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

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In summary, the problem wants the balance point of a weight hanging from a string. The balance point is found by using the formula Px/Py. Once the balance point is found, the center of mass (CM) can be determined by multiplying the weight by the string's tension.
  • #1
Alt+F4
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http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/sp05

Problem 21 Answer (D)

So i found the total kinetic energy of the whole system which is 119.25 J then what. Thanks these are practice exams so can u please just tell me how to solve them. Thanks
 
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  • #2
Mechanical energy (PE + KE) of the system is conserved. Figure out the change in gravitational PE.
 
  • #3
and how could question 9 be Postive i thought gravity always does negative work
 
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  • #4
Doc Al said:
Mechanical energy (PE + KE) of the system is conserved. Figure out the change in gravitational PE.
so MGH = 119.25 but that doesn't work out
 
  • #5
[tex]\Delta {PE} + \Delta {KE} = 0[/tex]
 
  • #6
Alt+F4 said:
and how could question 9 be Postive i thought gravity always does negative work
Why would you think that? If the force and the displacement are in the same direction, the work done by the force will be positive; if they are in opposite directions, the work is negative.
 
  • #7
Doc Al said:
Why would you think that? If the force and the displacement are in the same direction, the work done by the force will be positive; if they are in opposite directions, the work is negative.
according to the formula Wgrav = -mgy

Okay so if they ask if the ball swing from C to B is that still positve or would it be negative now
 
  • #8
Alt+F4 said:
according to the formula Wgrav = -mgy
But y can be positive or negative.

Okay so if they ask if the ball swing from C to B is that still positve or would it be negative now
You tell me.
 
  • #9
Doc Al said:
But y can be positive or negative.


You tell me.
hmmm i would have to say postive since like u said up there force and displacement would be in the same direction. So can i safely assume for these type of problems on exam gravity will always be positive then?
 
  • #10
K problem 21. i Have (8)(9.8)(X) + (6)(9.8)(X) + .5 (8)(3^2) + .5 (6)(3^2) + .5 (2)*(3/.4)^2 = 0
So where am i doing mistake or is this even right?
 
  • #11
Alt+F4 said:
hmmm i would have to say postive since like u said up there force and displacement would be in the same direction.
Right. If the ball falls, the work done by gravity is positive.


So can i safely assume for these type of problems on exam gravity will always be positive then?
Don't know what you mean. Do you mean will the work done by gravity always be positive? Of course not. (Balls can rise as well as fall.)

The way to understand that formula is as follows:
[tex]W = Fd[/tex]
[tex]W_{grav} = (-mg)(\Delta y)[/tex]

This uses the convention that down is negative. So when something falls, [itex]\Delta y [/itex] is negative.
 
  • #12
Alt+F4 said:
K problem 21. i Have (8)(9.8)(X) + (6)(9.8)(X) + .5 (8)(3^2) + .5 (6)(3^2) + .5 (2)*(3/.4)^2 = 0
So where am i doing mistake or is this even right?
The KE terms look good, but the PE needs some work. Realize that one mass goes down a distance h (what's its change in PE?), while the other goes up a distance h (what's its change in PE?).
 
  • #13
thanks alot, for your help. Question 24. This is a torque problem , so i make the fulcrum be zero and the equation would be -GL + -GL ( sin or cos). so (-1)(9.8)(1) + (X)(1) (sin/cos) I don't understand why Cos was used instead of sin. I had the answers narrowed down to D and E. Answer is E
 
  • #14
Alt+F4 said:
I don't understand why Cos was used instead of sin.
Depending on how you are taught to find the torque:

(1) Torque = Force X moment arm; moment arm is the perpendicular distance between the line of the force and the pivot point. That perpendicular distance is L cos(35)

(2) Torque = Force X distance sin(theta); the angle theta is the angle between the force (in this case downward) and the distance along the lever. Theta = 90 degrees + 35 degrees; mathematically, sin (90 + x) = cos (x).
 
  • #15
sorry for all these questions, i just wana know if i had the right logic for 15 on this exam. http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/sp04

Okay so i figured the PE of the top and set that equal Kinetic and got a velocity of 5.77. so i assumed that it will start at a low V since the height has to be lower than 1.7 and has to work up to 5.77 inorder to equal the potential energy up there so speed has to increase. Thanks
 
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  • #16
soory last problem of the day http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/sp04 QUestion 16. What is this problem wanting i know it wants the balance point so i am trying to find the center mass by using the formula so i get (50)((6.5+22)/2)) + (610 * 6.5) / (610+50).. Thanks for all your help.
 
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  • #17
ok i lied one more http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/fa03 11 and 12. Okay i know
Tan 15 = Px/Py.
i got the verticel and horizonatl of the after collision which is .77 and 2.89. Momentum is conserved. What do i do after to find the X and Y of Mass M. Thanks
 
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  • #18
Alt+F4 said:
sorry for all these questions, i just wana know if i had the right logic for 15 on this exam. http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/sp04

Okay so i figured the PE of the top and set that equal Kinetic and got a velocity of 5.77. so i assumed that it will start at a low V since the height has to be lower than 1.7 and has to work up to 5.77 inorder to equal the potential energy up there so speed has to increase. Thanks
Don't really know what you're doing here. In problem 14, you calculated the work done by the 30N force. The work done will equal the change in total mechanical energy.

