Solve Physics 101 Practice Exam 2 Problem 21: Total Kinetic Energy

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The discussion revolves around solving a physics problem related to total kinetic energy and mechanical energy conservation. Participants calculated the total kinetic energy of a system and discussed the implications of gravitational potential energy changes, emphasizing that work done by gravity can be positive or negative depending on the direction of force and displacement. Clarifications were sought regarding the correct application of formulas for work and torque, with specific attention to the use of sine and cosine in torque calculations. The conversation also touched on momentum conservation in collision problems, highlighting the importance of vector components in determining final velocities. Overall, the thread illustrates the complexities of applying physics concepts to problem-solving in a practice exam context.
  • #31
Alt+F4 said:
Question 22. http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/sp04

Okay so I = .5 (M)(R)^2 for cylinder
I = 112.5 M.

Work = F *D
Work = F * 80
Wnet = THe change in Kinetic Energy
and i don't know what to do from there,
This is an equilibrium problem: no need for rotational inertia or energy methods. Hint: What must be the net torque on the cylinder?
 
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  • #32
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/fa04 Question 17.

Wnet = Change in Kinetic Energy = (.50(2)(3^2) = 9
W = F*D * Cos (theta)

Ukmg * D * Cos (41) = 9

Why doesn't that work. Thanks
Two reasons:

(1) You must consider the net force along the incline; you considered friction, but ignored gravity.

(2) The angle in the formula W = F*D * Cos (theta) is between the force and the displacement. Since the displacement is along the incline, find the net force along the incline and then theta would be zero.
 
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  • #33
I've just realized my previous post was rather misleading, Doc Al has it covered though.
 
  • #34
Doc Al said:
Two reasons:

(1) You must consider the net force along the incline; you considered friction, but ignored gravity.

(2) The angle in the formula W = F*D * Cos (theta) is between the force and the displacement. Since the displacement is along the incline, find the net force along the incline and then theta would be zero.
Okay So

mgsin41 - Force of friction = (2)(9.8)(sin41) - (.27)(2)(9.8) = 7.56675 ( i am guessing this is the energy that got lost due to friction or the work that was done to overcome friction) so from there W = F *D Cos (theta) and i am lost.

Do i figure what the Kinetic energy at the top of the block was then subtract it from 9.
 
  • #35
Doc Al said:
This is an equilibrium problem: no need for rotational inertia or energy methods. Hint: What must be the net torque on the cylinder?
i know the sum of torques should add up to zero. so t= F*L t = F * 80. i have no idea any more hints. I did 15*9.8*50 / ( 80) and got the answer don't know if that was luck or i actually understood it
 
  • #36
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/fa02 Question 15, i did like 10000 of these why isn't this working

torque = Zero at Cable

-(85)(2)(9.8) - (15 * 9.8 *1 ) + F (.5) = 0

F = 3626 N
what am i doing wrong ? thanks
 
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  • #37
Alt+F4 said:
Okay So

mgsin41 - Force of friction = (2)(9.8)(sin41) - (.27)(2)(9.8) = 7.56675 ( i am guessing this is the energy that got lost due to friction or the work that was done to overcome friction) so from there W = F *D Cos (theta) and i am lost.

Do i figure what the Kinetic energy at the top of the block was then subtract it from 9.
(2)(9.8)(sin41) is correct for the component of gravity down the incline, but (.27)(2)(9.8) is not correct for the friction force. (Remember that F_f = \mu N.)

Once you find the correct force, use it to find the work done on the block.
The net work on the block equals the change in its KE. (It starts from rest.)
 
  • #38
Alt+F4 said:
i know the sum of torques should add up to zero. so t= F*L t = F * 80. i have no idea any more hints. I did 15*9.8*50 / ( 80) and got the answer don't know if that was luck or i actually understood it
That's all there is to it. The hanging mass exerts a counter-clockwise torque on the cylinder; the force on the handle exerts a clockwise torque. These must be equal (and opposite) for rotational equilibrium.
 
  • #39
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/fa02 Question 15, i did like 10000 of these why isn't this working

torque = Zero at Cable

-(85)(2)(9.8) - (15 * 9.8 *1 ) + F (.5) = 0

F = 3626 N
what am i doing wrong ? thanks
What you're doing wrong is jumping the gun and not reading the question. You immediately went to solve for the support force, but that was not the question.

