A 98.0 N box slides 25.0 m along a 35.0° incline. If the force of friction along the incline is 32.0 N, and the box starts from rest at teh top, what is the momentum of the box at the bottom?(adsbygoogle = window.adsbygoogle || []).push({});

This is how i attempted to solve this question. The answer in the book is 1.1 * 10^2 Kgm/s. Can you please tell me waht error i have made. Thank-you.

Eptop + Ektop = Epbot + Ekbot

mgh + 0 = 1/2mv^2 + 0

(9.81m/s^2)(25.0m*Sin35) = 1/2v^2

v=16.77 m/s

Fg=mg

m=98.0 N/9.81m/s^2

m=9.9898 Kg

p=mv

p=(9.9898 Kg)(16.77 m/s)

p=168 Kgm/s

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# Homework Help: What is the momentum of the box at the bottom?

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