Help finding distance using work and velocity

Click For Summary

Homework Help Overview

The original poster is investigating the distance a ski jump should be placed from the top of a hill to ensure a skier does not exceed a certain speed. The problem involves concepts from mechanics, specifically the work-energy theorem, kinetic energy, potential energy, and frictional forces.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply the work-energy theorem and considers the transformation of potential energy to kinetic energy as the skier descends. They question how to incorporate the work done by friction into their calculations.

Discussion Status

Participants are engaging in a productive discussion, clarifying the roles of different forms of energy and the importance of sign conventions in the equations. Some guidance has been provided regarding the need to account for gravitational potential energy and the effects of friction.

Contextual Notes

The original poster is working within the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is an emphasis on understanding the energy transformations involved in the skier's motion.

Salacious Crumb
Messages
3
Reaction score
0

Homework Statement


I'm looking to the distance (h) a ski jump should be placed from the top of the hill so that a skier with a mass of 85kg will not exceed 30.0m/s. I am given an initial velocity of 2.0m/s and the total work of frictional forces to 4000J

Homework Equations


How can I use total work to determine distance with the variables I'm given? I'm not looking for the answer I just need to be pointed in the right direction here.

The Attempt at a Solution


I used the work-energy theorem W(total) = Kinetic (final) - Kinetic(initial)
Kf = 1/2mv^2 = 1/2 * 85kg * (30.0m/s)^2 = 38250J
Ki = 1/2mv^2 = 1/2 * 85kg * (2m/s)^2 = 170J
Wtotal = 38250 - 170 = 38080J
subtract the frictional energy of 4000J
34080J and this is where I'm stuck. Honestly I don't even know if I'm on the right track.
 
Physics news on Phys.org
Hi SC, Welcome to Physics Forums.

A question is not a relevant equation. Questions should go in the problem statement. You should list equations you believe may be relevant to the type of problem being solved in the relevant equations section.

Before tackling the details of a problem like this it's a good idea to list all the places where energy is coming from or going. You've recognized the initial KE, final KE, and the frictional force losses. What other form of energy is in play here? When a skier descends a mountain, why does he gain speed?
 
Sorry, I guess I skimmed over that too quickly. I read it as questions.
The skier accelerates due to the vector of the gravitational force working along the slope of the hill.
So the potential energy (U) of the skier is transformed to K as he moves down the slope.
Kf - Ki = Uf - Ui + Efric -> Kf - Ki = mg(Yf - Yi) + Efric Yf-Yi is h
Am I on the right track now?
 
Yes, that's the idea. As the skier descends he trades gravitational PE for KE, but also loses KE to friction.

Make sure that you get your signs right to reflect gains or losses. For example, if the skier starts at the top of the hill at height Yi and ends lower down at height Yf, then mg(Yf - Yi) is going to be negative. While it's true that he's losing gravitational PE, his KE should be increased by this amount. Just spend a bit of time making sure that the signs you give the terms reflect what's taking place.
 
Thank you for your help, I think I have it.
 
Salacious Crumb said:
Thank you for your help, I think I have it.
Great! Glad to help.
 

Similar threads

Replies
8
Views
3K
  • · Replies 8 ·
Replies
8
Views
1K
Replies
6
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 56 ·
2
Replies
56
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
11
Views
3K