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Joshua2 2008-11-18 07:18

Physics problem Please
 
I know this is math, but I don't know any physics forum, and I've been a member here a long time and like these people. Anyway, I think it is mostly a math problem.

The lower block in the figure is pulled on by a rope with a tension force of 20 N. The coefficient of kinetic friction between the lower block and the surface is 0.30. The coefficient of kinetic friction between the lower block and the upper block is also 0.30.

What is the acceleration of the 2.0 kg block?

[url]http://jjoshua2.googlepages.com/a.p8.28.jpg[/url]
Thats what it looks like.

The answer is 1.77 m/s^2 the book says and that is what I got. Now on the test there is an almost identical problem, same picture. [url]http://jjoshua2.googlepages.com/knight_Figure_08_28.jpg[/url]

A rope pulls on the lower block in the figure with a tension force of 20 N. The coefficient of kinetic friction between the lower block and the surface is 0.37. The coefficient of kinetic friction between the lower block and the upper block is also 0.37. What is the acceleration of the 2.0 kg block?

It says the answer is 2.7, which I guessed since it was the lowest answer, but I think the answer should be lower than 1.77, I calculated about 0.6.

I can ask my teacher in a couple days, but she'll probably get mad at me if I bother her with what should be a non-issue.

Thanks so much! :smile:

hockmeng 2008-11-18 08:24

I tried the problem but I couldn't get 1.77 m s[tex]^{-2}[/tex] for the first question. Looks like I'm having some problem with my physics concepts. Would you mind showing me how you got that figure? Thanks.

axn 2008-11-18 09:07

[CODE](20 - 4*9.8*0.37)/2 = 2.748[/CODE]
Is this not how it should be calculated? :confused:

EDIT:- Probably should be (20 - 4*9.8*0.37)/[B]3[/B] = 1.8

davieddy 2008-11-18 14:45

Applying Newton's 2nd law to the top block we get
T - 9.8*1*0.37 = 1*a

For the bottom block we get
20 - T - 9.8*1*0.37 - 9.8*3*0.37 = 2*a

Adding the equations gives
20 - 9.8*5*0.37 = 3*a

David

Sorry to be too explicit for the homework help roolz :)

Joshua2 2008-11-18 18:31

20 - 2*9.8*1*mu - 9.8*3*mu = 3*a is how I thought of it, giving 0.62 for the 2nd and 1.77 for the first. So I think the 2nd answer is wrong, right?

davieddy 2008-11-18 20:41

[quote=Joshua2;149757]20 - 2*9.8*1*mu - 9.8*3*mu = 3*a is how I thought of it, giving 0.62 for the 2nd and 1.77 for the first. So I think the 2nd answer is wrong, right?[/quote]
Your formula agrees with mine.

Joshua2 2008-11-18 21:05

Right, I was agreeing with yours. I was saying that the tests answer of 2.7 isn't correct and doesn't agree with the similar problem with mu of 0.3 the answer being 1.77. :)

davieddy 2008-11-18 21:49

Yes. More friction means lass acceleration.
Note that we are assuming motion in the direction of the 20N force,

davieddy 2008-11-28 19:25

ou could make this "problem" more interesting
by speifying a moment of inertia, radius and frictional
torque to the pulley.
BTW the problem should have specified g
(which we have been taking as 9,8 m/s^2)

David

flouran 2009-03-02 03:39

[QUOTE=Joshua2;149704]I know this is math, but I don't know any physics forum[/QUOTE]
Ever tried [URL="http://www.physicsforums.com"]http://www.physicsforums.com[/URL]?


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