Another way to solve this is to find the acceleration of the block using Newton's 2nd law.
 
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  • #19
Alt+F4 said:
soory last problem of the day http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/sp04 QUestion 16. What is this problem wanting i know it wants the balance point so i am trying to find the center mass by using the formula so i get (50)((6.5+22)/2)) + (610 * 6.5) / (610+50).. Thanks for all your help.
All they want is the position of the center of mass of the beam. The beam is uniform, so where is its center of mass?
 
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  • #20
Alt+F4 said:
ok i lied one more http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/fa03 11 and 12. Okay i know
Tan 15 = Px/Py.
i got the verticel and horizonatl of the after collision which is .77 and 2.89. Momentum is conserved. What do i do after to find the X and Y of Mass M. Thanks
Total momentum is conserved. What is the momentum of the system before the collision? What's the x-component of that momentum? The y-component?

You can figure the x and y components of the momentum of 2M after the collision, since you are given its velocity. Subtract that from the x and y components of the total momentum to find the momentum of M, and then its velocity.

Remember: Momentum and velocity are vectors; direction and signs count.
 
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  • #21
Doc Al said:
Don't really know what you're doing here. In problem 14, you calculated the work done by the 30N force. The work done will equal the change in total mechanical energy.

Another way to solve this is to find the acceleration of the block using Newton's 2nd law.
for this one i was saying that take the height at the start of the ramp which is near 0 which will give a Velocity of 0. Now go all the way on top and PE = .5MV^2 V becomes 5.77. So if it strarts at 0 bottom and end at 5.77 on top that means the speed had to increase. So using this logic say if the reverse the ramp and ur going down would the speed of the block also decrease?
 
  • #22
Alt+F4 said:
for this one i was saying that take the height at the start of the ramp which is near 0 which will give a Velocity of 0. Now go all the way on top and PE = .5MV^2 V becomes 5.77.
Sorry, but this makes no sense. For one thing, energy is not conserved--there's a 30N force acting on the block. For another, PE increases as you go up the ramp, so (per energy conservation) KE should decrease. (It doesn't, due to the 30N force.) You can't ignore the work done by the 30N force; if it wasn't for that force, the block would slide down the ramp, not up.
 
  • #23
Doc Al said:
Total momentum is conserved. What is the momentum of the system before the collision? What's the x-component of that momentum? The y-component?

You can figure the x and y components of the momentum of 2M after the collision, since you are given its velocity. Subtract that from the x and y components of the total momentum to find the momentum of M, and then its velocity.

Remember: Momentum and velocity are vectors; direction and signs count.
okay so for this one. The Y component of 2m is 2.8977 m/s and the X component is .776 M/s. Momentum Before collision is (2M)(6) = 12M
12M = (2M)(2.897)+MX
X= 6.20 M/s.
Got that what i wana know is to find the X axis do u find the momentum of the X axis before collision which is 0 because i got 1.55 m/s or would u still use the 12M. Also on exam to double check will the angle that M bounces off be the same as the angle 2M bounces off thanks for ur help
 
  • #24
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/fa03 Question 19. Okay so to start this I knew that

Fwallx -Tx = 0
Fwally -Ty-Fg = 0
then what do i do from there. I have all the #'s needed, does torque = 0 at the midpoint and should i work from there?
 
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  • #25
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,
 
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  • #26
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
 
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  • #27
can somone help with these above problems
 
  • #28
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

Because not all of that energy is used up traveling down the slope because the block rebounds. Try considering forces instead.

-Hoot:smile:
 
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  • #29
Alt+F4 said:
okay so for this one. The Y component of 2m is 2.8977 m/s and the X component is .776 M/s.
OK, those are the components of 2M's velocity after the collision. Note that the x-component points to the left, so it should be negative.
Momentum Before collision is (2M)(6) = 12M
12M = (2M)(2.897)+MX
X= 6.20 M/s.
OK, what you've found is the y-component of M's velocity after the collision.
Got that what i wana know is to find the X axis do u find the momentum of the X axis before collision which is 0 because i got 1.55 m/s or would u still use the 12M.
Momentum is a vector and must be treated as such. The x-component of the total momentum is 0; that's what you need to use.

Also on exam to double check will the angle that M bounces off be the same as the angle 2M bounces off thanks for ur help
Figure it out and see. Since you know the x and y components of M's velocity after the collision, you have all the info needed to figure out the angle it bounces off at.
 
  • #30
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/fa03 Question 19. Okay so to start this I knew that

Fwallx -Tx = 0
Fwally -Ty-Fg = 0
then what do i do from there. I have all the #'s needed, does torque = 0 at the midpoint and should i work from there?
Identify all the forces on the rod: weight, cable tension, lamp, wall.

The torque about any point must be zero. If you choose the pivot point wisely, you can eliminate unknowns.
 
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  • #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
 

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