Question

I need help finding Applied force at which the block moves with constant velocity (f)

Answer 1

i can't figure out of its F = ma or if its another formula like Fk = µk N .

Answer 2

i found the value of kinetic friction by a spring and mass.

Answer 3

That second equation is perfect: F = µN

Answer 4

to find the Applied force at which the block moves with constant velocity (f)?

Answer 5

how did you get the value of kinetic friction? i just want to make sure that we mean the same thing.

at constant velocity, a=0 and this means that the net force F is zero. This would mean that the magnitude of the force applied = magnitude of kinetic friction, if this experiment is done on a horizontal flat surface.

Answer 6

i think the 2 values are the same

Answer 7

so which formula should i use? I'm confused because it has to deal with kinetic friction so i don't know if its F = ma or what.

Answer 8

Pompeii00 has addressed what you need to use to find \(\mu\). For the normal force N, you will need to use another application of "F=ma"

\[F_{y}=0=mg-N\]

Answer 9

Thank you so much !

Answer 10

but what other formula is it?

Answer 11

in replacement of f = ma

Answer 12

The forces can be broken down into 2 parts, one in the y direction, the other in the x direction. We will be using newton's 2nd law:
\[\Sigma F=ma\]
In the x direction,
\[(\Sigma F_{x}=F_{applied}-friction)=ma_{x}\]But \(a_{x}=0\), and \(friction=\mu N\)
where N is the normal force, and \(\mu\) is the coefficient of friction (kinetic friction in this case). thus,
\[(\Sigma F_{x}=F_{applied}-\mu N)=m\times 0 = 0 \]which reduces to
\[F_{applied}=\mu N\] But what is N? We can find N by considering the forces in the y direction:
\[(\Sigma F_{y}=N-mg)=ma_{y}\]
Since the block is not accelerating in the y direction, \(a_{y}=0\) and so: \[N=mg\]
Putting them together,
\[F_{applied}=\mu \times m \times g\]
where g is the acceleration of free fall.

Answer 13

thank you so much :D i understand it now