# Force of acceleration physics homework

a force of 40 N accerlerates a 5 kg block at 6 m/s^2 along a horizontal surface..

a) how large is the frictoinal force?
b) what is the coefficient of friction?

For Ans A, how would i find frictional force... 40N-5x6 = 10 N??

thanks

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also heres another question... If you use a horizontal force of 30N to slide a 12 kg wooden crate across a flood at a cnostant velocity, what is the coefficient of kinetic friction between the crate and the floor?

i cant seem to use MA=Fapp - Ff
i.e. ma = fpp - ukmg sice i dont have the value of a...

set up a FBD for the block
Use Newton's second Law for both horizontal and vertical directions
$$\vec{F} = m\vec{a}$$
They give the acceleration in the horizontal direction... what is the accel. in the vertical direction??
What are all the forces acting on the block?
The equation for the force of friction is
$$\vec{F}_f = \mu_kN$$
where $\mu_k$ is the coefficient of friction and $N$ is the Normal force
show your work for more help

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jai6638 said:
also heres another question... If you use a horizontal force of 30N to slide a 12 kg wooden crate across a flood at a cnostant velocity, what is the coefficient of kinetic friction between the crate and the floor?

i cant seem to use MA=Fapp - Ff
i.e. ma = fpp - ukmg sice i dont have the value of a...
The definition of acceleration is
$$\vec{a} = \frac{d\vec{v}}{dt}$$
ie: the change in velocity wrt time.
here they give you that the velocity is constant so that should tell you what the acceleration is. damn... this must be a bad day or somethin... im just blank! these questoins seemed easy to me the first time i did it a few weeks ago but now the night before my exam.... i just cant seem to do them..

F= MA
F/M=a

therefore , A= 30/12. = 2.5 m/s^2

am i correct?

The NET force = ma. The NET force is the sum of all the forces, in this case, friction and the applied force. MathStudent gave you what you need. acceleration is defined as the CHANGE in velocity over a period of time. Now, in the problem you are told that the velocity is CONSTANT. What does that make the acceleration?

jai6638 said:
damn... this must be a bad day or somethin... im just blank! these questoins seemed easy to me the first time i did it a few weeks ago but now the night before my exam.... i just cant seem to do them..

F= MA
F/M=a

therefore , A= 30/12. = 2.5 m/s^2

am i correct?
Not quite... I gave you a big hint in the previous post
-for one you need to draw a Free Body Diagram (FBD) for the crate
- identify All the forces acting on the crate

-Then you need to break newton's second law into vertical and horizontal components... do this and show your work (list the forces along with their directions dont try to draw the FBD) and I'll help you further.

as for the acceleration.. you need to realize that newtons 2nd law equates the NET force (the sum of all forces) ...you forgot about one

- since the velocity is CONSTANT that means that its not changing wrt time in the horizontal direction so the acceleration would be 0 right?

futb0l
jai6638 said:
damn... this must be a bad day or somethin... im just blank! these questoins seemed easy to me the first time i did it a few weeks ago but now the night before my exam.... i just cant seem to do them..

F= MA
F/M=a

therefore , A= 30/12. = 2.5 m/s^2

am i correct?
It's is at a CONSTANT VELOCITY. Therefore the acceleration is = ...

have a 12 kg crate....with 30 N force vector pointing toward right.. then Normal Force vector pointing upword and Force due to gravity pointing downwards...

ALso, just for my information, if there is constant velocity is Ffrictoin = F applied?

jai6638 said:
have a 12 kg crate....with 30 N force vector pointing toward right.. then Normal Force vector pointing upword and Force due to gravity pointing downwards...

ALso, just for my information, if there is constant velocity is Ffrictoin = F applied?
Don't forget about the friction force... what direction does it point in?

In this case yes since
$$\Sigma\vec{F}_x = \vec{F}_h +\vec{F}_f$$
$$m\vec{a} = 12*0$$

$$\Rightarrow \vec{F}_h +\vec{F}_f = 0$$
$$\Rightarrow \vec{F}_h = -\vec{F}_f$$
$$\Rightarrow F_f = -F_h$$

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Frictoinal force points to the left...

I gave you an example of using newtons 2nd law for the horizontal components ... why don't you try it now for the vertical components...
keep in mind what your trying to solve for

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i dont mean to sound like an ass but i still dont get it... /me bangs his head on the wall! ! ! ! !!

im gonna sleep since i am not accomplishing much by stayin awake as all the answers/equations are goin above my head .. makes me wonder if im fit for becoming an engineer... doh anyways... thanks for ur help though.. appreciate it :)