A skier slides horizontally along the snow for a distance of 22.3 m before coming to rest. The coefficient of kinetic friction between the skier and the snow is 0.0500. Initially, how fast was the skier going?

Answer 1

KE2 - KE1= A(friction).
KE2 - KE1=0-(m•v^2)/2= - (m•v^2)/2.
A(friction) = F(friction)•s•cosα = k•m•g•s•(-1) = - k•m•g•s.

(m•v^2)/2= k•m•g•s.
v=sqroot (2•k•g•s)=
=sqroot (2•0.05•9.8•22.3)=21.85 m/s.