A boat mass M and I fill it up with fuel mass m. When it moves with velocity υ the resistive force is F=kυ^2. 1kg of fuel gives λ Joules of energy*. How much fuel (kg) is needed to sail d meters? *Suppose no energy is lost.
2 Answers
Assuming the journey is made at constant velocity, fuel used in Kg is:
Explanation:
Newton's 2nd Law:
#(dp)/(dt) = (d(M + m) dot x)/(dt) #
The instantaneous Power delivered by the engine is:
If
# 1" kg fuel" equiv lambda " J"# ,
then:
# m" kg fuel" equiv m lambda " J"# ,
And, bearing in mind that the boat is losing mass so
#P = - dot m lambda " J/s"#
This is looking horrible. If we simplify by assuming that the boat travels at constant velocity
Integrating:
At constant velocity time
#tau = d/v#
So fuel used, in Kg, is:
# m_o - (m_o - (k dv^2)/(lambda + v^2)) = (k d v^2)/(lambda + v^2)#
Explanation:
Data:
- I will assume constant velocity
#u# , with units#m/s# . - The resistive force is
#F=kυ^2# , with units N. - 1 kg of fuel has
#lambda " Joules"# .
To sail a distance
This would be in Joules. (The units of the constant k would be
Now we need to convert the above Joules to kg of fuel. From the 3rd bullet under Data above, the conversion factor I will use is
I said above that
I hope this helps,
Steve