Conservation Of Mechanical Energy PROBLEM

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Homework Help Overview

The problem involves a novice skier sliding down a frictionless incline with a vertical height of 185 m and an angle of 35.00 degrees. The objective is to determine the skier's speed at the bottom of the incline.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of conservation of mechanical energy, questioning the correctness of the formula used. There are attempts to derive the final speed using energy principles, and some participants suggest using trigonometry to find the slope length and acceleration.

Discussion Status

The discussion is ongoing with various interpretations being explored. Some participants have provided guidance on the energy equation, while others are questioning the calculations and seeking clarification on the approach.

Contextual Notes

There is a request for a complete solution, indicating that participants are navigating the boundaries of providing help without directly solving the problem.

CoffinSupply
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Homework Statement



A novice skier, starting from rest, slides down a frictionless 35.00 incline whose vertical height is 185 m.How fast is she going when she reaches the bottom?


Homework Equations



Need the rest of the solution.

The Attempt at a Solution



PICTURE SCAN: http://img528.imageshack.us/img528/8666/homeworkgi6.jpg

Homework Statement





I Need to check if this is correct formula and need to do the rest by someone who knows. Thanks.
 
Last edited by a moderator:
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Welcome to PF!

Hi CoffinSupply! Welcome to PF! :smile:
CoffinSupply said:
A novice skier, starting from rest, slides down a frictionless 35.00 incline whose vertical height is 185 m.How fast is she going when she reaches the bottom?

Yes, that's right … mg(185) = (1/2)mvf2 …

so vf = … ? :smile:
 


tiny-tim said:
Hi CoffinSupply! Welcome to PF! :smile:


Yes, that's right … mg(185) = (1/2)mvf2 …

so vf = … ? :smile:



Thanks, PF a great place!

vf =49.5 m/s ?
 
CoffinSupply said:
Thanks, PF a great place!

vf =49.5 m/s ?

I don't think so …

how did you get that?
 
crossed out like terms. But how would you do it?
 
Can't you get the length of the slope using trigs and then then acceleration is g*sin35?
 
Can I get a complete solution to the problem done , if possible?
 

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