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Physics: Post your doubts here!

K

kitkat <3 :P

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sagar65265 said:
When a total amount of work W is done on any object, the kinetic energy of that object changes by a quantity W - if negative work is done on the object, the kinetic energy changes by a negative amount and if positive work is done on the object it's energy changes by a positive amount.

Suppose you throw an object upwards, gravity is the only force that does any work on the object. That work is negative, so the kinetic energy of the object decreases until the object comes to a stop. When it falls down, gravity does positive work on the system thus increasing it's kinetic energy.

In this case, taking the first situation, the force is constant at magnitude F, over the entire displacement s. Thus, the work done by that force is Fs. By the
Work-Kinetic Energy theorem, this is equal to the increase in kinetic energy of the system. This value is given as 4 Joules (8-4 = change in KE = +4 Joules).

In the upcoming situation, the force is 2F, displaced through a distance 2s. Thus, the total work done is 4Fs. This is also the change in kinetic energy of the object.
From above, we know that Fs is 4 Joules, so 4Fs = 4*4 = 16 Joules increase. From 4 Joules, the increase of 16 Joules takes it to 20 Joules = B.

Well, I just noticed unique111's solution, and I have to say as far as I know it's absolutely correct, (so no need for the disclaimer, unique111! it's totally right!)

Good Luck for all your exams!
Click to expand...

okay so u meanwhen work is done against gravity its negative and the kinetic energy decreases right :oops: just confirming (developing a concept actually :oops:)
 
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kitkat <3 :p said:
okay so u meanwhen work is done against gravity its negative and the kinetic energy decreases right :oops: just confirming (developing a concept actually :oops:)
Click to expand...

There's some very tight wording there, so just want to confirm:

When the force of gravity itself does negative work, the kinetic energy decreases. This is correct.

However, when work is done against the force of gravity, it implies there is some force opposing gravity (suppose you are lifting an object with your hands, you are doing work against gravity. Suppose you throw the object up: as long as it is in contact with your palm and it is moving upwards with you hand, the normal force you exert on it does work against gravity. When you let go of the object, no force opposes gravity so gravity does negative work on it until it stops and falls back down).

It all depends on the displacement. When an object is moving upwards, gravity is doing negative work because the displacement is in the opposite direction to the force (displacement is upwards, force acts downwards). This is negative work, and shows itself as the object slows down. When the object is falling down, gravity does positive work because the displacement (downwards) and the force of gravity (downwards) are in the same direction.

So when work is done against gravity, it implies some force aside from gravity doing that work (as far as i know - the distinction is important, because CIE examiners like to use the most specific wording in their problems and options). In that case, depending on the displacements, the work can vary.

If the object is falling, but the force opposing gravity is stronger than the force of gravity, then the object slows down (because force opposing gravity is upwards and displacement is downwards, that force does negative work. Because both gravity and displacement are downwards, gravity does positive work. However, since the opposing force is stronger than gravity, it does more negative work than gravity does positive work, so the change in kinetic energy is overall negative. In other words, the kinetic energy decreases).

Suppose the object is going upwards and the force opposing gravity is weaker than gravity. In that case also the object slows down (because force opposing gravity is upwards and displacement is upwards, that force does positive work. Because gravity acts downwards and displacement is upwards, gravity does negative work. However, since gravity is stronger than the opposing force, it does more negative work than the opposing force does positive work, so the change in kinetic energy is overall negative. In other words, the kinetic energy decreases).

Hope this helped!

Good Luck for all your exams!
 
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sagar65265 said:
There's some very tight wording there, so just want to confirm:

When the force of gravity itself does negative work, the kinetic energy decreases. This is correct.

However, when work is done against the force of gravity, it implies there is some force opposing gravity (suppose you are lifting an object with your hands, you are doing work against gravity. Suppose you throw the object up: as long as it is in contact with your palm and it is moving upwards with you hand, the normal force you exert on it does work against gravity. When you let go of the object, no force opposes gravity so gravity does negative work on it until it stops and falls back down).

It all depends on the displacement. When an object is moving upwards, gravity is doing negative work because the displacement is in the opposite direction to the force (displacement is upwards, force acts downwards). This is negative work, and shows itself as the object slows down. When the object is falling down, gravity does positive work because the displacement (downwards) and the force of gravity (downwards) are in the same direction.

So when work is done against gravity, it implies some force aside from gravity doing that work (as far as i know - the distinction is important, because CIE examiners like to use the most specific wording in their problems and options). In that case, depending on the displacements, the work can vary.

If the object is falling, but the force opposing gravity is stronger than the force of gravity, then the object slows down (because force opposing gravity is upwards and displacement is downwards, that force does negative work. Because both gravity and displacement are downwards, gravity does positive work. However, since the opposing force is stronger than gravity, it does more negative work than gravity does positive work, so the change in kinetic energy is overall negative. In other words, the kinetic energy decreases).

Suppose the object is going upwards and the force opposing gravity is weaker than gravity. In that case also the object slows down (because force opposing gravity is upwards and displacement is upwards, that force does positive work. Because gravity acts downwards and displacement is upwards, gravity does negative work. However, since gravity is stronger than the opposing force, it does more negative work than the opposing force does positive work, so the change in kinetic energy is overall negative. In other words, the kinetic energy decreases).

Hope this helped!

Good Luck for all your exams!
Click to expand...

thanks alot :)
but....in the last example you gave how can gravity act upwards like doesnt gravity always acts downwards?
 
