Josh starts his sled at the top of a 2.9-m-high hill that has a constant slope of 25∘. After reaching the bottom, he slides across a horizontal patch of snow. The hill is frictionless, but the coefficient of kinetic friction between his sled and the snow is 0.06. How far from the base of the hill does he end up?

S=48.29 m

Explanation:

Given that the height of the hill h = 2.9 m

Coefficient of kinetic friction between his sled and the snow μ = 0.08

Let u be the speed of the skier at the bottom of the hill.

By applying conservation of energy at the top and bottom of the inclined plane we get.

Potential Energy=kinetic Energy

mgh = (1/2) mu²

u² = 2gh

u²=2 (9.81) (2.9)

=56.89

u=7.54 m/s

a = - f / m

a = - μ*m*g / m

a = - μg

From equation of motion

v² - u² = 2 - μ g S

v=0 m/s

- (7.54) ²=-2 (0.06) (9.81) S

S=48.29 m