Lab Quiz 4
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Lab Quiz 4
Question 1: (1 point)
In your lab on the conservation of total mechanical energy you have a 24 gram weight (mass m) connected via a string with a 276 gram glider, which glides horizontally with negligible friction as shown above. When you release the mass m, it and the glider are accelerated due to the weight of the mass m. What is the change in potential energy PE (Ch6 sheet 12) of the mass m when the glider moves a distance of 111.4 cm? Indicate with a positive (negative) sign whether the change of the potential energy of the glider is nonzero (zero).
(How to solve):
Question 2: (1 point)
In a setup as shown above you neglect the mass of the string: What is the change in total kinetic energy of the 18.2 gram mass m and the 289 gram glider M combined, if the glider moves a distance of 93.6 cm? Indicate with a negative (positive) sign whether the inclusion of a nonzero mass of the string would decrease (increase) your calculated value. (Hint: Since the change in the velocity of the two masses is not given, you use conservation of the total mechanical energy (see Ch6 sheet 15) in your calculation.)
(How to solve):
Question 3: (1 point)
You check conservation of the total mechanical energy by comparing the change in potential energy of the mass m with the change in kinetic energy, delta KE, of the masses m and M combined, which is 304 grams. The change in kinetic energy is KE = 1/2(m+M)v2 (see Ch6 sheet , where v is the velocity of the glider when starting with zero initial velocity, v = 0.5 m/s. In your error calculations you have previously determined the error of the term [ 1/2(m+M) ] to be 0.965 %, and the relative error of the velocity v to be 0.00662 . What is the absolute error of KE? Indicate with a positive (negative) sign whether the expected slope of a plot of potential energy change PE versus kinetic energy change KE is 0 (1). (Hint: You use expression( in "Error and Uncertainty" ("EU") to get (v2)/v2 from v/v and expression(7) in "EU" to get the error of KE.)
(How to solve):
In your lab on the conservation of total mechanical energy you have a 24 gram weight (mass m) connected via a string with a 276 gram glider, which glides horizontally with negligible friction as shown above. When you release the mass m, it and the glider are accelerated due to the weight of the mass m. What is the change in potential energy PE (Ch6 sheet 12) of the mass m when the glider moves a distance of 111.4 cm? Indicate with a positive (negative) sign whether the change of the potential energy of the glider is nonzero (zero).
(How to solve):
Question 2: (1 point)
In a setup as shown above you neglect the mass of the string: What is the change in total kinetic energy of the 18.2 gram mass m and the 289 gram glider M combined, if the glider moves a distance of 93.6 cm? Indicate with a negative (positive) sign whether the inclusion of a nonzero mass of the string would decrease (increase) your calculated value. (Hint: Since the change in the velocity of the two masses is not given, you use conservation of the total mechanical energy (see Ch6 sheet 15) in your calculation.)
(How to solve):
Question 3: (1 point)
You check conservation of the total mechanical energy by comparing the change in potential energy of the mass m with the change in kinetic energy, delta KE, of the masses m and M combined, which is 304 grams. The change in kinetic energy is KE = 1/2(m+M)v2 (see Ch6 sheet , where v is the velocity of the glider when starting with zero initial velocity, v = 0.5 m/s. In your error calculations you have previously determined the error of the term [ 1/2(m+M) ] to be 0.965 %, and the relative error of the velocity v to be 0.00662 . What is the absolute error of KE? Indicate with a positive (negative) sign whether the expected slope of a plot of potential energy change PE versus kinetic energy change KE is 0 (1). (Hint: You use expression( in "Error and Uncertainty" ("EU") to get (v2)/v2 from v/v and expression(7) in "EU" to get the error of KE.)
(How to solve):
gguest Guest
Question #1 and #2
Question 1: (1 point)
In your lab on the conservation of total mechanical energy you have a 24 gram weight (mass m) connected via a string with a 276 gram glider, which glides horizontally with negligible friction as shown above. When you release the mass m, it and the glider are accelerated due to the weight of the mass m. What is the change in potential energy PE (Ch6 sheet 12) of the mass m when the glider moves a distance of 111.4 cm? Indicate with a positive (negative) sign whether the change of the potential energy of the glider is nonzero (zero).
