Two WorkDone/Power Potential Energy Questions

First one is a Work Done / Power question.

If the air resistance to the motion of an airliner at speed $vms^{-1}$ is given by $kv^2$ newtons at ground level, then at $6000$ meters the corresponding formula is $0.55kv^2$, and at $12000$ meters it is $0.3kv^2$. If an airliner can cruise at $220ms^{-1}$ at $12000$, at what speed will it travel at $6000$ meters with the same power output from the engines.

I've done this bit, the answer being $180ms^{-1}$. It is the second part I cannot do:

Suppose that $k=2.5$ and that the mas of the airliner is $250$ tonnes. As the airliner takes off its speed is $80ms^{-1}$ and it immediately starts to climb with the engines developing three times the cruising power. At what angle to the horizontal does it climb?


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Second is this Potential Energy question:

Two particles of mass $0.1$ kg and $0.2$ kg are attached to the ends of a light inextensible string which passes over a smooth peg. Given that the particles move vertically after being released from rest, calculate their common speed after each had traveled $0.6$ meters.

Done this bit, answer being $1.98ms^{-1}$. It is the next two bits I cannot grasp:

Deduce the work done on the lighter particle by the string, and use this to calculate the tension in the string.

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I would really appreciate an explanation of the second one especially, including an in-depth description of included forces and why we use whatever you used to get the answers.

Thanks for everyone' time.

First one is a Work Done / Power question.

If the air resistance to the motion of an airliner at speed $vms^{-1}$ is given by $kv^2$ newtons at ground level, then at $6000$ meters the corresponding formula is $0.55kv^2$, and at $12000$ meters it is $0.3kv^2$. If an airliner can cruise at $220ms^{-1}$ at $12000$, at what speed will it travel at $6000$ meters with the same power output from the engines.

I've done this bit, the answer being $180ms^{-1}$. It is the second part I cannot do:

Suppose that $k=2.5$ and that the mas of the airliner is $250$ tonnes. As the airliner takes off its speed is $80ms^{-1}$ and it immediately starts to climb with the engines developing three times the cruising power. At what angle to the horizontal does it climb?


--------------------------------


Second is this Potential Energy question:

Two particles of mass $0.1$ kg and $0.2$ kg are attached to the ends of a light inextensible string which passes over a smooth peg. Given that the particles move vertically after being released from rest, calculate their common speed after each had traveled $0.6$ meters.

Done this bit, answer being $1.98ms^{-1}$. It is the next two bits I cannot grasp:

Deduce the work done on the lighter particle by the string, and use this to calculate the tension in the string.

--------------------------------

I would really appreciate an explanation of the second one especially, including an in-depth description of included forces and why we use whatever you used to get the answers.

Thanks for everyone' time.

For q1a is the speed 180 ms per sec?

I get the following:

At ground level the air resistance is kv^2
At 6000m the air resistance is 0.55kv^2
At 12000m the air resistance is 0.3kv^2

When at 12000m and travelling at 220 m/s the air resistance is
0.3*k*220^2 = 14520k

Since the plane is cruising there is no resultant force so the engines thrust = air resistance

So engines thrust = 14520k

Now at 6000m we are told the plane is cruising again. And we are told that the thrust is the same.

So:

Resultant force = 0
Therefore Engines thrust = Air resistance
14520k = 0.55kv^2
14520/0.55 = v^2
v = 162.48 m/s

The book gave the answer 180. I got the same answer by:

Power = Fv (Force x Velocity)
Engine thrust = 14520K (same way you got it)

Power = Fv = 14520k * 220 = 3194400k

Same power output at 6000m

F=0.55kv^2

Fv=0.55kv^3

Power = 0.55kv^3

So 3194400k=0.55kv^3

Cancel the K's and solve for v.

V= 179.7527...

I think your different answer is that you've done it having the same thrust from the engines, not power (which would be F*v = Thrust*220)

I see the error of my ways

I'll still try part b though.

For q2 is T=2.2887N ?

The answer from the book is 0.784 J for the work done - and 1.31 N for the tension.

Damn these tricky mechanics questions!

I retire lol.

Are they M2 questions?

I'm going to have lunch, refocus and come back here with answers (the right answers)

The answer from the book is 0.784 J for the work done - and 1.31 N for the tension.

Damn these tricky mechanics questions!