When to use energy method and when to use SUVAT: MEI M2 vs OCR NW

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Primus2x
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In question 2 (b) (i) of the OCR Physics Newtonian World June 2014 Paper, I am asked to find the minimum distance a jet aircraft travels in order to accelerate to a take-off speed of 56 ms-1, provided the engine outputs a thrust of 28 kN, the jet aircraft has a mass of 6200, and that gravitational field strength is a constant -9.81 ms-2. Friction is neglected (and we have to write an explanation why the distance actually bigger, taking friction and air resistance into consideration in part (ii)). So the resultant force simply acts horizontally with a magnitude of 28 kN

I used Newton's second law in the form F=ma. SUVAT equation v^2=u^2+2as.
I got an answer of 347.2 m exactly and that rounds up to 350 m, which is what they have in the mark scheme.

In question 2 (a) (ii) of the MEI Mathematics Mechanics 2 June 2009 paper, I found the resultant force to have a magnitude of 147 N (with gravitional field strength being a constant -9.8 ms-2 exactly) and used the energy method to find the distance the block travels to reach a velocity of 1.5 ms-1.
I got an answer of x=\frac{1875}{4900} which is approximately 0.38265 m, they have that in the mark scheme. But when I check the answer using a SUVAT, identical to the one I used in the NW question, I get half of 0.38265 m instead of 0.38265 m, why does that happen?

I went through the question again using calculus and a more rigorous definition of work and I get the correct answer too.

\mathbf{F}=147\mathbf{i} \text{N},\mathbf{v}=\mathbf{v}_0  +t\mathbf{a}=\frac{147t}{25} \mathbf{i}

\\

\mathbf{v}=1.5\mathbf{i} \text{m}\text{s}^{-1}\Rightarrow t=\frac{25}{98} \text{s}

\\

W=\int_{0}^{25/98}\mathbf{F}\cdot\mathbf{v}dt= \int_{0}^{25/98}147\frac{147t}{25}dt=\left[ {\left( {\frac{147t}{5}} \right)^2} \right]_{0}^{25/98}=56.25 \text{J}

\\

W=\mathbf{F}\cdot\mathbf{s}= \left| {\mathbf{F}} \right| \left| {\mathbf{s}} \right| \cos{0} \Rightarrow \left| {\mathbf{s}} \right|=\frac{1875}{4900} \approx 0.382653 \text{m}

http://www.mei.org.uk/files/papers/m209ju_jk32.pdf

Why does SUVAT work in the NW question but not the M2 question? And presumbly, using the energy method to solve the NW question will give me an answer of 694.4, but no, if I use the energy method to solve the NW question, I get the correct answer:
\frac{1}{2} 6200 (56^2)=\left| {28000} \right| \left| {\mathbf{s}} \right \cos{0} \Rightarrow \left| {\mathbf{s}} \right|=\frac{9721600}{28000} =347.2 m
Why are both the SUVAT and Energy methods valid for the NW question but only the Energy method valid for the M2 question (furthermore, why does using SUVAT for that give me half the correct answer)? Both feature constant force, mass and acceleration and in both situations the initial velocity is 0.
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Joinedup
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(Original post by Primus2x)
In question 2 (b) (i) of the OCR Physics Newtonian World June 2014 Paper, I am asked to find the minimum distance a jet aircraft travels in order to accelerate to a take-off speed of 56 ms-1, provided the engine outputs a thrust of 56 kN, the jet aircraft has a mass of 6200, and that gravitational field strength is a constant -9.81 ms-2. Friction is neglected (and we have to write an explanation why the distance actually bigger, taking friction and air resistance into consideration in part (ii)). So the resultant force simply acts horizontally with a magnitude of 28 kN,
how so?

I used Newton's second law in the form <math>F=ma</math> SUVAT equation <math>v^2=u^2+2as</math>.
I got an answer of 347.2 m exactly and that rounds up to 350 m, which is what they have in the mark scheme.

In question 2 (a) (ii) of the MEI Mathematics Mechanics 2 June 2009 paper, I found the resultant force to have a magnitude of 147 N (with gravitional field strength being a constant -9.8 ms-2 exactly) and used the energy method to find the distance the block travels to reach a velocity of 1.5 ms-1.
I got an answer of <math>x=\frac{1875}{4900}</math> which is approximately 0.38265 m, they have that in the mark scheme. But when I check the answer using a SUVAT, identical to the one I used in the NW question, I get half of 0.38265 m instead of 0.38265 m, why does that happen?
http://www.mei.org.uk/files/papers/m209ju_jk32.pdf

Why does SUVAT work in the NW question but not the M2 question? And presumbly, using the energy method to solve the NW question will give me an answer of 694.4, why wouldn't the energy method be valid there? Both situations feature constant acceleration and the initial velocity is 0 in both situations tool
first question
F=ma
a=F/m
a=28000/6200
a=4.52

v2=2as
3136=2 x 4.52 x s
s=3136/9.03
s=347

why do you think you should half the resultant force?
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Primus2x
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(Original post by Joinedup)
how so?


why do you think you should half the resultant force?
Sorry, that was a typo, the force the engine outputs is 28 kN,
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Primus2x
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I believe the problem was I assumed the rope could be extended or compressed and that I didn't account for the two masses where the two weight forces originate. It's really one system with a total mass of 5+25+20=50 kg and there is a force of 5*9.8 pulling it one way and a force of 20*9.8 acting the other way. The entire system moves at 1.5 ms-1 not just the 25 kg block.

The resultant force is
\mathbf{F}=147\mathbf{i} \text{N}

Thus acceleration is
\mathbf{a}(t)=\frac{147}{50} \mathbf{i} \text{ms}^-1

So time is
\mathbf{v}=\mathbf{v}_0+\mathbf{  a}=0+\frac{147}{50} t \mathbf{i} 

\mathbf{v}=1.5\mathbf{i} \Rightarrow t=\frac{25}{49} \text{s}

Therefore displacement is
\mathbf{s}=\int_0^{25/49} \mathbf{v} dt =\mathbf{s}=\int_0^{25/49} \frac{147}{50} t \mathbf{i} dt = \left[ {  \frac{147}{100} t^2 \mathbf{i} } \right]_0^{25/49}

\\

=\dfrac{75}{196} \mathbf{i} \text{m} \approx 0.382653 \mathbf{i} \text{m}
And that is the correct answer.
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