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Chapter 5: Work and Energy

Re: Chapter 5: Work and Energy

Postby goldstanda3269 » Sun Dec 20, 2015 10:19 am

ilbe wrote:Queries with regards to Q& A 5.

I thought this is simple harmonic motion, thus E(W)=1/2 (kX^2). As this equation doesn't show mass, thus I thought this is not related to mass... But I am wrong..


RE: Q&A Page 1/2

Quote from the following link which deals with simple harmonic motion: "Use Energy Conservation (often the easiest approach)":

http://www.cscamm.umd.edu/people/facult ... h14-15.pdf
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Re: Chapter 5: Work and Energy

Postby goldstanda3269 » Sun Dec 20, 2015 10:27 am

ilbe wrote:Queries with regards to Q10

I used kinematic equation, V^2 = V0^2+2as. This is because no friction, under the gravity.
Thus, V^2 = 0 + 2as.
So Whatever mass are, the speed will be same. Am I right?


The explanation shows:

mgh = (1/2)mV^2

Cancel mass

gh = (1/2)V^2

Multiply both sides by 2:

2gh = V^2

Because the acceleration a is the same as gravity g in this problem, the equation in the Explanation is identical to the conclusion you made using the equation of kinematics. Often, there are several different ways to solve the same problem.
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Re: Chapter 5: Work and Energy

Postby goldstanda3269 » Sun Dec 20, 2015 11:26 am

laura.cp897117 wrote:For Q6:

6) If you lift a 10 kg box 1 meter in 2 seconds, then your friend does the same thing but it takes 6 seconds, which of the following is true about the work done on the box?

....

Explanation: Work is force times distance. In both cases, the force is about 100 newtons (i.e. mg) and the distance is 1 meter so there is no change. The fact that the box is moved faster in one case compared to the other is a function of the power (i.e. the rate of work or W/t).


I used F = ma instead of F=mg, giving me for the first object:
W = kg.m/s^2 = 10x1/(2)^2 = 10/4
For the second object:
W = kg.m/s^2 = 10x1/(6)^2 = 10/36

This is clearly different than using F = mg. Why do we use F = mg rather than F = ma in this case?


Well, let's take a closer look at the statement in the Explanation "the force is about 100 newtons ".

Because the truth is, your calculation was partly, very reasonable. However, think of raising a box. To move it upwards, you absolutely must act against gravity, or more specifically, you must counteract the weight mg of the objective. If you don't act against mg, then the object accelerates downwards. So in fact you are doing 2 things, just by holding it, you counteract mg AND by accelerating it, you are causing ma.

So the sum of forces F = mg + ma

But mg is about 100 and ma in both cases is less than 3! So given the answer choices, the results are approximately the same.
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Re: Chapter 5: Work and Energy

Postby goldstanda3269 » Sun Dec 20, 2015 11:28 am

goldstanda3269 wrote:
laura.cp897117 wrote:For Q6:

6) If you lift a 10 kg box 1 meter in 2 seconds, then your friend does the same thing but it takes 6 seconds, which of the following is true about the work done on the box?

....

Explanation: Work is force times distance. In both cases, the force is about 100 newtons (i.e. mg) and the distance is 1 meter so there is no change. The fact that the box is moved faster in one case compared to the other is a function of the power (i.e. the rate of work or W/t).


I used F = ma instead of F=mg, giving me for the first object:
W = kg.m/s^2 = 10x1/(2)^2 = 10/4
For the second object:
W = kg.m/s^2 = 10x1/(6)^2 = 10/36

This is clearly different than using F = mg. Why do we use F = mg rather than F = ma in this case?


Well, let's take a closer look at the statement in the Explanation "the force is about 100 newtons ".

Because the truth is, your calculation was partly, very reasonable. However, think of raising a box. To move it upwards, you absolutely must act against gravity, or more specifically, you must counteract the weight mg of the object. If you don't act against mg, then the object accelerates downwards. So in fact you are doing 2 things, just by holding it, you counteract mg AND by accelerating it, you are causing ma.

