Posted by **Claire Holt** on Friday, June 15, 2012 at 9:09pm.

A skier traveling 11.0 m/s reaches the foot of a steady upward 19.6° incline and glides 13.8 m up along this slope before coming to rest. What was the average coefficient of friction?

- physics -
**Anonymous**, Tuesday, April 5, 2016 at 11:45pm
Use equation mgh + mgh(initial) = -1/2mv^2 - 1/2 mv(initial) + W

- physics -
**Khai Lim**, Saturday, October 22, 2016 at 10:11pm
KE = PE + W_friction

1/2 * mv^2 = mgh + mu*mgcos(theta)d

----- m is cancel out

--- sin(theta) = heigh/ distance( inclide) ===> h = d*sin(theta)

1/2 v^2 = g (d*sin(theta)) + mu* (g*cos(theta))*d

now you can solve for mu ^^ Good luck.

## Answer This Question

## Related Questions

- Physics - A skier traveling 12.0 m/s reaches the foot of a steady upward 18° ...
- PHYSICS!!! - A skier traveling 12.0 m/s reaches the foot of a steady upward 18 ...
- physics - A skier travelling at 14 m s-1 reaches the foot of a steady upward 14...
- Uni Physics - A skier travelling at 13 m s-1 reaches the foot of a steady upward...
- physics - A skier starts from rest at the top of a frictionless incline of ...
- Physics - A skier slides horizontally along the snow for a distance of 12.1 m ...
- physics - A skier slides horizontally along the snow for a distance of 20.8 m ...
- physics - A skier slides horizontally along the snow for a distance of 22.3 m ...
- Physics(Please help) - A skier slides horizontally along the snow for a distance...
- physics - Find the horizontal distance the skier travels before coming to rest ...

More Related Questions