(How to solve):
Neglect the mass of the string (276 grams)
W = PE = mgh *(remember to convert to SI units!)
PE = (.024kg)(9.81m/s^2)(1.114m)
PE = 0.26228 J
(the answer is negative)
Question 2: (1 point)
In a setup as shown above you neglect the mass of the string: What is the change in total kinetic energy of the 18.2 gram mass m and the 289 gram glider M combined, if the glider moves a distance of 93.6 cm? Indicate with a negative (positive) sign whether the inclusion of a nonzero mass of the string would decrease (increase) your calculated value. (Hint: Since the change in the velocity of the two masses is not given, you use conservation of the total mechanical energy (see Ch6 sheet 15) in your calculation.)
(How to solve):
Neglect the mass of the string (291 grams)
W = PE = mgh *(remember to convert to SI units!)
PE = (.0182kg)(9.81m/s^2)(.936m)
PE = 0.167115 J
(answer is positive)
Does anyone know how to do #3
(the answer is negative)
In your lab on the conservation of total mechanical energy you have a 24 gram weight (mass m) connected via a string with a 276 gram glider, which glides horizontally with negligible friction as shown above. When you release the mass m, it and the glider are accelerated due to the weight of the mass m. What is the change in potential energy PE (Ch6 sheet 12) of the mass m when the glider moves a distance of 111.4 cm? Indicate with a positive (negative) sign whether the change of the potential energy of the glider is nonzero (zero).
(How to solve):
Neglect the mass of the string (276 grams)
W = PE = mgh *(remember to convert to SI units!)
PE = (.024kg)(9.81m/s^2)(1.114m)
PE = 0.26228 J
(the answer is negative)
Question 2: (1 point)
In a setup as shown above you neglect the mass of the string: What is the change in total kinetic energy of the 18.2 gram mass m and the 289 gram glider M combined, if the glider moves a distance of 93.6 cm? Indicate with a negative (positive) sign whether the inclusion of a nonzero mass of the string would decrease (increase) your calculated value. (Hint: Since the change in the velocity of the two masses is not given, you use conservation of the total mechanical energy (see Ch6 sheet 15) in your calculation.)
(How to solve):
Neglect the mass of the string (291 grams)
W = PE = mgh *(remember to convert to SI units!)
PE = (.0182kg)(9.81m/s^2)(.936m)
PE = 0.167115 J
(answer is positive)
Does anyone know how to do #3
(the answer is negative)
periwinkle Posts : 22
Join date : 20080917
Re: Lab Quiz 4
Okay the question actually gives us the formula for Question 3, you just have to keep going back and watch your work.
Watch out for SI units. You want kg!
∆KE = (1/2)*KE*(V^{2}
V = 0.5m/s^{2} (Note: this should be constant, but make sure, it is given to you in the question outright)
∆V = 0.0xxx (Again this is given to you and is a variable with each try)
%∆ = 0.965 (Again this is given to you and is a variable with each try)
N = The value you get from the error formula.
Take the percentage and times it by 100. We do this because of our equation X%=((∆X/X)*100); we are just reversing the process.
For this problem in particular:
√((0.038/0.304)^{2}+(0.00662/0.5)^{2})=N
√((0.125)^{2}+(0.01324)^{2})=N
√((0.015625)+(0.0001752976))=N
N=0.1256992347
______________________________________________
N/(.965*100)
0.1256992347/(.965*100)
0.1256992347/96.5
=0.0013025827
0.0013025827 is the final answer and can be rounded to the nearest thousandth: 0.001
You can probably get away with taking the test repeatedly and entering 0.001 for Question 3 until it comes up right. Other wise the answer should be that small, most likely 0.002 and perhaps on occasion 0.003 but I doubt it will even be that high.
Watch out for SI units. You want kg!
KE = .304kg (This is given to you and is a variable with each try)√((∆KE/KE)^{2}+(∆V/V)^{2}) = N ; N/(%∆*100)
∆KE = (1/2)*KE*(V^{2}
V = 0.5m/s^{2} (Note: this should be constant, but make sure, it is given to you in the question outright)
∆V = 0.0xxx (Again this is given to you and is a variable with each try)
%∆ = 0.965 (Again this is given to you and is a variable with each try)
N = The value you get from the error formula.
Take the percentage and times it by 100. We do this because of our equation X%=((∆X/X)*100); we are just reversing the process.
For this problem in particular:
√((0.038/0.304)^{2}+(0.00662/0.5)^{2})=N
√((0.125)^{2}+(0.01324)^{2})=N
√((0.015625)+(0.0001752976))=N
N=0.1256992347
______________________________________________
N/(.965*100)
0.1256992347/(.965*100)
0.1256992347/96.5
=0.0013025827
0.0013025827 is the final answer and can be rounded to the nearest thousandth: 0.001
You can probably get away with taking the test repeatedly and entering 0.001 for Question 3 until it comes up right. Other wise the answer should be that small, most likely 0.002 and perhaps on occasion 0.003 but I doubt it will even be that high.
Guest01 Posts : 133
Join date : 20080919
number 3
I'm trying to do number 3 as well but its not correct
barbie admin Admin
 Posts : 1
Join date : 20080917
Re: Lab Quiz 4
Still having trouble? Can you post your calculations and work?
Guest01 Posts : 133
Join date : 20080919
Question 3
So the question goes:
You check conservation of the total mechanical energy by comparing the change in potential energy of the mass m with the change in kinetic energy, delta dKE, of the masses m and M combined, which is 329grams . The change in kinetic energy is dKE = 1/2(m+M)v2 (see Ch6 sheet , where v is the velocity of the glider when starting with zero initial velocity, v = 0.5 m/s. In your error calculations you have previously determined the error of the term [ 1/2(m+M) ] to be 0.9%, and the relative error of the velocity v to be 0.00925. What is the absolute error of dKE ? Indicate with a positive (negative) sign whether the expected slope of a plot of potential energy change PE versus kinetic energy change dKE is 0 (1). (Hint: you use expression( in "Error and Uncertainty" ("EU") to get D(v2)/v2 from Dv/v and expression(7) in "EU" to get the error of dKE)
The answer: 0.000846
But I got a different answer every time I do it ...This is my work:
dKE = 1/2*KE*v^2
dKE = 1/2*.329*.5^2
dKE = 0.041125
N = √((0.041125/0.329)^2+(0.00925/0.5)^2)
N = √((0.125)^2+(0.0185)^2)
N = √((0.015625)+(0.00034225))
N = 0.1263615844
______________________________________________
N/(.9*100)
0.1263615844/(90)
= 0.0014040176
I'm really lost on this...
You check conservation of the total mechanical energy by comparing the change in potential energy of the mass m with the change in kinetic energy, delta dKE, of the masses m and M combined, which is 329grams . The change in kinetic energy is dKE = 1/2(m+M)v2 (see Ch6 sheet , where v is the velocity of the glider when starting with zero initial velocity, v = 0.5 m/s. In your error calculations you have previously determined the error of the term [ 1/2(m+M) ] to be 0.9%, and the relative error of the velocity v to be 0.00925. What is the absolute error of dKE ? Indicate with a positive (negative) sign whether the expected slope of a plot of potential energy change PE versus kinetic energy change dKE is 0 (1). (Hint: you use expression( in "Error and Uncertainty" ("EU") to get D(v2)/v2 from Dv/v and expression(7) in "EU" to get the error of dKE)
The answer: 0.000846
But I got a different answer every time I do it ...This is my work:
dKE = 1/2*KE*v^2
dKE = 1/2*.329*.5^2
dKE = 0.041125
N = √((0.041125/0.329)^2+(0.00925/0.5)^2)
N = √((0.125)^2+(0.0185)^2)
N = √((0.015625)+(0.00034225))
N = 0.1263615844
______________________________________________
N/(.9*100)
0.1263615844/(90)
= 0.0014040176
I'm really lost on this...
periwinkle Posts : 22
Join date : 20080917
Re: Lab Quiz 4
Hmmm... your calculations are correct. Try entering just 0.001 maybe?
Guest01 Posts : 133
Join date : 20080919
Re: Lab Quiz 4
No, that doesn't work for me, I tried several times...Did it work for you on mapleTA?
periwinkle Posts : 22
Join date : 20080917
Re: Lab Quiz 4
Yes. 0.001.
I do not know if you actually did calculations or not. After a while I noticed it never went higher then 0.002, so I would enter 0.001 until I got it right because I ran into inconsistency problems as well.
How many tires have you entered? Have you only entered in 0.001? Have you gotten anything like 0.00149? it might just look at it as 0.002, or perhaps 0.0015 maybe? I think MapleTA will correct up to the third decimal place, so anything longer really won't matter, I believe that though, not 100% sure.
I do not know if you actually did calculations or not. After a while I noticed it never went higher then 0.002, so I would enter 0.001 until I got it right because I ran into inconsistency problems as well.
How many tires have you entered? Have you only entered in 0.001? Have you gotten anything like 0.00149? it might just look at it as 0.002, or perhaps 0.0015 maybe? I think MapleTA will correct up to the third decimal place, so anything longer really won't matter, I believe that though, not 100% sure.
Guest01 Posts : 133
Join date : 20080919
Re: Lab Quiz 4
I've tried it around 9 times and 0.001 hasn't worked for me. Thank you for trying to help though.
periwinkle Posts : 22
Join date : 20080917
Re: Lab Quiz 4
any new insights on question three... tried applying the suggested formula and hasnt worked. Tried .001 several times (upwards of ten) and that hasnt worked either...
guesto Guest
Hurray #3 has arrived!
This is my data from the CD practice question:
279g = 0.279kg
error of KE = 1.427% convert to relative error 1.427/100 = 0.01427
relative error of v = 0.01125
v = 0.5
relative error of v^2 = 0.01125 * 2 = 0.0225
use equation 7: sqrt 0.01427^2 + 0.0225^2 = 0.026644
find the change in KE 1/2 * 0.279* 0.5^2 = 0.034875
final answer: 0.026644 * 0.034875 = 0.000929
I used the proff's blog to help me out and this is the exact answer on the practice CD. I am going to try it on MapleTA now.
Let me know if this works for you.
279g = 0.279kg
error of KE = 1.427% convert to relative error 1.427/100 = 0.01427
relative error of v = 0.01125
v = 0.5
relative error of v^2 = 0.01125 * 2 = 0.0225
use equation 7: sqrt 0.01427^2 + 0.0225^2 = 0.026644
find the change in KE 1/2 * 0.279* 0.5^2 = 0.034875
final answer: 0.026644 * 0.034875 = 0.000929
I used the proff's blog to help me out and this is the exact answer on the practice CD. I am going to try it on MapleTA now.
Let me know if this works for you.
super Mo Guest
to periwinkle
After all the help everyone gives me it feels so good to give back  it worked for my MapleTA too!
super Mo Guest
Re: Lab Quiz 4
number 3 didn't work for me  is it supposed to be positive or negative? i thought i did all the calculations right too,i got something around the numbers you guys did
confused Guest
Re: Lab Quiz 4
so for number 3 you guys put in 0.001 and it worked? because it still isnt working for me
thb1026 Posts : 8
Join date : 20080927
question number 3
is the answer supposed to be positive or negative? i've tried both ways and it still doesn't work for me, can anyone help me?
guest 00 Guest
Re: Lab Quiz 4
The answer should be negative for Question 3. Do NOT use my calculations, it has been proven wrong. The correct answer should be posted after mine, as well on Professor's Dawber's Blog
Guest01 Posts : 133
Join date : 20080919
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