# Skier Going Uphill - Intro Physics Problem

In summary, at the bottom of the hill the skier has kinetic energy and no potential energy. When he goes up the 30 degree incline, eventually his velocity goes to zero and his kinetic energy goes to zero. However, he will have gravitational potential energy. So what you can do is say that his kinetic energy before going up the 30 degree incline is equal to his gravitational potential energy when he stops on the incline. You should have all the numbers to do this and solve for the h, which is his height (altitude). From there, for part B, just consider where he is on the 30 degree right triangle. You know his height h (which becomes the y coordinate, and you know the degree of the incline

## Homework Statement

You've skied down a slope and are going 20m/s when you hit the base of a slope that is 30 degree incline.?

Assuming friction and drag are negligible, how much altitude (y) will you gain as you’re slowing down?

How far up the 30 degree slope (∆x along the incline) will you coast before you stop?

I know the following variables:
vi = 20 m/s
vf = 0 m/s
angle is 30°

I know the skier's energy is kinetic changing to potiential gravitational energy.

## Homework Equations

Ki → Ug

KE = 1/2*m*v^2
PE = m*g*h

For a)
It seems like I need to use 20tan30, but I'm not sure.

For b)
need ma = mgsin(theta) to solve for a = gsin(theta)

and vf^2 = vi^2+2*a*d → d = -vf^2 + vi^2 + 2a

## The Attempt at a Solution

For a) 20*tan(30°) = 17.32

For b) (-9.8)sin(30°) = -4.9

d = 0 + (20)^2 + 2(-4.9)
d = 40 + (-9.8)
d= 30.2

For part A just consider the conservative forces. At the base of the hill the guy has a kinetic energy and no potential energy (relative to the base of the hill that is). When he goes up the 30 degree incline, eventually his velocity goes to zero (he stops) and his kinetic energy goes to zero. But, he will have gravitational potential energy.

So what you can do is say that:
$$\frac{1}{2}mv^{2} = mgh$$
That is, his kinetic energy before going up the 30 degree incline, is equal to his gravitational potential energy when he stops on the incline. You should have all the numbers to do this and solve for the h, which is his height (altitude).

From there, for part B, just consider where he is at on the 30 degree right triangle. You know his height h (which becomes the y coordinate, and you know the degree of the incline, so using basic trigonometry you can find the hypoteneuse (the distance).

Ok, that makes sense. Makes part B easy to find h in part A.

So I want to rearrange $$\frac{1}{2}mv^{2} = mgh$$ to
$$h = \frac{1}{2} \frac {mg}{mv^{2}}$$

So mass will cancel out, leaving me with the known variables. If that is the correct rearranged form, I know enough trig to solve for part b. Thanks.

EDIT: I realized my rearranging is wrong.

$$h = \frac{1}{2} g-v^{2}$$

$$h = \frac{1}{2} (9.8)-(20)^{2}$$ = 195.1 m

Part B: 195.1sin(30°)= 97.55 m

Last edited:
Ok, that makes sense. Makes part B easy to find h in part A.

So I want to rearrange $$\frac{1}{2}mv^{2} = mgh$$ to
$$h = \frac{1}{2} \frac {mg}{mv^{2}}$$

So mass will cancel out, leaving me with the known variables. If that is the correct rearranged form, I know enough trig to solve for part b. Thanks.

EDIT: I realized my rearranging is wrong.

$$h = \frac{1}{2} g-v^{2}$$
Yikes! : How did you manage to work a new term into a proportional relationship?
Also note that the units of g are not the same as the units of v2, so something MUST be wrong...

Try again to rearrange the expression. What happens if you divide both sides by mg?

I was rearanging it again last night since the new arrangement didn't seem right either.

$$h = \frac{1}{2} \frac {mv^{2}}{mg}$$

I noticed that I had mv^2 on the bottom, but it should be on the top, while dividing by mg.

Thats correct, so what's next?

$$h = \frac{1}{2} \frac {400}}{9.8} = 20.4 m$$

Knowing that mass cancels out, I get 20.4 m = y

If I use 20.4sin30 = 10.2. But I think that's too short.

You've got the right answer for y. But your answer for the distance traveled up the slope is not correct. Draw a diagram, so you can see how the lengths are related.

## What is the "Skier Going Uphill - Intro Physics Problem"?

The "Skier Going Uphill - Intro Physics Problem" is a physics problem that involves a skier starting from rest at the bottom of a hill and skiing up a slope with a constant acceleration due to gravity. This problem is often used to introduce students to the concepts of force, motion, and energy in physics.

## What are the main forces acting on the skier in this problem?

The main forces acting on the skier in this problem are gravity and friction. Gravity is pulling the skier down the hill, while friction is acting in the opposite direction, slowing down the skier's motion.

## How is acceleration calculated in this problem?

Acceleration in this problem is calculated using Newton's second law of motion, which states that the net force on an object is equal to its mass multiplied by its acceleration. In this case, the net force is the difference between the force of gravity and the force of friction, and the acceleration is the change in velocity over time.

## What is the significance of the skier's velocity and displacement in this problem?

The skier's velocity and displacement are important in this problem because they help us understand how the skier's motion changes over time. The skier's velocity is the rate at which their displacement is changing, while their displacement is the total distance covered in a given direction. These values can be used to calculate the skier's acceleration and the work done by the skier's movement.

## How does the skier's motion change if the slope of the hill is increased?

If the slope of the hill is increased, the skier's motion will also change. The steeper the slope, the greater the force of gravity and the greater the acceleration of the skier. This means the skier will reach higher velocities and cover greater distances in a shorter amount of time. Additionally, the force of friction may also increase, making it more difficult for the skier to move up the hill.

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