Question

A skier of mass 50 kg starts from rest and slides downhill. What will be the speed of the skier if he drops by 20 meters in vertical height? Let the work done by the air resistance when the skier goes from AA to BB along the given hilly path be -2000 J. The work done by air resistance is negative since the air resistance acts in the opposite direction to the displacement.

Answer 1

`17.67(m)/(s)`

Step-by-step explanation:

By work-energy theorem we can relate total work and the change on kinetic energy in the next way:

`W_(tot)=K_f-K_i` (1)

with Wtot the total work, Kf the final kinetic energy and Ki the initial kinetic energy. The total work is the sum of the work done by gravity:

`W_(gra)=mgd=(50)(9.81)(20)=9810J` (2)

(with m the mass of the skier, g the gravitational acceleration and d the vertical distance of the movement) and the work done by air:

`W_(air)=-2000J` (3)

Note that the sign of Wgra is positive because weight acts in the direction of the displacement and Wair is negative because the air resistance acts in the opposite direction to the displacement. So, by (3) and (2) on (1):

`W_(tot)=W_(gra)+W_(air)=9810 -2000 = K_f-K_i` (4)

Kinetic energy is
`(mv^2)/(2)`

with m the mass and v the speed so (4) is:

`7810=(mv_f^2)/(2)-(mv_i^2)/(2)=(50v_f^2)/(2)-(50(0)^2)/(2)`

`7810=(50v_f^2)/(2)`

solving for vf:

`v_f=\sqrt{(2*7810)/(50)}=17.67(m)/(s)`