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

Youngsters sled down a 43 -long hill inclined at 24 . On the backside, the slope ranges out.

PE(preliminary) = mgh = mgLsin = mg43sin24* = 17.49mg J
W(friction) = f(friction) x L = x N x L = x mgcos x L = 0.1 x mgcos24* x 43 = 3.93mg J
=>By the regulation of power conservation:-
=>KE(remaining) = PE(remaining) W(friction)
=>KE(remaining) = 17.49mg 3.93mg = 13.56mg J
=>Let youngsters journey s meter on the extent, By work power relation at stage:-
=>W(friction) = KE
=>F(friction) x s = KE
=> x mg x s = 13.56mg
=>s = 13.56/0.12
=>s = 113 m

This post is last updated on hrtanswers.com at Date : 1st of September – 2022

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