A ski starts from rest and slides down a 28∘ incline 100 m long. If the coefficient of friction is 0.090, what is the ski's speed at the base of the incline? If the snow is level at the foot of the incline and has the same coefficient of friction, how far will the ski travel along with the level? Use energy methods.

Explanation:

Frictional force acting on incline = μ mg cosθ

μ is coefficient of friction , m is mass of object , θ is incline

= .09 x m x 9.8 x cos 28

= .78 m

work done by friction

= frictional force x displacement

= - .78m x 100

= -  78m

Potential energy of sli at height

= mgh

= m x 9.8 x 100 sin 28

= 460.08 m

net energy at the base

= 460.08m - 78 m

= 382.08 m

 This will be in the form of kinetic energy .

1/2 m v² = 382.08 m

.5 x v² = 382.08

v = 27.64 m/s

After that it travels on plane surface .

Let the distance travelled be d

work done by frictional force

= μ mg x d

= .09 x m x 9.8 x d

This will be equal to kinetic energy at the base

.09 x m x 9.8 x d  = 382.08 m

d = 433.2 meter .

layer a is the exosphere and layer b is the thermosphere.   is the answer i think but i hope this .