Question

(Newton's Law of Motion) The skier has just begun descending the 30 degrees slope. Assuing the coefficient of kinetic friction is 0.10, (a) calculate her acceleration and (b) the speed she will reach after 4s.

Answer 1

mass of the skier =m
angle of slope= 30 degree
coeff of kinetic friction=0.10
formula is
\[mg \sin 30-f_{k}=F_{net}\]
rewriting this formula to suit our problem
\[mg \sin 30-f_{k}=ma\]
\[f_{k}=\mu_{k}N\]
N= normal force=mg cos 30
therefore,\[mg \sin 30-\mu_{k}mg \cos 30=ma\]
\[a=m \sin 30-\mu_{k} \cos 30\]
\[a=9.8\times(\sin 30-(0.1)\cos 30)\]
a=4.0512m/s^2
since she has just began to decend
initial velocity, u=0m/s
final velocity at time t=4s
v=u+at
=0+4.0512*4=16.0608m/s
please check it for yourself...

Answer 2

oops i made a small mistake
the equation for acceleration is
\[a=g(\sin 30-\mu_{k}\cos 30)\] not a=msin30...