1.) A skier coasts down a very smooth, 10-m-high. If the speed of the skier on the top of the slope is 5.0 m/s, what is his speed at the bottom of the slope?

Results 1 to 3 of 3

- Aug 1st 2007, 02:44 PM #1

- Joined
- Jun 2007
- Posts
- 18

- Aug 1st 2007, 03:48 PM #2
Use $\displaystyle U_1 + K_1 + W_{\text{other}} = U_2 + K_2 $ (conservation of energy)

Or use kinematic equations (under the assumption that we have**constant**acceleration). It would probably be better to use energy methods.

$\displaystyle v_f = v_0 + at $

$\displaystyle x-x_0 = \frac{1}{2}(v_0 + v_f)t $

$\displaystyle x-x_0 = v_{x0}t + \frac{1}{2}at^2 $

$\displaystyle v_{f}^2 = v_{0}^2 + 2a(x-x_0) $

- Aug 1st 2007, 05:19 PM #3
The use of the kinematic equations depends on the object having a constant acceleration. Since we aren't given any information about what the slope is like. (Yes, I know "slopes" are typically represented by lines, but I don't feel we can make that assumption in this case. We also don't have any information about the grade of the slope. As it happens the answer doesn't depend on these details, but that is a consequence of the energy theorem.)

To make a long story short, use energy methods.

-Dan