So the sum of forces F = mg + ma

But mg is about 100 and ma in both cases is less than 3 (by your calculations)! So given the answer choices, the results are essentially the same.
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Re: Chapter 5: Work and Energy

Postby dechof » Tue Dec 03, 2019 10:38 am

Why is the mass of the roller coaster irrelevant? - 5.5.1
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Re: Chapter 5: Work and Energy

Postby goldstanda3269 » Tue Dec 03, 2019 2:44 pm

dechof wrote:Why is the mass of the roller coaster irrelevant? - 5.5.1


"Intuition" might want to argue: Should 2 different masses have different velocities as they fall?

The Physics answer is: No, if there is no friction then there is no difference in velocity.

But for GAMSAT purposes, that's not the most important point: In 5.5.1, it shows how mass m cancels from the equation to solve velocity (manipulating equations is very important for this exam). This is proof positive that mass is not a factor in the calculation for velocity in Conservation of Energy problems with no friction.
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Re: Chapter 5: Work and Energy (Q9)

Postby HunterEd » Fri Apr 03, 2020 12:45 am

9) What velocity would Batman have just before he strikes the platform? (Assume negligible air resistance)
A) 15.3 m s-1
B) 23.4 m s-1
C) 9.8 m s-1
D) 32.8 m s-1

Explanation:
Using the Principle of conservation of energy and Kinetic Energy = 1/2 x Mass x (velocity)2
8 232 J = 1/2 x 70 kg x v2 {see the previous question}
v2 = 235.2 = 225 approximately

v = square root 225 = 15 m s-1

Where did you get the quantity for v2 from? it mentions to 'see previous question' but I can't seem to find it/work it out.

Thank you.
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Re: Chapter 5: Work and Energy

Postby goldstanda3269 » Sun Apr 05, 2020 4:48 am

HunterEd wrote:Explanation:
Using the Principle of conservation of energy and Kinetic Energy = 1/2 x Mass x (velocity)2
8 232 J = 1/2 x 70 kg x v2 {see the previous question}
v2 = 235.2 = 225 approximately

v = square root 225 = 15 m s-1

Where did you get the quantity for v2 from? it mentions to 'see previous question' but I can't seem to find it/work it out.


"Using the Principle of conservation of energy": This means that the maximum potential energy at the top (highest point while not moving) is equal to the maximum kinetic energy at the very bottom.

In the previous question, we calculated the potential energy to be 8 232 J. So now we use that value as the kinetic energy for this question.

Thus:

Kinetic energy = 1/2 m v2
8 232 J = 1/2 x 70 kg x v2
8 232 = 35 x v2
8 232 / 35 = v2
235 = v2
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Re: Chapter 5: Work and Energy

Postby horbowiec.7527 » Mon Apr 26, 2021 4:32 pm

This is regarding 5.5.1. Conservation of Energy Problem (SI units) calculations

Forgive me but the book calculations (PHY-64-65) are not clear for me to see what exactly has happened step by step in this equation. Why does m cancel out and how exactly? Since it is on both sides of the addition equation --> 1/2mv^2 + mgh? It is a sum of products first but then later you minus the products under square root, is it because it has been put under root square?
Please kindly give me some more detailed explanation as I am lost here:)

Regards,
Ania
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Re: Chapter 5: Work and Energy

Postby goldstanda3269 » Mon Apr 26, 2021 6:13 pm

horbowiec.7527 wrote:This is regarding 5.5.1. Conservation of Energy Problem (SI units) calculations

Forgive me but the book calculations (PHY-64-65) are not clear for me to see what exactly has happened step by step in this equation. Why does m cancel out and how exactly? Since it is on both sides of the addition equation --> 1/2mv^2 + mgh? It is a sum of products first but then later you minus the products under square root, is it because it has been put under root square?
Please kindly give me some more detailed explanation as I am lost here:)

Regards,
Ania


From PHY-65:

The total energy ET = 300(m) = 1/2 m v^2

So notice that this is an equation, both sides are equal to each other:

300(m) = 1/2 m v^2

since mass m is a multiple on both sides of the equation, you can cancel it out. Another way to say it is this: Divide both sides by m and then m cancels, so we get:

300(m)/m = 1/2 m v^2/m

300 = 1/2 v^2

Now there is only one variable left (the velocity v). We must isolate that variable (i.e. get it alone on one side of the equation so we can determine its value). To isolate v, multiply both sides by 2 so you can get rid of the 1/2:

600 = v^2

Now take the square root of both sides to isolate v.

