Problem 1 A 10-kg wood crate is being pushed across a wooden floor with a horizontal force of 350 N. What is the net force acting on the crate given the coefficient of friction, µbetween the crate and the wooden floor to be 0.3? Given: m = 10 kg Hy = 0.3 Find: F net F applied = 350 N Vector representation

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Chapter5: Newton's Law Of Motion
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EXAMPLES:
Problem 1
A 10-kg wood crate is being pushed across a wooden floor with a horizontal
force of 350 N. What is the net force acting on the crate given the coefficient of friction,
Hbetween the crate and the wooden floor to be 0.3?
Given:
m = 10 kg
H, = 0.3
Find: F.
net
F
applied
= 350 N
Vector representation
F = W
F.
applied
F,
Solutions:
a.) W = mg
b.) F; = 4, FN
%3D
net
applied
W = (10kg)9.8m/s²)
F, = (0.3)(98N)
F = 350N – 29.4N
%3D
net
W = 98N
F = 29.4N
F.
= 320.6N
%3D
net
101
Transcribed Image Text:EXAMPLES: Problem 1 A 10-kg wood crate is being pushed across a wooden floor with a horizontal force of 350 N. What is the net force acting on the crate given the coefficient of friction, Hbetween the crate and the wooden floor to be 0.3? Given: m = 10 kg H, = 0.3 Find: F. net F applied = 350 N Vector representation F = W F. applied F, Solutions: a.) W = mg b.) F; = 4, FN %3D net applied W = (10kg)9.8m/s²) F, = (0.3)(98N) F = 350N – 29.4N %3D net W = 98N F = 29.4N F. = 320.6N %3D net 101
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