Posted by **Hannah** on Sunday, June 3, 2012 at 11:52am.

A skier slides horizontally along the snow for a distance of 11.9 m before coming to rest. The coefficient of kinetic friction between the skier and the snow is 0.0458. Initially, how fast was the skier going?

I do not know how to start this. Thank you.

Physics(Please help) - Elena, Saturday, June 2, 2012 at 8:59am

ΔKE =KE2 –KE1 = 0 -m•v²/2.

W(fr) =μ•m•g•s•cosα,

where α is the angle between the friction force and displacement.

α =180º, cos α = -1.

-m•v²/2 = - μ•m•g•s,

v = sqrt(2•μ• g•s).

So for v would I do sqrt(2*11.9*9.8*0.0458) ?

## Answer this Question

## Related Questions

- Physics(Please help) - A skier slides horizontally along the snow for a distance...
- 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 - A skier starts from rest at the top of a hill that is inclined at 9.8...
- physics - A skier starts from rest at the top of a hill that is inclined at 10.3...
- physics - A skier starts from rest at the top of a hill that is inclined at 9.5...
- physics - A skier starts from rest at the top of a hill that is inclined at 9.5...
- physics - A skier starts from rest at the top of a hill that is inclined at 11.0...
- physics...again :/ - A skier starts from rest at the top of a hill that is ...
- physics - A skier starts from rest at the top of a hill that is inclined 18.5° ...

More Related Questions