Asked by Maredith
a 36.1 N crate is resting on the floor. When you push horizontally on the crate with a force of 16.9 N, it accelerates at 0.15 m/s2. How large is the frictional force acting on the crate?
Answers
Answered by
Damon
F = m a
F = (36.1/9.8)* .15
F = .552 N
16.9 - friction force = .552
friction force = 16.35 N
(I suspect a typo in your problem statement by the way)
F = (36.1/9.8)* .15
F = .552 N
16.9 - friction force = .552
friction force = 16.35 N
(I suspect a typo in your problem statement by the way)
Answered by
matt
sum of forces-> F=ma
F(friction)= ufn* you don't need this
F(applied)-F(friction)=ma
f(a)-f(f)=ma
f(a)/ma=f(f)
add numbers
F(friction)= ufn* you don't need this
F(applied)-F(friction)=ma
f(a)-f(f)=ma
f(a)/ma=f(f)
add numbers
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