A worker pushed a 27 kg block 9.2m along a level floor at constant speed with a force directed 32 degrees below the horizontal. If the coefficient of kinetic friction between block and floor was 0.20, what was the work done by the worker's force?



Okay what I did was that I found out what Fk was by multiplying 0.20 * 27 kg * 9.8m/s^2. I received 52.9 N.
Now I need to get the force in the horizontal distance so I did 52.9/cos(32) and I received 62.4 N.
Since Work = Force * Distance, I multiply 62.4 N with 9.2m to get 574 J.
However, this answer seems to be wrong for some reason as the book has 560 J. I cannot comprehend how they are getting this answer, can you guys please help me?

Try this: kinetic friction is not simply mass * G * coefficient of Kinetic Friction, but instead it is the Normal force times the coefficient of kinetic friction. In this case part of your normal force is equal to the force of gravity on the block (M*G) PLUS the vertical component of force from the worker, so you will actually end up with a higher force of friction. If you were to draw a force diagram this would be clear.

FK = .20 * (27kg * 9.8 M/S^2 + Fw*sin(32))

Since the block has constant speed:

Fw*cos(32) = .20 * (27kg * 9.8 M/S^2 + Fw*sin(32))

Which results in a force of ~71 N

Taking just the Horizontal component Fw*cos(32), and you get ~60.5 N

Then you get work by Force * Distance, so ~60.5 N* 9.2 M = ~556 J which is good enough for government work.

(all back of the envelope calculations...)

Now, go do your own homework, and let me do mine…

Damn, I could have answered this one

Oh well.

I could have answered, too. Darn you, west wind!

Thank you so much West Wind. Wow I didn't know Massassi was that eager to solve physic problems, guess maybe I should consider this place my second line of defense if I can't solve it.

42

Understand the diagram and the rest is cake.

Does anyone else here have another physics problem we can do?

It's just, its something easy I know how to do, which is uncommon here. I really suck at all the other math problems that get posted

Derive E = mc^2

required by my quantum physics professor by tomorrow.

A 5.0-m-diameter merry-go-round is initially turning with a 4.0s period. It slows down and stops in 20s.

Before slowing, what is the speed of a child on the rim?

How many revolutions does the merry-go-round make as it stops?

How about mine from a course i'm doing in my masters...

"If there were 400 days in the year in the Devonian period 400 million years ago, what has been the average rate of dissipation of rotational energy of the Earth over the past 400 million years?

Assume a uniform rigid Earth with a mass of 5.976 x 10^24 kg. The moment of inertia of a uniform rigid sphere is (2ma^2)/5 where m is the total mass and a its radius"

Thats one of three questions I had to do for my planetary atmosphere's course...lovely bloody questions...

I do kinda like them though, they don't really give you any place to start from, i.e. no real hint of what formula to use or anything, the one I've quoted is probably the easist to think about and get formula for, you've also got to get the numbers in a form you can use (ie seconds not years) and you also have to figure out what to do...i.e. do we include the Earth's rotation into the question? or just stick to the Earth going round the Sun?

You kinda have to draw a line and say I'm not making it any more complicated than that...

I'd also like to point out that so far in the course there has been not one mention of any mechanics...its all been about atmospheric *stuff* hence the class were expected to either know how to do this or figure it out for themselves, plus I'm the only one doing it from my college....rant...rant....

/rant

[edit] I suppose I should point out that its not all that difficult...[/edit]

On the topic of Physics. Does anyone here belong to, or know anything about Sigma Xi ?

I resently threw a random Undergraduate Research Proposial at the Physics department, and got a recemndation that I talk to some people about working with Sigma Xi. I have checked out their Website, but I am curious if anybody knows anything about them?

[QUOTE=Connection Problem]Understand the diagram and the rest is cake.[/QUOTE]
Truth. Always always always draw out force diagrams. They will make your life much easier than trying to mentally picture the scenario.

You could, you know, just ask your teacher...

[QUOTE=Connection Problem]A 5.0-m-diameter merry-go-round is initially turning with a 4.0s period. It slows down and stops in 20s.

Before slowing, what is the speed of a child on the rim?

How many revolutions does the merry-go-round make as it stops?[/QUOTE]

The circumference of the rim is 5pi meters.

With M = a point on the rim, dM/dt at t=0 is 5pi m / 4 s = 5pi/4 m/s.

This is the initial speed of that child.

