Finding the Formula for Speed and Distance on an Icy Hill

[artireiter]

Homework Statement

Maxine (mass m) slides down an icy slope then along a flat surface before coming to a stop.
The slope has angle θ and height h.

a. Derive a formula for speed at the bottom of the hill in terms of the angle of the hill θ, the height from which she starts h, and the coefficient of friction μ.
b. Now assuming she begins the flat part of the trip with this velocity, Derive a formula for the distance traveled at the bottom of the hill before coming to a stop.
c. If her friend (mass 2m) gets in the sled and they repeat the whole trip how much does the answer to b change?

Homework Equations

I believe I am supposed to use
v_f² = v_i² + 2ad

The Attempt at a Solution

tan θ = h/x
x= h/ tan θ
v_i (the top of the slope) = 0

a. v_fy = √(|2(-g)h|)
That was what I was doing but it seemed odd because I was finding the square root of a negative. But I am looking for speed so would I just take the absolute value? Even so I am unsure how I would take this any further if it is correct.

 

[jbriggs444]

That was what I was doing but it seemed odd because I was finding the square root of a negative

How did you arrive at the formula $v_{fy} = \sqrt{2(-g)h}$ ? One has to expect a sign error somewhere. In particular, what value are you using for "g"? 9.8m/sec^2 or -9.8m/sec^2?

 

[artireiter]

How did you arrive at the formula $v_{fy} = \sqrt{2(-g)h}$ ? One has to expect a sign error somewhere. In particular, what value are you using for "g"? 9.8m/sec^2 or -9.8m/sec^2?

I'm using -9.8m/s² for g

 

[jbriggs444]

I'm using -9.8m/s² for g

Then you are taking the square root of a positive. -g is +9.8m/s².

 

[artireiter]

Then you are taking the square root of a positive. -g is +9.8m/s².

Sorry I was unclear. I put the negative in front of the g to represent -9.8
Am I just using the wrong formula maybe?

 

[jbriggs444]

Sorry I was unclear. I put the negative in front of the g to represent -9.8

Then we are back to the previous question. How did you derive the formula? The derivation certainly has a sign error. [My bet: The h should be negative as well]

 

[artireiter]

Then we are back to the previous question. How did you derive the formula? The derivation certainly has a sign error.

I was solving vf2 = vi2 + 2ad for the y components. But I realize now I forgot to account for the fact that this is on a slope so the y acceleration would not be exactly -g.

So after solving for the y component of g I got g_y = -gsin(90-θ)
Still that doesn't really seems to help me in finding the formula. Am I approaching this problem entirely wrong?

 

[jbriggs444]

I was solving vf2 = vi2 + 2ad for the y components. But I realize now I forgot to account for the fact that this is on a slope so the y acceleration would not be exactly -g.

In the formula $v_f^2 = v_i^2 + 2ad$, what is a and what is d? What makes you consider that a is negative (-9.8m/s²) while d is positive?

So after solving for the y component of g I got g_y = -gsin(90-θ)
Still that doesn't really seems to help me in finding the formula.

That approach is one possible way of tackling the problem. It will work. You have correctly calculated the acceleration parallel to the slope. But you still have to figure out the sign and magnitude of the distance traversed along the slope. What are those?

Edit: To elaborate a bit on the square root of a negative concern, let us modify the original problem. Suppose that instead of starting at rest on the top of an incline, the object had started at rest at the bottom. We are then asked "what velocity will it have when it reaches the top"?

The correct answer is that it will not reach the top. If you try to do the algebra to solve for the velocity with which it reaches the top, you will find yourself taking the square root of a negative number even though you have performed every step in the algebra correctly. That is a clue that you can't get there from here.

 

[artireiter]

In the formula $v_f^2 = v_i^2 + 2ad$, what is a and what is d? What makes you consider that a is negative (-9.8m/s²) while d is positive?

That approach is one possible way of tackling the problem. It will work. You have correctly calculated the acceleration parallel to the slope. But you still have to figure out the sign and magnitude of the distance traversed along the slope. What are those?

I know the height of the slope is h and I found the length of the slope to be x=h/tanθ
The hypotenuse would be hyp= √(h²+x²)
or hyp= √(h²+(h/tanθ)²)
so assuming the hypotenuse is aligned with my x-axis that would be my displacement.
I think then the acceleration for that would be g_x = gcos(90-θ)

So using
v²_f=v²_i+2ad
v_f²=0²+2(gcos(90-θ))(√(h²+(h/tanθ)²))
v_f=√(2(gcos(90-θ))(√(h²+(h/tanθ)²)))

Sorry it's a mess and I'm not sure it's right. That also doesn't account for the friction. How would that be accounted for?

 

[jbriggs444]

I know the height of the slope is h and I found the length of the slope to be x=h/tanθ

How sure are you about that?

Also, it is not the length of the slope that you are trying to find. Length is always positive. You are concerned with the displacement along the slope. Displacement is a vector. It can be positive or negative.

 

[artireiter]

How sure are you about that?

Also, it is not the length of the slope that you are trying to find. Length is always positive. You are concerned with the displacement along the slope. Displacement is a vector. It can be positive or negative.

Sorry the x I found wasn't the hypotenuse. It was one side of the triangle I phrased that wrong. The displacement along the hypotenuse I found to be
d= √(h²+x²)
or d= √(h²+(h/tanθ)²)

I know how to find these values. But I'm confused on putting them into the equation to find the actual velocity. Especially since I am given the coefficient of friction to include as well. How would that effect my velocity?

 

[jbriggs444]

Sorry the x I found wasn't the hypotenuse. It was one side of the triangle I phrased that wrong. The displacement along the hypotenuse I found to be
d= √(h²+x²)
or d= √(h²+(h/tanθ)²)

There is a simpler expression. What is the ratio of the opposite side over the hypotenuse?

I know how to find these values. But I'm confused on putting them into the equation to find the actual velocity. Especially since I am given the coefficient of friction to include as well. How would that effect my velocity?

The coefficient of friction of the icy slope is zero. Actually, I take that back. On re-reading the question, it seems that the coefficient of friction on icy slope and ground are both taken to be non-zero.