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
Ultrilliam
Jun 1, 2018

Assuming the journey is made at constant velocity, fuel used in Kg is:

# (k d v^2)/(lambda + v^2) #

Explanation:

Newton's 2nd Law:

#sum bb F = underbrace(F)\_(" force from Engine") - underbrace(k dot x^2)\_(resistance)= (dp)/(dt) #

  • #(dp)/(dt) = (d(M + m) dot x)/(dt) #

# = (M + m) ddot x + dot m dot x #

#implies F = (M + m) ddot x + dot m dot x + k dot x^2#

The instantaneous Power delivered by the engine is:

#P = F * dot x = (M + m) dot x ddot x + dot m dot x^2 + k dot x^3#

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 #dot m lt 0#:

  • #P = - dot m lambda " J/s"#

#implies - dot m lambda = (M + m) dot x ddot x + dot m dot x^2 + k dot x^3#

This is looking horrible. If we simplify by assuming that the boat travels at constant velocity #v#, then :

# - dot m lambda = dot m v^2 + k v^3#

# dot m = - (k v^3)/(v^2 + lambda )#

Integrating:

# m = m_o - t (kv^3)/(lambda +v^2) #

At constant velocity time #tau# taken for the whole journey is:

  • #tau = d/v#

# m(tau) = m_o - d/v * (kv^3)/(lambda + v^2) #

# =m_o - (k dv^2)/(lambda + v^2)#

So fuel used, in Kg, is:

  • # m_o - (m_o - (k dv^2)/(lambda + v^2)) = (k d v^2)/(lambda + v^2)#
Steve J.
Jun 1, 2018

#"fuel to sail d meters" = (kυ^2*d)/lambda " kg of fuel"#

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 #d# (meters), the work required to maintain velocity #u# against the resistive force #F=kυ^2# would be

#"work" = F*d = kυ^2*d#

This would be in Joules. (The units of the constant k would be #"kg"/m"# for that to work out in Joules.)

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 #((1 "kg of fuel")/(lambda " Joules")) #

#"fuel to sail d meters" = kυ^2*d * ((1 "kg of fuel")/(lambda " Joules")) #

I said above that #kυ^2*d# would have units of Joules. Therefore, the Joules of the expression #kυ^2*d# would cancel the Joules in the denominator above. And we have the result:

#"fuel to sail d meters" = (kυ^2*d)/lambda " kg of fuel"#

I hope this helps,
Steve

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