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Student12 said:
Questions 4 (A) , 13 (D) ,14 (A) ,16 (C) ,22 (B), 32 (D) Help pls. in these questions :(http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_11.pdf
Click to expand...

Q4)

On an oscilloscope screen, the height (or y-gain) displays the voltage of the signal and the x-axis/ x-length displays time.

So, taking the wave X, we can see that the peak of the wave is 3 squares above the base-line (the point right in between the crest and the trough).

We are also given that the peak voltage represented by those 3 unit squares is 12 Voltage. Therefore, if 3 squares represent 12 Volts, 1 square represents 12/3 = 4 Volts.

Taking waveform Y, we can see that the maximum voltage above the base-line (again, the point in between the crest and the trough of the waveform) is found 1 square unit on either side. This, we know from above, represents 4 Volts, therefore the maximum Voltage is 4 Volts.

Now for the period. Taking a look at the waveform X, we can see that the number of squares needed to complete a full wave is equal to 8 squares. Since time period = 1/frequency = 1/50Hz = 0.02 seconds, we can see that 8 squares represent 0.02 seconds and therefore 1 square represents 0.0025 seconds.

For waveform Y, we notice that the same number of squares (on the same scale) are needed to complete 1 cycle of that waveform. Since 1 square represents 0.0025 seconds, 8 squares will represent 0.02 seconds. We know that the time period is defined as the time required to complete one cycle, therefore we can say that the time period is 0.02 seconds. Multiplying this by 1000 to convert from seconds to milliseconds, we find that the period is 20 ms and the maximum voltage is 4 Volts. The only option that has both these values = A.

I'll do the remaining questions after some time, sorry for the delay.

Hope this helped!
Good Luck for all your exams!
 
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Azee said:
please help.
View attachment 44170View attachment 44171
Click to expand...

Let us take the components of the forces acting on the door in the vertical and horizontal directions:

In the vertical direction, the force of gravity acts downwards (written as W), the vertical component of the hinge force H acts downwards (component written as Hv) and the vertical component of the tension force T acts upwards(component written as Tv). These components have to balance out for the door to remain in equilibrium, so we can write:

Tv = Hv + W

Since the vertical component of Tv is greater than the entire vertical component of W (and the magnitude of T is also greater than Tv, giving us W<Tv<T) we can say that T is always greater than W. This necessarily means that we can eliminate A and B, since those say W is greater in magnitude than T.

Now moving on to the horizontal components of the forces involved, we see that gravity has no horizontal component, H has a component to the right (H[h]) and Tension T has a component to the left (T[h]). Balancing these out, we get

T[h] = W[h]

Now we have to look at the angles involved - the diagram may not necessarily be to scale, but the angle that H makes with the horizontal (say we write this as θ) is definitely smaller than the angle that T makes with the horizontal (say we write this as α).
Writing out the components in the horizontal direction:

T * cos(α) = H * cos(θ)

So T = H * cos(θ)/cos(α)

Since α is greater than θ, we can say that the cosine ratio for α is smaller than the cosine ratio for θ (as you go from 0 to 90 degrees, the cosine function keeps decreasing with increasing angle). So, we can also say that cos(θ)/cos(α) is greater than 1.

Therefore, T = (something greater than 1) * H.
Therefore, T>H.

Lastly, there is no doubt that H is greater than W - it balances out the horizontal component of T, which is itself proof that it has a large horizontal component. However, it has a vertical component to add to that, so that makes it even stronger. Therefore, W<H<T = C.

Hope this helped!
Good Luck for all your exams!
 
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_Ahmad said:
but there is another option 2.6, why it is not the answer?
Click to expand...

Since the area under the curve on the Force-extension graph is the energy stored in the spring, you can split the graph into two parallelograms (as in the image attached), and using the formula for calculating area of parallelogram (area = 0.5 * distance between parallel sides * (sum of lengths of parallel sides)) can find out the area as follows:
Graph.JPG
Area = 0.5 * 6 * (0.2 + 0.1) + 0.5 * 0.1 * (6 + 17) = 0.9 + 1.15 = 2.05 = approximately 2.00 Joules = A.

Hope this helped!
Good Luck for all your exams!
 
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_Ahmad said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_12.pdf

Q22,26
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22)
because since k is low for the first spring, the box will prevent it from breaking by over extending. Hence A
26)
We know that distance between two minima is given by λ/2 (example, distance b/w T and S is λ/2)
We also know that Microwave travels with the speed of light c i.e is 3 * 10^8 m/s
Given distance b/w each minima is 15 mm so total distance from T to P would be 15 * 4 = 60 mm

Now distance b/w:
T and S = λ/2,
S and R = λ/2
R and Q = λ/2
Q and P = λ/2
and them you get 2λ
so 2λ = 60*10^-3
hence λ = 0.03 m

We know that c = f * λ
So f = c / λ
f = ( 3 * 10^8 ) / ( 0.03 )
f=10GHz
 
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Hijab said:
Can someone plz solve this question... I would really be grateful :)
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Got it,
Since R of variable resistor is increased, it’s voltage also increases as it will require more energy for current to pass. So the voltage share of the other fixed resistor decreases, and voltmeter reading decreases as well (The voltmeter is connected across the fixed resistor not the variable). Current is unchanged. The ammeter will give the total current in the circuit. You were right about saying I increases but which I? It’s the current of the fixed resistor not the TOTAL current. Because the total voltage in the circuit remains the same (It’s just divided between components according to the ratio of their Resistances. ) so TOTAL current is unchanged as well.
 
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