A 25 kg wood box is sliding across a wood surface, experiencing a coefficient (u) of kinetic friction of 0.20. The box is experiencing an applied force of +63 Newtons. What is the force of friction acting on the box? (Ff=u x Fn), (Fn=-Fg), (Fg3mass x -9.81). P =mx v p= momentum m=mass V= velocity

A 25 kg wood box is sliding across a wood surface, experiencing a coefficient (u) of kinetic friction of 0.20. The box is experiencing an applied force of +63 Newtons. What is the force of friction acting on the box? (Ff=u x Fn), (Fn=-Fg), (Fg3mass x -9.81). P =mx v p= momentum m=mass V= velocity

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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BIUA
12
+
三
1
I will break you all into breakout rooms but each of you need to submit a copy
Unit 3: Interactions and Conservation of Momentum
1. Students will show works for this problem to receive credits:
A 25 kg wood box is sliding across a wood surface, experiencing a coefficient (u) of
kinetic friction of 0.20. The box is experiencing an applied force of +63 Newtons. What is
the force of friction acting on the box? (Ff=u x Fn), (Fn=-Fg), (Fg=mass x-9.81).
P = mx v
p= momentum
m=mass
V= velocity
Directly proportional vs Inversely Proportional:
When two variables are on different sides of the equal sign ( the equation),
-if they are both on top of the fraction or both on the bottom of the fraction–
directly proportional (Fc and m)( Fc and Vt)
!!!

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