Help on a problem with multiple parts, please!

[keeper54]

hey i was wondering if anybody could show me how to do a problem that contains multiple parts:

55. A 3.00 kg block starts from rest at the top of a 30.0° incline and accelerates uniformly down the incline, moving 2.00 m in 1.50 s.

a. Find the magnitude of the acceleration of the block.
b. Find the coefficient of kinetic friction between the block and the incline.
c. Find the magnitude of the frictional force acting on the block.
d. Find the speed of the block after it has slid a distance of 2.00 m.

I know that the initial velocity=0, Fg (weight)=29.4N, Fn (normal force)=14.7N, and I think that the falling force is 21.6N, but I'm not sure if i did all this right. Just show me what I need to do, and I'll do the rest! thanks for your help.

[Pyrrhus]

The force making it slide is gravity, and there should be only two forces acting on it [normal and gravity] if there is no friction of course.

Normal will be equal to mgcos\theta = n because there is no movement on the y-axis.

Now for mgsen\theta = ma which will become gsen\theta = a because there's movement on the x-axis.

[keeper54]

there is definetly friction involved, because one of the questions is asking for the coefficient of friction.

[Pyrrhus]

there is definetly friction involved, because one of the questions is asking for the coefficient of friction.

oh i didn't read the questions :smile:

Well another force :cool:

F_{f} = \mu N

so changing the x-axis equation it will be if we take positive aceleration the way as the block slides it will be

mgsen\theta - \mu N = ma
mgsen\theta - \mu mgcos\theta = ma
gsen\theta - \mu gcos\theta = a

[keeper54]

I plugged everything in so that it was a=9.8sin30=4.9m/s/s. This is wrong, though. The back of the book says it should be 1.78m/s/s. What's wrong?

[Pyrrhus]

I plugged everything in so that it was a=9.8sin30=4.9m/s/s. This is wrong, though. The back of the book says it should be 1.78m/s/s. What's wrong?

I did the first analysis without friction...

[keeper54]

so how would i find mu?

[Pyrrhus]

so how would i find mu?

Kinematics! :smile:

The problem states uniform acceleration which is
gsen\theta - \mu gcos\theta = a

[keeper54]

but mu is a value, isn't it (and one that I don't have, at that)? So how do I find accel. without mu? I'm sorry if I'm being difficult, but as you can see, this isn't my strongest subject :).

[Pyrrhus]

Well the block starts from rest

info:
V_{o} = 0 m/s
X_{o} = 0 m
X = 2 m
t = 1.5 s

I think this equation will help

X - X_{o} = V_{o}t + \frac{1}{2}at^2
X = \frac{1}{2}at^2

X = \frac{1}{2}(gsen\theta - \mu gcos\theta)t^2

[keeper54]

still not answering the mu thing, though. Sorry.....lol. d=1/2 (9.8sin30 - [mu]9.8cos30)

[Pyrrhus]

Are you understanding? or should we go back?

[Pyrrhus]

still not answering the mu thing, though. Sorry.....lol. d=1/2 (9.8sin30 - [mu]9.8cos30)

Of course it is!, just get \mu alone

\frac{\frac{2X}{gt^2} - sen\theta}{-cos\theta} = \mu

Plug in known values and voila!

[keeper54]

Understanding everything except for that darn mu thing. I'm really sorry I'm making this so hard.

[Pyrrhus]

Understanding everything except for that darn mu thing. I'm really sorry I'm making this so hard.

\mu is just the greek letter given to friction quotient.

[keeper54]

OK. got that. so... 2(2)/9.8(1.5*1.5)- sin30 / -cos30 = mu. mu=-.48

[Pyrrhus]

I guess you can finish it now...

[Pyrrhus]

OK. got that. so... 2(2)/9.8(1.5*1.5)- sin30 / -cos30 = mu. mu=-.48

No, you're calculating it wrong. \mu = 0.3678

[keeper54]

don't know where i went wrong....

[keeper54]

i did it again and still got .48308

[Pyrrhus]

Use enough parenthese in the calculator or do it by pieces.

If you put it all together it should be

(((2*2)/(9.8*1.5^2)) - sin(30))/(-cos(30))

Edit: Oops one parentheses left out

[keeper54]

got it. thanks so much for your help. sorry once again for making things so difficult. I'm goin' to bed.

[Pyrrhus]

got it. thanks so much for your help. sorry once again for making things so difficult. I'm goin' to bed.

It was good to be of help, good night!