Powerfully fast
Problem
The power-to-weight ratio (or specific power) of an engine is defined to be the maximum power an engine can produce divided by the mass of the engine.
A certain jet engine has a mass of $4000\textrm{ kg}$ and a power to weight ratio of $12000\textrm{ Js}^{-1}\textrm{ kg}^{-1}$. An aeroplane of unloaded weight $140$ tonnes contains two of these engines.
What is the overall power to weight ratio of the plane?
Neglecting friction, how long would it take for the plane to accelerate from $0$ to $60$ miles per hour? How long to accelerate from $120$ to $180$ or $180$ to $240$?
A spaceship is made with the same weight and power as the aeroplane. Its boosters can burn for 8 hours without refuelling and contain $120000$ litres of fuel, of density $0.81715\textrm{ kg l}^{-1}$. If the spaceship is launched from in space (so that we can ignore friction and gravity), what is the maximum velocity that the spaceship can achieve? (Note: the variation of mass in this part is
significant. To start, make an approximation assuming constant mass. You might then consider the variable mass problem, which will involve harder calculus).
Watt found by experiment in 1782 that a 'brewery horse' (presumably a large draft horse) was able to produce $32400$-foot-pounds per minute. Estimate the power to weight ratio of a horse. How many such horses would be needed to power the aeroplane above? Assuming constant production of power, how long would it take such a horse to accelerate from $0$ to $60$ miles per hour?
Investigation: Explore the power to weight ratios of various interesting objects such as the spaceshuttle, Concorde, Ferraris, sprinters, cheetahs, milkfloats and tortoises.Which produce the best acceleration over various ranges of velocities?
NOTES AND BACKGROUND
There are various parts to this problem, which all use realistic numbers. Conversion of units will be required at various points. Be sure to apply common-sense checks at each point to be sure that you have not made numerical errors -- many engineering disasters have occurred because of problems with conversion of units!
Getting Started
Thus, if all of the work is converted into kinetic energy then we will have
$$ P = \frac{d}{dt}\left(\frac{1}{2}mv^2\right) $$
Note that you can integrate this expression!
Student Solutions
The specific power of the 'plane is
$$\frac{2 \times 12000 \mathrm{\ W\ kg^{-1}} \times 4000 \mathrm{\ kg}}{140000 \mathrm{\ kg}} = 685.714 \mathrm{\ W\ kg^{-1}}$$
For the next part, equate the energy used to the kinetic energy.
$$Pt = \frac{1}{2}mv^2 \quad \therefore \quad t = \frac{mv^2}{2P}$$
$$60 \mathrm{\ mph} = 26.8224 \mathrm{\ m s^{-1}}\Rightarrow t = 0.5246 \mathrm{\ s}$$
Then we look at the change in energy.
120 mph to 180 mph is $3 \times 60 \mathrm{\ mph} - 2 \times 60 \mathrm{\ mph}$, and energy is proportional to the square of velocity, so this requires $(9-4) \times 0.5246 \mathrm{\ s} = 2.623 \mathrm{\ s}$ at the constant power.
180 mph to 240 mph by the same logic requires $(16 - 9) \times 0.5246 \mathrm{\ s} = 3.672 \mathrm{\ s}$
For the rocket,
$$\textrm{[Energy]} = 8 \mathrm{\ hr} \times 3600 \mathrm{\ s\ hr^{-1}} \times 2 \textrm{[engines]} \times 12000 \mathrm{\ W\ kg^{-1}} \times 4000 \mathrm{\ kg \ \textrm{[engine]}^{-1}} = 2.7648 \times 10^{12} \mathrm{\ J}$$
Equate this to $\frac{1}{2}mv^2$, to find $V_{max} = 7509.4 \mathrm{\ m\ s^{-1}}$
For the horse, find a foot-pound in Joules, which the energy requires to lift a pound of mass through one foot, which equates to $1.356 \mathrm{\ J}$. 32400 of these divided by 60 seconds gives $732.4 \mathrm{\ W}$. Divide this through by about $500 \mathrm{\ kg}$, which is approximately the mass of a large horse, to get $1.465 \mathrm{\ W\ kg^{-1}}$.
To find the time taken to accelerate to the given speed, apply the same equations as with the aircraft to find that $t = 4 \mathrm{\ min}\ 6 \mathrm{\ s}$.
Extension: find the formula for acceleration in terms of time for a mass m accelerated with a constant power $P$. How large is the force applied to the mass when the acceleration starts?
Extension solution:
$$Pt = \frac{1}{2}mv^2 \quad \therefore \quad v = \sqrt{\frac{2Pt}{m}}$$
$$a = \frac{\mathrm{d}v}{\mathrm{d}t} = \sqrt{\frac{P}{2mt}}$$
When $t = 0$, the acceleration, and so the force, is theoretically infinite. We can get close to constant power acceleration, but in reality it is not reached, and thus forces are never infinite. For example a hard insect hitting a car windscreen can be thought of as a near constant-power acceleration, because the speed of the car does not change measurably during the collision. The glass is very hard, and so a bug will thus splat at even moderate speeds. The non constant-power element in that example comes from the fact that the insect is not infinitely hard, and so takes some time to compress during the collision.