Question

A skier of mass 59.0 kg starts from rest at the top of a ski slope of height 62.0 m .

b)  Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is μk = 0.250. If the patch is of width 66.0 m and the average force of air resistance on the skier is 180 N , how fast is she going after crossing the patch?

c)  After crossing the patch of soft snow, the skier hits a snowdrift and penetrates a distance 2.60 m into it before coming to a stop. What is the average force exerted on her by the snowdrift as it stops her?

Answer 1

The velocity of the skier at the skier at the bottom of the slope=v₀=squareroot(2gh)=squareroot(2*9.8*62)=35 m/s.

b)The frictional force on the skier at the bottom of the slope=(coefficient of friction)*N=(coefficient of friction)*mg=0.25*59*9.8=145 N.

The total force opposing the skier's motion=F_f+force due to air resistance=145+180=325 N.

Deceleration of the skier=325/59=5.5 m/s².

Let the velocity of the skier after crossing the snow patch be v₁ m/s.

Then applying the kinematical equation v₁²=v₀²-2a₁s₁=35²-2*5.5*66=1225-726=499.

Therefore v₁=squareroot(499)=22 m/s.

This is the velocity of the skier after crossing the patch.

c)Let the deceleration of the skier after hitting the snowdrift be a₂ m/s².

Then v₂²=v₁²-2a₂s₂=0.i.e. v₁²=2*2.6*a₂=5.2a₂.

a₂=v₁²/5.2=22²/5.2=93 m/s².

Thus the average force exerted by the snowdrift=ma₂=59*93=5491.5 N.