If the coefficient of friction is 0.12, how far do the children slide on the level ground?

Children sled down a 43 -long hill inclined at 24 . At the bottom, the slope levels out.

1 Answer

  • PE(initial) = mgh = mgLsinθ = mg43sin24* = 17.49mg J

    W(friction) = f(friction) x L = µ x N x L = µ x mgcosθ x L = 0.1 x mgcos24* x 43 = 3.93mg J

    =>By the law of energy conservation:-

    =>KE(final) = PE(final) - W(friction)

    =>KE(final) = 17.49mg - 3.93mg = 13.56mg J

    =>Let children travel s meter on the level, By work energy relation at level:-

    =>W(friction) = ∆KE

    =>F(friction) x s = ∆KE

    =>µ x mg x s = 13.56mg

    =>s = 13.56/0.12

    =>s = 113 m

Leave a Reply

Your email address will not be published. Required fields are marked *

Related Posts