Toboggan down hill, FIND ANGLE?

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Homework Help Overview

The problem involves a girl tobogganing down a hill with a specified height, mass, and final speed, while considering the effects of friction. The objective is to find the angle of the slope based on energy principles and forces acting on the system.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between kinetic energy, potential energy, and work done by friction. There are attempts to equate energies and derive the angle using trigonometric relationships, with some questioning the use of specific equations and terms.

Discussion Status

The discussion is active, with participants exploring different interpretations of the problem and clarifying the relationships between the variables involved. Some guidance has been offered regarding the use of trigonometric ratios and the calculation of work done, but no consensus has been reached on the correct approach or calculations.

Contextual Notes

Participants are navigating through the constraints of the problem, including the need to find the hypotenuse in relation to the height and angle, and the implications of friction on energy conservation. There is an acknowledgment of potential misapplications of trigonometric functions in the context of the problem.

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Homework Statement



A girl starting from rest, toboggans down a hill of height 15m. The mass of the girl and toboggan is 42kg. The speed of the girl and toboggan at bottom of hill is 2.9m/s. If force of friction is 112 N, what is the slope of the hill? Ignore air resistance.

Homework Equations



KE=PE, 15sin(theta), F=mg

The Attempt at a Solution



I started by equating initial and final energies by saying KE final is equal to PE initial plus the force of friction. (.5mv)squared = mgh - (mu)Fn
I ended up with 0.1429=sin(theta) which equals 8.2 degrees. The answer was 16 degrees, where did I go wrong?
 
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What is 15sin([tex]\vartheta[/tex]) meant to be?
 
I am trying to figure out the distance down the slope that she travels so I can use that along with the height (15m) to find the angle.
 
15sin(theta) will not give you the distance traveled down the slope.
 
Well I suppose I am stuck then.
 
You know what to do, you just have your trig relations mixed up a bit. If sintheta = height/hypotenuse, then 15sintheta = height x height/hypotenuse...
 
I should know this, but that still leaves me with unknowns for theta and hypotenuse.
I am drawing a blank on how to find the hypotenuse with what is given.
Thanks for the replies by the way.
 
Sintheta = height/hypotenuse where hypotenuse = the distance traveled down the slope. Since we know the height, we can state the hypotenuse in terms of the height and theta, that is hypotenuse = height/sintheta. This way you can use 15m/sintheta in the expression for the work done by friction, and then solve for theta, which was your idea to begin with. You just misused the sin ratio!
 
W=Ff*d


5997.4/112N=d

d=53.6m
 
  • #10
Beautiful. Thanks much!
 
  • #11
So how did you determine W was 5997.4?
 
  • #12
I have to disagree with Squeezebox, the work done is not as quoted. To calculate the work done, you can apply conservation of energy.
 

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