Inclined Plane Problem: Understanding the Acceleration of a Skier on a Slope

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Homework Help Overview

The problem involves a skier transitioning from horizontal motion to descending an inclined plane at a 10-degree angle. The skier's initial and final speeds are provided, and the discussion focuses on understanding the acceleration down the slope, particularly the reasoning behind the expression for acceleration as a function of gravitational acceleration and the incline angle.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between gravitational acceleration and the incline angle, questioning why acceleration is expressed as gsin(θ). There is confusion regarding the geometric interpretation of the forces acting on the skier and the components of gravity along the slope.

Discussion Status

The discussion is active, with participants providing insights into the geometric relationships involved. Some participants have raised concerns about the physical meaning of the calculated acceleration, noting that it should not exceed gravitational acceleration. There is an ongoing exploration of how to correctly interpret the triangle formed by the incline and the forces acting on the skier.

Contextual Notes

Participants are grappling with the definitions and assumptions related to the angle of the incline and the components of gravitational force. There is a mention of a solutions page that provides a specific value for acceleration, which has led to further questioning and attempts to visualize the problem correctly.

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Homework Statement


A skier is gliding along at 3m/s on horizontal frictionless snow. He suddenly starts down a 10 degree incline. His speed at the bottom is 15 m/s.
a) what is the length of the incline?
b) how long does it take for him to reach the bottom?
we know:
Vo=3m/s
Vf=15m/s
t=?
x=?
a=? [gsin10 -- but why?]

Homework Equations


V^2=Vo^2 +2ax


The Attempt at a Solution


okay...i don't understand why a=gsin(10) [g=9.8]
because from the triangle i draw ..g is not the hypotenuse? I am confused to see WHY a=9.8sin(10).

http://img125.imageshack.us/img125/6060/53872528yt8.jpg
 
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I don't think it is!

You're correct that g is not the hypotenuse. g acts vertically downwards, and so if you want the component of the acceleration due to gravity that acts down the slope you need to resolve this into components parallel and perpendicular to the slope. To do this, g is the vertical side of the triangle you've drawn, and you want to find the hypotneuse, x say. Hence, sin(10)=g/x and so x=g/sin(10)
 
thanks for replying so fast cristo!
okay so if i do that:
sin(10)=g/a and so a=g/sin(10)

a = 9.8/(sin10)
a=56.4m/s^2?
that just seems way out of wack...its incorrect...according to my solutions page it says a=gsin10. I just cannot seem to figure how that is visually according to what i see on the triangle. Any ideas?
 
It is incorrect, since for it to be physically meaningful [itex]a\leq g[/itex]. I think the question is defining the angle from the vertical (10 degrees from the horizontal is a pretty small incline for a ski slope).
 
yes...a bunny slope i suppose?
but even so
why is:
a=gsin(theta) --> from the vertical: a=gsin(80) frm the horizontal a=gsin(10)?
as u said it makes sense for a< or equal to g..but how is it that we get that as the trigonometric function we use...i just don't see the triangle...hopefully that makes sense...if someone could draw it out for me how we get a=9.8sin(10)=1.7m/s^2
 

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