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

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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

 

[Doc Al]

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.)

 

[Doc Al]

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.

 

[Doc Al]

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.

 

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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

 

[Doc Al]

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.

 

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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. 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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okay this is what i think it should be but it is not working, G = Force of friction = Uk Normal

Uk = G/Normal

 

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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.

 

[Doc Al]

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. 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. $f_s <= \mu_s N$.

 

[Doc Al]

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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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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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

 

[Doc Al]

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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Edit: you i just realized that when i was solving that i have T1 and T2 which will get set to I A

 

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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.

 

[Doc Al]

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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k thanks alot, i just took my physics exam and Aced it i got an A, thanks for all ur help

 

[Doc Al]

Awesome! That's what I like to hear.