Exam III Practice Problems (liquids and pressures)

In summary: Well it is released and the rest of the energy is converted to heat. What was the force of friction then well i have no ideaIn summary, the conversation is about a student preparing for an exam and asking for help with various physics problems. They discuss using the Bernoulli equation, drawing free body diagrams, and solving for tension and forces. They also discuss using multiple choice questions to find the correct equations and using the formula for Archimedes' principle. The conversation also includes multiple questions about specific problems and the process of solving them, including using conservation of energy.
  • #1
Alt+F4
306
0
hi, i got an exam soon and i need help with a couple of problems

http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/sp05 Question 7.


How would i find the tension i know that i must use Bernoulli equation but what would i be solving for
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
What are the forces acting on the block?

Can you draw a free body diagram of the block?

The Bernoulli equation is for fluid flow so I'm not sure why (or how) you would use it here.
 
  • #3
dav2008 said:
What are the forces acting on the block?

Can you draw a free body diagram of the block?

The Bernoulli equation is for fluid flow so I'm not sure why (or how) you would use it here.
ooo well all i have been doing for the past 4 weeks is using that eqaution for HW and stuff so i thought it must be it. Anywyas, You have the Water pushing up, Wood Pushing down and String pulling down
 
  • #4
Right. You have the buoyant force acting upwards, the weight of the block acting downwards and tension acting downwards.

Can you see how to solve for tension now?
 
  • #5
Edit:Never mind
 
  • #6
You are given the volume of the block and its density. Using those facts you can find its mass.

As far as the forces, like you said earlier there are only 3 forces: Tension down, weight down, and buoyancy up.

Recall Archimedes' principle that the buoyancy force is equal to the weight of the displaced fluid.
 
  • #7
thanks, I'll be back with more in 5 min. :)
 
  • #8
dav2008 said:
You are given the volume of the block and its density. Using those facts you can find its mass.

As far as the forces, like you said earlier there are only 3 forces: Tension down, weight down, and buoyancy up.

Recall Archimedes' principle that the buoyancy force is equal to the weight of the displaced fluid.
ya i got it, there are too many stuff that are same like V= Velocity but then it is Volume in that equation. Crazy
 
  • #9
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/sp05 Question 15. ANy trick to finding the quick answer to these. Thanks
 
Last edited by a moderator:
  • #10
Well given that it's a multiple choice you can just see which one of those three equations fits your given information.

You're given that at t=0 s v=11.1 m/s.
 
  • #11
Thanks alot, i need to ace this exam so sorry for all the questions.

http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/sp05 Question 16

Can i use one of these equations for 15 to just plug in 1.5 and get the V which i can get Kinetic energy from
 
Last edited by a moderator:
  • #12
Well try it and see if you get the right answer.
 
  • #13
dav2008 said:
Well try it and see if you get the right answer.
:( nope well i was thinking of using X(t) = A cost(omega t) but what would A be hmmmm
 
  • #14
Im stupid. I got it
 
Last edited:
  • #15
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/sp05 Question 19. I am not getting why Pa > Pb. Thinking of it B is a more tighter area so there will be more pressue on it
 
Last edited by a moderator:
  • #16
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa04 Question 7. SO what i did was

Force of Friction - Force of compression of spring
UkMg - -kx = 0
(X)(20*9.8) = 270 * .2

What am i doing wrong.
 
Last edited by a moderator:
  • #17
Question 25 http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa04

Okay so i got the answer to 24 which was 600. FOR 25 i am going to use this formula h = (M + m)/ (Density* A)

D= M/v
600 = M/ (2*.15) or is it (2*.25)
little m is just 20*3 = 60
Density on bottom is 1000
A = .25*2
plug that it and i am not getting the answer
 
Last edited by a moderator:
  • #18
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa05 Question 10. hmmmmmmm Well wat i did was

(1000)(9.8)(.3) = (X)(9.8)(.6)
Which gave me 500 which is wrong.
 
Last edited by a moderator:
  • #19
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa05 Question 19. I think this should be it for a while, Thanks for everyones help.

So what i did was

Patm = P1 + .5(1000)(2^2) + (1000*9.8*2)
Which gives me

Patm - PB = 2.2*10^4 (not the answer)

Answer: 1.8*10^4
 
Last edited by a moderator:
  • #20
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa04 Question 7. SO what i did was

Force of Friction - Force of compression of spring
UkMg - -kx = 0
(X)(20*9.8) = 270 * .2

What am i doing wrong.
I drew out free body diagram and i still can't get it
 
Last edited by a moderator:
  • #21
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa04 Question 7. SO what i did was

Force of Friction - Force of compression of spring
UkMg - -kx = 0
(X)(20*9.8) = 270 * .2

What am i doing wrong.
For that one


Would it be

Force of friction - (The Total Work of spring) = 0

so

Mu mg - .5KX^2 = 0

but i can't get the answer
 
Last edited by a moderator:
  • #22
You can't subtract work from force. It's a meaningless expression

You need a statement of conservation of energy. What energy does the block have when it is pulled .2m, and where does that energy go?
 
  • #23
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa05 Question 10. hmmmmmmm Well wat i did was

(1000)(9.8)(.3) = (X)(9.8)(.6)
Which gave me 500 which is wrong.
Think about what you are doing. What you said in your expression is that the pressure due to the column of substance b is equal to the pressure of the half column of substance A on the left side. Why did you assume that is true?

