A force of 36.ON acclerates a 6.0kg block at 7.0m/s2 along a boriztonal surface. How large is the friction force?

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I have a question on physics. I searched for the solution to this problem:
A force of 36.ON acclerates a 6.0kg block at 7.0m/s2 along a boriztonal surface. How large is the friction force?

and opened two solutions. It actually has the same answers but I am wondering if fand fk are the same?
If ff and fare the same, then Friction Force and Kinetic Friction are the same?
Thank you.

Given
* The force applied on the block is F36N.
* The mass
of block is m = 6
The acceleration of block is a ==7 m/s².
The free body diagram of block is,
fret
m=6kg.
= ma
mg
The net force to calculated as
fret
F-ff=
= ma
2
36N-F₂ = 6kg x 7 m/s ²
36N-42N =
F=36N
ff
f = -6 N
Thus, the friction force is GN
[&R foration force]
[from Diagram, force in]
x dixchon & F-f
[-re sign shows only the
dixchien of friction
which is opposite of what
take ]
ное
Transcribed Image Text:Given * The force applied on the block is F36N. * The mass of block is m = 6 The acceleration of block is a ==7 m/s². The free body diagram of block is, fret m=6kg. = ma mg The net force to calculated as fret F-ff= = ma 2 36N-F₂ = 6kg x 7 m/s ² 36N-42N = F=36N ff f = -6 N Thus, the friction force is GN [&R foration force] [from Diagram, force in] x dixchon & F-f [-re sign shows only the dixchien of friction which is opposite of what take ] ное
Dat a
given F = 36 N
}
ㅠ
ma
m =
But
a
6.к.д.
69
Since, the block is moving,
F-fk
=) |fk
= 7m/s2
Now; from.
fk
F-ma
6 kg
fk.
—
= UKN
from equation 1
58.8μK
икту
P
= ukx6x9.8
Transcribed Image Text:Dat a given F = 36 N } ㅠ ma m = But a 6.к.д. 69 Since, the block is moving, F-fk =) |fk = 7m/s2 Now; from. fk F-ma 6 kg fk. — = UKN from equation 1 58.8μK икту P = ukx6x9.8
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