Answer to Question #99904 in Mechanics | Relativity for Charles Masenya

Question

Two boats are identical, but one carries more people and so has a total mass that is twice as big as the other. If the two boats have identical initial kinetic energies and the frictional force from the water is the same in both cases, how far will the heavier boat travel before stopping, compared to the lighter boat?

Answer 1

We apply the kinetic energy theorem

$\Delta E_k=F \cdot S$

For the first boat we write

$\frac{m_1 \cdot v^2}{2}-\frac{m_1 \cdot v_{0}^2}{2}=-F \cdot S_1$

or

$\frac{m_1 \cdot 0^2}{2}-\frac{m_1 \cdot v_{0}^2}{2}=-F \cdot S_1$

where from

$\frac{m_1 \cdot v_{0}^2}{2}=F \cdot S_1$ (1)

similarly, we get the equation for the second boat

$\frac{m_2 \cdot v_{0}^2}{2}=F \cdot S_2$ (2)

divide the second equation into the first

$\frac{\frac{m_2 \cdot v_{0}^2}{2}}{\frac{m_1 \cdot v_{0}^2}{2}}=\frac{F \cdot S_2}{F \cdot S_1}$

$\frac{m_2 }{m_1}=\frac{ S_2}{S_1}$

by condition

$m_2=2 \cdot m_1$

Then we get the equation

$\frac{2 \cdot m_1 }{m_1}=\frac{ S_2}{S_1}$

where will we write

$S_2=2 \cdot S_1$

A boat with a larger mass will go to a stop distance twice as large.