Think about which pressures have to be equal.
 
Last edited by a moderator:
  • #24
dav2008 said:
You can't subtract work from force. It's a meaningless expression

You need a statement of conservation of energy. What energy does the block have when it is pulled .2m, and where does that energy go?
okay so Potential Energy is .5KX^2 = 5.4 J

Energy then is converted to Kinetic which is also 5.4 J
 
  • #25
You're right that's converted to kinetic but you forgot about friction.

Compare the two states that the block is in:

State 1: all potential energy equal to .5*k*x2

State 2: No potential energy (x=0), no kinetic energy (they told you it's not moving), so the energy must have gone into friction.

Do you know how to find work done by a force over a distance?
 
  • #26
Honestly I think you need to think a bit more about the concepts behind the problems instead of just trying to find an equation to blindly plug numbers into and then posting on here when that equation doesn't work.
 
  • #27
dav2008 said:
You're right that's converted to kinetic but you forgot about friction.

Compare the two states that the block is in:

State 1: all potential energy equal to .5*k*x2

State 2: No potential energy (x=0), no kinetic energy (they told you it's not moving), so the energy must have gone into friction.

Do you know how to find work done by a force over a distance?
okay so are u saying that 5.4J = the energy of friction

So what then do i use to get an asnwer of .14

casue i did 5.4 = X(20*9.8)
 
  • #28
dav2008 said:
Honestly I think you need to think a bit more about the concepts behind the problems instead of just trying to find an equation to blindly plug numbers into and then posting on here when that equation doesn't work.
i would if we had a book, but the lectures as u can see on the site are basics so i just gota do it by trial and error that is why there are so many exams given
 
  • #29
Look at what you're doing.

x(20)(9.8) gives you the force of friction.

That force acts over a certain distance.

Now you want to find the work done by that force.

Do you think you can find it given a force and the distance over which it acts?
 
  • #30
dav2008 said:
Look at what you're doing.

x(20)(9.8) gives you the force of friction.

That force acts over a certain distance.

Now you want to find the work done by that force.

Do you think you can find it given a force and the distance over which it acts?
oooooooooooooooooooo Thanks so much, i forgot totally about W = F*D
 
  • #31
Alt+F4 said:
Question 25 http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa04

Okay so i got the answer to 24 which was 600. FOR 25 i am going to use this formula h = (M + m)/ (Density* A)

D= M/v
600 = M/ (2*.15) or is it (2*.25)
little m is just 20*3 = 60
Density on bottom is 1000
A = .25*2
plug that it and i am not getting the answer





Forget what i did upthere

this is what i did

the buyoant force has to balance the weight

Fb = m*9.8
(1000)(9.8)(X) = (300+60)(9.8)

X=.36

.15-.36 = .21 m which is not answer
 
Last edited by a moderator:
  • #32
Alt+F4 said:
Forget what i did upthere

this is what i did

the buyoant force has to balance the weight

Fb = m*9.8
(1000)(9.8)(X) = (300+60)(9.8)

X=.36

.15-.36 = .21 m which is not answer
Two problems:

(1) Why do you think the mass of the wood is 300 kg?

(2) The "X" in your equation stands for volume displaced, not depth.
 
  • #33
Doc Al said:
Two problems:

(1) Why do you think the mass of the wood is 300 kg?

(2) The "X" in your equation stands for volume displaced, not depth.
cause i got a density of 600

So what i did was

D= M/V

600 = M/ (2*.25) which gave me 300
 
  • #34
Doc Al said:
Two problems:

(1) Why do you think the mass of the wood is 300 kg?

(2) The "X" in your equation stands for volume displaced, not depth.
well i was thinking since it is already at .15, whatever the number i will get will have the .15 added to it
 
  • #35
Alt+F4 said:
cause i got a density of 600

So what i did was

D= M/V

600 = M/ (2*.25) which gave me 300
Volume requires all three dimesions. You multiplied length by height, but what about width? (If you include units in your calculations, you'll be less likely to make this kind of error.)
 

Similar threads

Solve Physics 101 Practice Exam 2 Problem 21: Total Kinetic Energy
    • Introductory Physics Homework Help
2
Replies
53
Views
5K
Final Exam Q6: Finding Problem with Torque Equation
    • Introductory Physics Homework Help
4
Replies
112
Views
9K
A question about exam problems for Engineering students
    • STEM Academic Advising
Replies
2
Views
698
C/C++ Preparing for Exam with Programming Problem Practice
    • Programming and Computer Science
Replies
1
Views
1K
Is Walter Lewins Loop Trick. Correct or False?
    • Introductory Physics Homework Help
Replies
18
Views
1K
Find the maximum AC current flowing through the generator.
    • Introductory Physics Homework Help
Replies
2
Views
4K
Can I Use My Calculator for the Upcoming Physics Exam?
    • Introductory Physics Homework Help
2
Replies
66
Views
6K
UIUC Phys 102: Exam 3 Prep - Questions 4-6
    • Introductory Physics Homework Help
Replies
10
Views
2K
Planning on studying math for the college entrance exam, help needed
    • STEM Academic Advising
Replies
4
Views
1K
How to use extra information in physics exams
    • STEM Academic Advising
Replies
5
Views
973

Hot Threads

Recent Insights

Back
Top