
FBD of Skier

MAD of Skier
|
|
Using the principle of conservation of energy, the velocity of the skier at the end of the jump can be found. Energy is a simple
way to find the velocity of any object if you know how far it falls. Of course, this
method neglects air and surface friction which is lost energy. The principle of conservation was introduced in your basic physics course, and will
studied in detail in the section on
conservative forces and potential energy.
Equating the potential of the skier at the top of the jump with the kinetic energy at the bottom of the jump gives
Upotential = Ukinetic
hmg = 0.5mv2
If the velocity at the bottom of the jump is noted as v2,
then

The free-body and mass-acceleration diagrams are shown at the left. Summing the forces in the normal direction gives
ΣFn = man
N - mg = m v2/ρ
To solve for the normal force, N, the radius of curvature, ρ, must be determined at the
end of the jump. From calculus, the radius of curvature for any continuous line
function, y = f(x), is
For this problem,
 |

Velocity of Skier
|
|
Thus, at x = 0,
Using the radius of curvature, the normal force exerted on the skier at the end of the jump is
N = m v2/ρ + mg
= 90 (17.15)2/20 + 90(9.8) = 2,210 N
|

Acceleration of Skier
|
|
To find the total acceleration of the skier, sum the forces in the tangential direction,
ΣFt = mat
0 = mat
Thus, the acceleration at the end of the jump in the tangential direction is zero. The total acceleration is
a = an en + at et
= v2/ρ
en + 0 et = 17.152/20 en
= 14.71 en
m/s2
|