Question

A skier starts from rest at the top of a 45.0-m-high hill, skis down a 30° incline into a valley, and continues up a 40.0-m-high hill. The heights of both hills are measured from the valley floor. Assume that you can neglect friction and the effect of the ski poles. How fast is the skier moving at the bottom of the valley? What is the skier’s speed at the top of the next hill? Do the angles of the hills affect your answers?

Answer 1

1) The speed of the skier at the bottom of the valley is 29.7 m/s

2) The speed of the skier at the top of the second hill is 9.9 m/s

3) No, the angle of the hills does not affect the result

Explanation:

1)

We can solve this problem by using the law of conservation of energy. In fact, in absence of friction, the total mechanical energy of the skier is conserved:

$E=U+K$

where

E is the total mechanical energy

U is the gravitational potential energy

K is the kinetic energy

At the top of the hill, K = 0 since the skier is at rest, so all its energy is potential energy:

$E=U=mgh$ (1)

where

m is the mass of the skier

$g=9.8 m/s^2$ is the acceleration of gravity

h = 45.0 m is the height of the hill

As the skier descends the hill, the potential energy is converted into kinetic energy. At the bottom, all the mechanical energy has been converted into kinetic energy:

$E=K=\frac{1}{2}mv^2$

where

v is the speed of the skier at the bottom of the hill

Since the total energy is conserved,

$U=K\\mgh=\frac{1}{2}mv^2$

And so we find

$v=\sqrt{2gh}=\sqrt{2(9.8)(45)}=29.7 m/s$

2)

The top of the next hill is located at a height of

h' = 40.0 m

So the total mechanical energy at the top of the second hill is

$E=U'+K' = mgh'+\frac{1}{2}mv'^2$

where v' is the speed at the top of the second hill.

Since the total mechanical energy must be conserved, we can equate this energy to mechanical energy at the beginning (eq. 1), so we have

$mgh = mgh' + \frac{1}{2}mv'^2$

and we can now solve for v':

$v=\sqrt{2g(h-h')}=\sqrt{2(9.8)(45-40)}=9.9 m/s$

3)

As we saw from the previous equations, the angles of the hill does not enter at all the calculations, so it does not affect the value of the speed of the skier.

The reason for this is that the gravitational potential energy of the skier depends only on the height of the hill, h and h', and not from the length of the path along the hill (and so, not on the angle of the hill), and therefore the angle does not enter the calculation.

Learn more about kinetic energy and potential energy:

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