Please help me!

Question

A novice skier, starting from rest, slides down a frictionless 13.0^oincline whose vertical height is 150 m.  How fast is she going when she reaches the bottom?

Answer 1

NOTE: I will not be following any significant figure rules. Make sure you know whether your instructor requires them.

Before we start, please keep in mind these hints for physics problems like this one:

-Draw a picture, and make a free-body diagram (FBD) when applicable. It will help you figure out what to do, and some teachers will give you at least partial credit if you do this.

-Make sure you understand all of the given quantities in the problem. Values for given quantities aren't always explicit (like "v_i = 0"), and are often implicit instead (like "starting from rest" which means the same thing)

-Find a formula that has variables for the quantities you need as well as the quantities you have. If there are only formulae which require some quantities you don't have, this means you will need to calculate one or more of the other quantities.

When you draw the FBD (do it!) of the skier on the frictionless slope, you will see that the only forces are gravity (F_g) and the slope's normal force (F_N).

We know that the skier will accelerate down the slope, so we will

-Make the starting point (0,0) and the direction down the slope the positive-x direction, for convenience (perpendicularly away from the slope can by positive-y)

-Break down all of our force vectors into components which point along one of our defined axes

The initial velocity of the skier is 0 and the initial position on the x-axis is 0 (we defined it to be so). We know the force acting on a mass is related to its acceleration, so we should be able to figure out the acceleration. We know the height and angle of the slope, so we can calculate the length of the slope. We know a formula that has what we need (final velocity) and some of these other values...

a = (v_f - v_i) / t_f

or, since v_i = 0

a = v_f / t_f

v_f = at_f

Close, but we don't know t_f, the time to reach the final velocity v_f. That means there is another step: finding t_f. We know another formula that will help here:

x(t) = 1/2at² + v_it + x_i

Since v_i and x_i are 0, and we are looking for t_f:

x(t_f) = 1/2at_f² + (0)t_f + (0)

x_f = 1/2at_f²

Rearrange to solve for t_f:

t_f = sqrt(2x_f / a)

NOTE: Remember that these formulae only work when acceleration is constant.

So if we can find a and x_f (the length of the slope surface), we can calculate how much time it will take to get to the bottom.

The normal force always points perpendicularly away from a surface, so it is pointing in the positive-y direction. It is exactly opposing the component of gravity acting in the opposite direction: INTO the surface, the negative-y direction. This means the gravity in that direction is cancelled by the normal force (this is what the normal force ALWAYS does-- it cancels all other forces which point perpendicularly into a surface, because it represents the surface's resistance to breaking due to being pushed down on). The only remaining force acting on the skier is the gravity component pointing to positive-x. Since gravity is constant, we are allowed to use the formulas we found above!

NOTE: It may sound strange to say there is gravity in any direction besides down. But any 2D vector can be broken down into 2 orthogonal components at any angle which is less than 90 degrees to the original.

We can figure out the magnitude of gravity pointing to positive-x by drawing the gravity vector ("down") and the two component vectors (one pointing toward negative-y and one pointing toward positive-x) as a right triangle with the gravity vector as the hypotenuse. Now we use some easy trigonometry to find the vector we want, because one of the angles is given.

When everything is drawn correctly, the positive-x component (I will call it F_g-x) will be the leg of the right triangle which is opposite the given angle. We started with the hypotenuse, which is F_g. So we consult our trig chant to figure out which trig ration has the side we know as well as the side we want:

sOH cAH tOA

The winner is sOH, which is just shorthand for

sin(θ) = O/H

We started with H and are trying to find O, so we can rearrange it:

O = H sin(θ)

or, if we rename the legs and plug in the angle:

F_g-x = F_g sin(13º)

Since the force due to gravity is

F_g = mg = 9.8m

we can substitute:

F_g-x = F_g sin(13º)

F_g-x = 9.81m sin(13º) N/kg

This is the force acting on the skier. But what we really want in order to calculate the answer was the acceleration. To relate force and acceleration, we just use

F_g-x = ma

or

a = F_g-x/m

Substitute in the above expression for F_g-x:

a = [9.81 m sin(13º) N/kg] / m

a = 9.81 sin(13º) m/s²

a = 2.21 m/s²

We need to know the length of the slope travelled by the skier. We know the vertical height of the starting point (h) to be 150m and that it is elevated above the horizontal by 13º, so we can use trig ratios again. This is just like the above trig ratio calculation:

sOH

or

sin(θ) = O/H

H = O / sin(θ)

which means

x_f = h / sin(13º) = 150 / sin(13º) = 666.81 m

We can now find the time it takes for the skier to slide to the bottom with our above formula:

t_f = sqrt(2x_f / a) = sqrt[2(666.81) / (2.21)] s = 24.57 s

Sanity check: Is this a reasonable number? It seems like a really short time to cover more than half a kilometer. But, in this problem, there is no friction or air resistance, meaning the skier will keep accelerating even up to crazy speeds. So in these made-up conditions, it makes sense.

With the value of t_f known, we can calculate the final speed with our above formula:

v_f = at_f = (2.21)(24.57) m/s = 54.30 m/s

Sanity check: 54.30 m/s is over 120 mph. This high speed is again because there is no friction or air resistance. In reality, the speed will be much lower, since friction will result in a lower net force and the low net force will result in a pretty low terminal velocity.

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One more example.

A box starting from rest slides down a ramp inclined at 30º without friction or air resistance. The horizontal distance travelled by the box is 2.6 m. How fast is the box travelling when it reaches the bottom of the ramp?

FBD reveals that the normal force form the ramp cancels the component of gravity perpendicular to the ramp. Now the only force is the component of gravity along to the ramp. Gravity is constant, so we can use constant-a formulae. Taking down the ramp to be positive x and the starting x-position to be 0:

v_f = v_i + at_f = at_f

and

x_f = 1/2at_f² + v_it + x_i = 1/2at_f²

t_f = sqrt(2x_f / a)

in the x-direction.

To find a, we need x_f, the length of the ramp. This can be found by

cos(30º) = (2.6 m) / x_f

x_f = 2.6 / cos(30º) m = 3.00 m

To find the component of gravity F_g-x, we use

sin(30º) = F_g-x / F_g

F_g-x = F_g sin(30º)

or

F_g-x = mg sin(30º) = 9.81 m sin(30º) N/kg

Because

F_g-x = ma

or

a = F_g-x / m

we can now substitute

a = [9.81 m sin(30º) N/kg] / m = 9.81 sin(30º) m/s² = 4.91 m/s²

Now we can find t_f:

t_f = sqrt(2x_f / a) = sqrt[2(3.00) / 4.91] s = 1.11 s

We can now find v_f in the x-direction:

v_f = at_f = (4.91)(1.11) m/s = 5.45 m/s