Toboggan down hill, FIND ANGLE?

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In summary: F=mg and KE=PE. So W=Ff*d=mg*d+PE. Since W is a negative number, the work done is opposite of what is quoted.
  • #1
j_suder2
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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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  • #2
What is 15sin([tex]\vartheta[/tex]) meant to be?
 
  • #3
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.
 
  • #4
15sin(theta) will not give you the distance traveled down the slope.
 
  • #5
Well I suppose I am stuck then.
 
  • #6
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...
 
  • #7
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.
 
  • #8
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!
 
  • #9
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.
 

FAQ: Toboggan down hill, FIND ANGLE?

How do I calculate the angle for tobogganing down hill?

To calculate the angle for tobogganing down hill, you will need to know the height and length of the hill. Then, you can use the tangent function (tan) to find the angle. The formula is tan(angle) = height/length. Use a scientific calculator or online calculator to find the inverse tangent (arctan) of the ratio to get the angle in degrees.

What factors can affect the angle for tobogganing?

The angle for tobogganing can be affected by several factors, including the steepness of the hill, the type of toboggan used, the snow conditions, and the weight of the rider. A steeper hill will require a larger angle for a safe ride, while a heavier rider may require a smaller angle. The type of toboggan and the snow conditions can also impact the angle needed for a smooth and safe ride.

How does the angle affect the speed of the toboggan?

The angle of the hill can greatly affect the speed of the toboggan. A steeper angle will result in a faster ride, while a smaller angle will slow the toboggan down. This is because the angle determines the force of gravity pulling the toboggan down the hill. The steeper the angle, the greater the force and the faster the toboggan will go.

Is there an optimal angle for tobogganing down hill?

The optimal angle for tobogganing down hill will vary depending on the individual and the conditions. Generally, a safe and enjoyable angle for tobogganing is between 20 and 30 degrees. However, it is important to always assess the conditions and adjust the angle accordingly for a safe ride.

Can I use the same angle for tobogganing on different types of hills?

The angle for tobogganing may need to be adjusted for different types of hills. A steeper hill will require a larger angle, while a gentler slope may require a smaller angle. It is best to always assess the conditions and adjust the angle accordingly for a safe and enjoyable ride.

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