Please let me know if anything is unclear.
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Re: Chapter 5: Work and Energy

Postby horbowiec.7527 » Tue Apr 27, 2021 9:26 am

What you have explained is situation (b) at ground level where h = 0 and that part is absolutely clear to me mathematically and what happens to the equation step by step.

I should have been more clear and say situation (a) where h = 20m is what I need clarification on.

300m = 1/2mv^2 + mgh, than you say m cancels... but which m cancels? we have 2m in the right side of equation... And than this immediately goes to square root with all the numbers and minus sign, where earlier it included sum of the products, is it minus because it has been put under square root? I just can't follow how one becomes another in situation (a)

Regards,
Ania
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Re: Chapter 5: Work and Energy

Postby goldstanda3269 » Tue Apr 27, 2021 2:02 pm

horbowiec.7527 wrote:What you have explained is situation (b) at ground level where h = 0 and that part is absolutely clear to me mathematically and what happens to the equation step by step.

I should have been more clear and say situation (a) where h = 20m is what I need clarification on.

300m = 1/2mv^2 + mgh, than you say m cancels... but which m cancels? we have 2m in the right side of equation... And than this immediately goes to square root with all the numbers and minus sign, where earlier it included sum of the products, is it minus because it has been put under square root? I just can't follow how one becomes another in situation (a)

Regards,
Ania


No problem!

300m = 1/2mv^2 + mgh

1) Since m is included in all factors, we can simply cancel m thus:

300 = 1/2v^2 + gh

2) Some prefer to add one step by isolating m on the right side of the equation, then dividing both sides by m (in the book: GM 3.1 and 3.2):

300m = 1/2mv^2 + mgh
300m = m(1/2v^2 + gh)
300 = 1/2v^2 + gh

3) Given:

300 = 1/2v^2 + gh

... and knowing that we are given the values in SI units of g = 10 and h = 20, we need to isolate the only variable, v (it is our only unknown value). Subtract gh from both sides:

300 - gh = 1/2v^2

Multiply through by 2:

600 - 2gh = v^2

Plug in the values for g and h:

600 - 2(10)(20) = v^2
600 - 400 = v^2
200 = v^2

Take the square root of both sides to finally, completely isolate v:

ROOT(200) = v
10 x ROOT 2 = 14 = v

The real exam won't have 75 questions like this, but there will usually be at least a handful of units with a similar degree of equation manipulation (with or without logs).
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Re: Chapter 5: Work and Energy

Postby horbowiec.7527 » Thu Apr 29, 2021 1:09 pm

Thank you so much that is so clear now!

I guess I get "intimidated" (haha) when I see so many steps combined together and all this numbers under square root.
This way is much clearer and obvious than what shown in a book. Much appreciated, thank you!

Regards,
Ania
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Re: Chapter 5: Work and Energy

Postby ohwaseng8484 » Tue Dec 28, 2021 1:43 am

Hi, for question 18, are we allowed to just assume that the weight of the cases are the same? Did not see any indication of the weight in the question. Thanks
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Re: Chapter 5: Work and Energy

Postby goldstanda3269 » Tue Dec 28, 2021 2:19 am

ohwaseng8484 wrote:Hi, for question 18, are we allowed to just assume that the weight of the cases are the same? Did not see any indication of the weight in the question. Thanks


You are correct.

"Consider an object being lifted ... in the time periods provided" implies that the object is the same in both cases, and thus the mass is the same.
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