All they wanted was the torque exerted by the man's weight as measured about the support point. Too easy. :smile:
 
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  • #40
Okay, Friction force = mgcos41 = 3.9999
Force of gravity = 12.8587
Wnet = 9

Force of gravity - Force of friction = 12.8587-3.9999 = 8.8597

W = F* D = 9
D = (9)/ (8.8597) = 1.01 = 1m for the answer, is this how its done
 
  • #41
Alt+F4 said:
Okay, Friction force = mgcos41 = 3.9999
Force of gravity = 12.8587
Wnet = 9

Force of gravity - Force of friction = 12.8587-3.9999 = 8.8597

W = F* D = 9
D = (9)/ (8.8597) = 1.01 = 1m for the answer, is this how its done
Yes, that's how it's done.
 
  • #42
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp05 Question 19. My exam is in 2 hrs so can i have strong hints. :smile: Thanks. Okay so 3 forces acting on car, Inward Centripital, normal and Gravity. Centripetal Acc = 7.5 m/s^2 how do i find friction thanks
 
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  • #43
okay this is what i think it should be but it is not working, G = Force of friction = Uk Normal

Uk = G/Normal
 
  • #44
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp04 Question 25 +26. Sorry bout all these question but my town was hit by the tornado so i had no power till an hour ago. kay to find Anglular frequency wat do i do.
 
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  • #45
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp05 Question 19. My exam is in 2 hrs so can i have strong hints. :smile: Thanks. Okay so 3 forces acting on car, Inward Centripital, normal and Gravity. Centripetal Acc = 7.5 m/s^2 how do i find friction thanks
"centripetal" is a direction, not a kind of force. The three forces are: friction, normal force (what is that?), and weight.

Use Newton's 2nd law to find the force; then realize that the only force acting centripetally is the friction force. f_s &lt;= \mu_s N<br />.
 
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  • #46
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp04 Question 25 +26. Sorry bout all these question but my town was hit by the tornado so i had no power till an hour ago. kay to find Anglular frequency wat do i do.
Hint: What must be the same about the two wheels, since the belt doesn't slip?
 
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  • #47
okay last problem, for the day i hope

http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/sp04 Question 2. I know this is prob a torque problem

so Torque = I (alpha a)
Alpha A = (6.7/.12)= 55.83
So how would i got bout solvin torque to get I so i can find mass. Thanks sooooooooooooooooooooooooooooooooo much
 
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  • #48
Doc Al said:
Hint: What must be the same about the two wheels, since the belt doesn't slip?
ooo man i am so stupid. I wish i had hints like these durin exams hahaa i was making it more complicated than it is. thanks
 
  • #49
Alt+F4 said:
okay last problem, for the day i hope

http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/sp04 Question 2. I know this is prob a torque problem

so Torque = I (alpha a)
Alpha A = (6.7/.12)= 55.83
So how would i got bout solvin torque to get I so i can find mass. Thanks sooooooooooooooooooooooooooooooooo much
You have all the info needed to figure out T_1 & T_2.
 
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  • #50
Edit: you i just realized that when i was solving that i have T1 and T2 which will get set to I A
 
  • #51
Okay, Last 2 i promise, http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/sp05 Okay so i drew a free body diagram and i have Friction froce - Tension = Ia ?? Right or am i missing somethin cause i am not getting the asnwer. and http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa03 6. I tried doing everything, anyhint i also drew free body diagram.
 
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  • #52
Alt+F4 said:
Okay, Last 2 i promise, http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/sp05 Okay so i drew a free body diagram and i have Friction froce - Tension = Ia ?? Right or am i missing somethin cause i am not getting the asnwer.
I assume you're talking about problem 4. Don't confuse force with torque. Hints: What is \alpha? (Key: rolling without slipping.) What force produces the torque about the center of mass?

and http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa03 6. I tried doing everything, anyhint i also drew free body diagram.
One way to solve this is to use conservation of energy. Hint: Realize that the cylinder will have both rotational and translational KE. (The two are related, since it rolls without slipping.)
 
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  • #53
k thanks alot, i just took my physics exam and Aced it i got an A, :smile: thanks for all ur help
 
  • #54
Awesome! That's what I like to hear.
 

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