Ask Question
29 November, 14:35

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?

0
Answers (1)
  1. 29 November, 16:16
    0
    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
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “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 ...” in 📙 Physics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers