Ch 6.3 Help

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Ch 6.3 Help

Post  Guest01 on Sat Oct 04, 2008 10:12 pm

Question 1
A 50 kg person starting with an initial downward velocity of 1.2048 m/s dips into a 2.395m deep pool. If 699.4J kinetic energy are lost due to friction what is the arrival speed at the bottom of the pool (see sheet 15) (If you know about "Buoyancy" disregard it.)? Indicate with a negative (positive) sign whether the potential energy of the person does (does not) change while sinking in the pool.

How to solve
(Answer is negative)

First Calculate the KE and PE.

KE = .5*50*1.20482
KE = 36.288576

PE = 50*9.81*2.395
PE = 1174.7475

Now add KE and PE together

W = KE+PE
W = 36.288576+1174.7475
W = 1211.036076

Now some energy is lost due to friction so subtract that from the the amount of Work we just totaled
KE2 = W-Ffr
KE2 = 1211.036076-699.4
KE2 = 511.636076

Now since we are looking for velocity we arrange our KE formula to ind it.
v = √((2*KE)/mass)
v = √((2*511.636076)/50)
v = √(1023.272152/50)
v = √(20.46544304)
v =
Spoiler:
4.523874782

Question 2
NOTE! Use the numbers given in the lecture notes for the radius and the mass of the earth.
A 447.1 kg object is lifted from the surface of the earth to a height of 782.4 km above the surface of the earth. Calculate the increase in gravitational potential energy using the exact expres​sion(see sheet 24) and compare it with the approximate expression using g=9.81 (see sheet 12,24'). Give the magnitude of the difference between the two expressions divided by the exact result (that is the relative error) as your answer. Make sure you use at least 6 significant digits when calculating the difference. Indicate with a positive (negative) sign whether your relative error is less than (greater than) a 1% accuracy .

How to solve
(Answer is negative)

PEGravity = PEExact-PEApprox

PEExact = ((-G*m*ME)/(RE+h))-((-G*m*ME)/(RE))
PEExact = ((-6.67e-11*447.1*6.0e24)/(6.4e6+782400))-((-6.67e-11*447.1*6.0e24)/(6.4e6))
PEExact = ((-1.7892942e17)/(7182400))-((-1.7892942e17)/(6.4e6))
PEExact = (-2.49122048e10)-(-2.79577219e10)
PEExact = 3045517100 J

PEEstimate = m*g*h
PEEstimate = 447.1*9.81*782400
PEEstimate = 3431646302

PEGravity = (|Exact-Estimate|)/Exact
PEGravity = (|3045517100-3431646302|)/3045517100
PEGravity = 386129202/3045517100
PEGravity =
Spoiler:
0.126786089

Question 3
NOTE! Use the numbers given in the lecture notes for the radius and the mass of the earth.
A 61.67 kg piece of a satellite left over from an explosion with zero orbital velocity in space falls from a distance of 163.08 km above the surface of the earth down toward earth. It arrives with 16.56 MJ (M="mega") kinetic energy on earth. How many MJ energy have been lost due to friction in the earth's atmosphere? Use the exact expres​sion(see sheet 24) for the gravitational potential energy. Make sure you use at least 6 significant digits when calculating the change in potential energy. Indicate with a negative (positive) sign whether the kinetic energy at arrival would be larger (smaller) than the 16.56 MJ without friction in the atmosphere.

How to solve
(Answer is negative)

KEMJ = PEExact-KE

First find PEExact:
PEExact = ((-G*m*ME)/(RE+h))-((-G*m*ME)/(RE))
PEExact = ((-6.67e-11*61.67*6.0e24)/(6.4e6+163080))-((-6.67e-11*61.67*6.0e24)/(6.4e6))
PEExact = ((-2.4680334e16)/(6563080))-((-2.4680334e16)/(6.4e6))
PEExact = (-3760480445)-(-3856302188)
PEExact = 95821743

To make it a little easier I'm turning into MJ now by dividing the number by 1,000,000
95821743/1000000 = 95.821743

Now just subtract the given KE (given in MJ) from the PE (which is now in MJ. Since they wan the answer in MJ we can just enter it right in and don't have to convert it at the end).

KEMJ = PEExact-KE
KEMJ = 95.821743-16.56
KEMJ =
Spoiler:
79.261743

Thanks guys for the help on Question 3!

Question 4
Calculate the escape velocity for a planet with the moon's mass (7.4x1022 kg) and a mean radius of 1.744 x106 m (see sheet 26,27)? Indicate with a positive (negative) sign whether the value for the earth is larger (smaller) than your result.

How to solve
(Answer is positive)

V=√((2*G*Mmoon)/(Rmoon)))
V=√((2*6.67e-11*7.4e22)/(1.744e6))
V=√((9.8716e12)/(1.744e6))
V=√(5660321.101)
V=
Spoiler:
2379.142934


Last edited by Guest01 on Thu Oct 09, 2008 10:44 pm; edited 4 times in total

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Questions 1-3

Post  sedwards on Wed Oct 08, 2008 12:12 am

anyone know how to do the first three?

sedwards
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Re: Ch 6.3 Help

Post  Guest01 on Wed Oct 08, 2008 4:43 pm

I'm having problems myself with the first three, somewhere in my calculations something goes wrong, I'm not 100% sure if I'm using the formula correctly or if I missed something in it. That blog doesn't help much either.

If anyone has gotten this please let me know.

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wtf

Post  Guest2 on Thu Oct 09, 2008 1:22 am

worst help ever!

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Re: Ch 6.3 Help

Post  Guest01 on Thu Oct 09, 2008 12:44 pm

Figured out where I went wrong with Question 1. The way to do it is posted.

Still having trouble with Question 2 and Question 3.

Guest2 thanks for your help.

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Re: Ch 6.3 Help

Post  thb1026 on Thu Oct 09, 2008 6:08 pm

ANYONE HAVE ANY IDEA HOW TO DO 2 and 3?? They are completely confusing and I'm not sure where to even start. Please help if you have any idea! Thanks!

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2 and 3 so far

Post  lanthony on Thu Oct 09, 2008 6:26 pm

I don't have the answers yet, but I'm posting what I have gathered so far, hoping this will spark something in you guys, these two questions suck.

2. Calculation of Exact Delta PE: -G*(mMe/r+h)-(-G*(mMe/r)
Calculation of estimated Delta PE: use mgh with g as gravity constant and h as given height in km from problem
Subtract: Exact - estimated PEs = error
Error/Exact (1st) caculation SHOULD give the relative error.
I have done this easily 10 times, 10 different ways and it still doesn't work. I believe I am on the right track based on the professor's comments on the help board.

2. I think you end up using Ffr = Delta KE/distance. You obviously need to use delta PE (I think it uses the formual I typed above for exact), but how do you get from calculated PE final to KE original? If you could do that, you could just subtract that from the KE final given in the problem and use it in Ffr above.

Maybe this will help someone else also struggling on these. I hope so, I'm so sick of retyping all this crap into my calculator.

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Re: Ch 6.3 Help

Post  thb1026 on Thu Oct 09, 2008 6:32 pm

but for question 2 they are giving you one mass and one height. so for PE=-GmM/r what are you adding into the M? the m=your 297.7kg or whatever you're given. im' not sure how this all joins into the equations we're supposed to use for this question

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Re: Ch 6.3 Help

Post  lanthony on Thu Oct 09, 2008 6:40 pm

The M's are the earths mass. The only difference between the two equations is the denominator: the first one is r + h, the earths radius + the height of the object and the second is JUST the earths redius. You are subtracting one from the other to get the difference in PE when the object is on the ground (no h) subtracted from the object up high (r + h) giving the increase in PE as the height has increased.

Everything else in the in the first equation is exactly the same on both sides.

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Re: Ch 6.3 Help

Post  c342 on Thu Oct 09, 2008 7:45 pm

2. Use this equation to find the exact PE:

G*m*M(-(1/(r + h)) (1/r))

G=6.67*10^-11
M=6*10^24
r=6.4*10^6

The approximate PE = mgh

The answer will be |exact - approximate|/exact

3. Find the exact PE(same as above).

The answer is the difference between the PE and The KE(given).

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#2/3

Post  lanthony on Thu Oct 09, 2008 8:02 pm

C342 can you show any more details with what you did on those two problems? I keep trying what you suggested and it just isn't working for me so I must be missing something.

thanks!

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Re: Ch 6.3 Help

Post  guest121 on Thu Oct 09, 2008 8:21 pm

Sure, here's an example for #3.

Given:
mass(m) 68.66kg
height(h): 156.02km = 156020m
KE on earth: 15.38 MJ

G*m*M(-(1/(r + h)) (1/r))

(6.67*10^-11)(68.66)(6*10^24)(-(1/(156020+ 6.4*10^6) + (1/6.4*10^6))

= 102174132.9 J = 102.174 MJ

102.17 - 15.38 = 86.79 MJ

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questions 2 and 3

Post  sedwards on Thu Oct 09, 2008 8:22 pm

Can someone please give a more detailed description of 2 and 3 cuz its not workin for me either?

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Re: Ch 6.3 Help

Post  lanthony on Thu Oct 09, 2008 8:34 pm

In one example you have G*m*M(-(1/(r + h)) (1/r)) and in the next the (1/r) is added to the 1/r+h?

Is it multiplied or added? Something is wrong it still isn't working for me either, and I have tried it both ways..

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Re: Ch 6.3 Help

Post  hjjhjkl on Thu Oct 09, 2008 8:37 pm

add

hjjhjkl
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Re: Ch 6.3 Help

Post  thb1026 on Thu Oct 09, 2008 9:25 pm

I keep getting 2 and 3 wrong, and i'm going by the guidelines that you guys are putting up..so something is wrong, can someone please tell me the precise way to calculate 2 and 3?

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Re: Ch 6.3 Help

Post  guesto on Thu Oct 09, 2008 9:38 pm

uhm guest121 i just typed in your expression to my calculator exactly and did not get the same answer as you have shown...

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Re: Ch 6.3 Help

Post  Guest01 on Thu Oct 09, 2008 9:56 pm

I have the calculations done for number two, but I believe using either 6.24e24 or 6.0e24 will change the value. I have another model to go by and my calculation seems close but not exact. I'm about to try and change the values.

Question 3 has now been added in detail.

I haven't submitted these values though so I'm not 100% sure they will work. Hopefully I'll be able to report back soon.

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Re: Ch 6.3 Help

Post  dsafdsf on Thu Oct 09, 2008 10:01 pm

Sorry..I may have been a little unclear and I did round some numbers.(although it didn't affect the answer)

copy and paste this into google

(6.67*10^-11)(68.66)(6*10^24)(((-1/(156020+(6.4*10^6)) + (1/(6.4*10^6)))

should come up with a very similar answer...

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Re: Ch 6.3 Help

Post  Guest01 on Thu Oct 09, 2008 10:50 pm

All questions have been answered and full credit obtained. Question 2's calculations should be correct but I had to redo it since I forgot the minus sign.

Please note originally I missed the True/False question included so I did not include that the answer was negative but I have corrected it.

Thanks to those who have helped including the Professor Matt Dawber's blog

Also I used the value MEarth = 6x1024 in my calculations.

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CH6.3

Post  mleung on Sat Oct 08, 2011 1:17 am

There is another problem in CH6_2:

I tried many times but I couldn't solve it

A 1,711 kg car travels with constant speed of 111.6 km/hr up an incline which makes an angle of 100 with the horizontal. The frictional coefficient is 0.1159. What is the power delivered by the motor in hp (horse powers) (see Ch 6 sheet 29) ? Indicate with a negative (positive) sign whether the hp is (is not) an S.I.unit .

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Re: Ch 6.3 Help

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