 kg child slides down a long slide in a playground. She starts from rest at a height h1 of 17.00 m. When she is partway down the slide, at a height h2 of 7.00 m, she is moving at a speed of 7.50 m/s. Calculate the mechanical energy lost due to friction (as heat, etc.).


Homework Equations



I used mg(h1)-mg(h2)+.5mv^2

The Attempt at a Solution



I inputted 27(9.81)17-27(9.81)(7) and got 2648.7 and added it to .5(27)(7.5^2) which was 759.375 and came up with the answer 3408.075 J.. this seems to be wrong.. What am I doing wrong here?
 
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bbreezy said:

Homework Statement


A 27.0 kg child slides down a long slide in a playground. She starts from rest at a height h1 of 17.00 m. When she is partway down the slide, at a height h2 of 7.00 m, she is moving at a speed of 7.50 m/s. Calculate the mechanical energy lost due to friction (as heat, etc.).

Homework Equations



I used mg(h1)-mg(h2)+.5mv^2

The Attempt at a Solution



I inputted 27(9.81)17-27(9.81)(7) and got 2648.7 and added it to .5(27)(7.5^2) which was 759.375 and came up with the answer 3408.075 J.. this seems to be wrong.. What am I doing wrong here?
Let K be kinetic energy and U be potential energy. In a perfect world, we have

K_1 + U_1 = K_2 + U_2

But in an imperfect world, there will be a change in the sum of kinetic and potential energy, and a simple way to account for that change is to blame it on friction. Thus,

W_{FO} = K_2 - K_1 + U_2 - U_1 = \Delta K + \Delta U

with W_{FO} being the work of friction done to the object. It will be negative, but your answer should be positive due to the phrasing, so change its sign.

You seem to be doing K_2 - K_1 and U_1 - U_2, causing your error.
 
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I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
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