At t=20, dM/dt = 0. Thus, dM/dt = -((5pi/4)/20)t+ 5pi/4 = -(pi/16)t + 5pi/4 m/s.

The total distance covered by the point M can then be expressed as -(pi/32)t^2 + (5pi/4)t + C. We will define the reference frame such that C is 0. Thus, in 20 seconds, M has traveled -(pi/32)(20)^2 + (5pi/4)(20) m, or ~39.27 m. Divide the exact distance by the circumference (5pi) and you get 2.5 revolutions.

Is that right? I pulled that out of my ***

I wish I could go back to mechanics.

You got it Pommy. I solved it a different way, but I like your solution much more...

Edit: Plenty more where that came from if ya'll are still bored

I can give you guys some Fourier Series and Fourier Transform problems if you like...

Anyone want a linear circut problem involving an Op-Amp?

Your mission should you choose to accept it.

[QUOTE=Connection Problem]You got it Pommy. I solved it a different way, but I like your solution much more...

Edit: Plenty more where that came from if ya'll are still bored :D[/QUOTE]

Yay


What was your way? My physics intuition is subpar and I can't think of another =_=

****, this thread should have died after the second post.

You bunch of dirty nerds.

Okay how about this problem:
Tarzan, who weighs 652 N, swings from a cliff at the end of a convenient vine that is 25 m long (Figure 8-37). From the top of the cliff to the bottom of the swing, he descends by 3.2 m. The vine will break if the force on it exceeds 950 N.


What is the greatest force on the vine during the swing?



Hopefully I'll be able to solve it tomorrow morning and before visiting Massassi during the afternoon.

So far this is what I have...and probably all you're going to get. Not sure if it's right. It's been a while

I may be mistaken, but I think all you are looking for here is the force of tension in the vine, which would be equal to the force of gravity acting on Tarzan, or, his weight.

Nice picture though.

I believe CP is correct.

What about the additional force on the vine nessesary to keep tarzan in a circular trajectory?

Hmm...I just saw something regarding me picture. At the lowest point, the vector of Tarzan moving is perpendicular to weight vector. So W = 0?

At the bottom of the swing (when the force will be maximum)… tension on the vine will be equal to the force of Tarzans weight Plus the force of centripetal acceleration.

Ca = v^2/e

Force equals mass * acceleration so

Fc = m* v^2 /r

Now, we need the velocity of Tarzan and the bottom of his swing.

When Tarzan starts his swing, all of his energy is assumed to be Potential energy (assuming he starts his swing with a velocity of zero), at the bottom of this swing, all the potential energy has been converted to Kinetic energy. Since Potential energy is strictly gravitational in this case,

U = m*g*y.

Kinetic energy is of course

K = 1/2 m * v^2,

And since U = K

m*g*y= ½ m * v^2

which works out to be.

v = sqrt(2 * g * y)

So, plugging all this in…

Fc = m * (sqrt(2 * g * y))^2 / r

Which is

Fc = m * 2 * g * y / r

Plug and chug, and we get a Fc of ~167 N

Tension on the Vine, as stated earlier, is Centripetal force plus the force of gravity, so T = Fc + Fg

T = Fc + Fg

T = 167 + 652
T = ~819 N

So, Tanzan is safe for today…

(Speed Physics)

I'm so jealous. This type of thing just doesn't come natural to me.

Can you build this type of physics intuition? What I do is just do problems till I'm 'good enough', and I pass tests (barely though), but often times I won't find the solution, or won't see it as clearly or elegantly as some of you guys on here obviously do.

The answer is 42!!!! LOLOLOL I AM SO ORIGINAL AND WITTY LOL!

West Wind, may I see what numbers you are using? I have done the same steps as you did, but I keep on getting a large number for the centripetal force. The numbers I'm using are:

U (initial) = K (final)
mgh = 1/2 * m * v^2
gh = 1/2 * m * v^2
v = sqrt ( 2 * g * h)
v = sqrt ( 2 * 9.8m/s^2 * 25)
v = 22.1 m/s

Fc = m (v^2 / R)
Fc = 66.5 kg ( (22.1 m/s) ^ 2 / 28.2m)
Fc = 1152N ??????

Wrong h. Use 3.2m. You are using the rope length. Tarzan doesn't drop 25m. He drops 3.2m. You use that to calculate the potential energy in U=mgh.

Ohh okay, thanks.

Physics is like sex: sure, it may give some practical results, but that's not why we do it. - Richard Feynman

I assume you learned dimensional analysis with units. That helped me alot.