A box sits at rest at the top of a frictionless hill, a heightof 45.04 m above t

Question

A box sits at rest at the top of a frictionless hill, a heightof 45.04 m above the level ground. The box slides down thehill, and then continues across a rough patch of ground(&muk, coefficient of kinetic friction, =0.44) until it stops. What distance will the box slide across therough, level ground before coming to a stop?
I know it relates to PE=KE but Im not sure how to finddelta x A box sits at rest at the top of a frictionless hill, a heightof 45.04 m above the level ground. The box slides down thehill, and then continues across a rough patch of ground(&muk, coefficient of kinetic friction, =0.44) until it stops. What distance will the box slide across therough, level ground before coming to a stop?
I know it relates to PE=KE but Im not sure how to finddelta x

Explanation / Answer

height h = 45.04 m velocity of the box at horizontal surface v = [ 2gh ]                                                               = 29.71 m / s coefficient of kinetic friction = 0.44 accleration a = -g                     =- 4.312 m / s ^ 2 final velocity V = 0 m / s from the kinematics relation V ^ 2 - v ^ 2 = 2 a S from this required distance S = [ V ^ 2 - v ^ 2 ] / 2 a                                            = 102.35 m

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