Thursday
March 23, 2017

Post a New Question

Posted by on Monday, February 27, 2012 at 2:58pm.

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?

  • physics - , Monday, February 27, 2012 at 5:00pm

    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.

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question