Question

Two forces act on a 1250 kg sailboat as it moves through the water with an initial velocity of 11 m/s. The forward force of the wind exerts 3.90 X 10 3N while the resistive drag force of the water exerts 4.35 X 10 3N. Calculate the velocity of the boat 15 seconds later

Answer 1

Answer:

The velocity of the boat 15 seconds later is 5.6 meters per second.

Explanation:

We assume that sailboat can be modelled as particle, so that we use solely translations equations. It is noticed that the sailboat is move by action of the wind and drag force of the water is opposed to such force, but the last force has a greater magnitude than the first one, meaning that net force is less than zero.

From Newton's Laws we have the following equation of equilibrium for the sailboat:

$\Sigma F = F - f = m\cdot a$ (Eq. 1)

Where:

$F$ - Force from the wind exerted on the sailboat, measured in newtons.

$f$ - Drag force of the water, measured in newtons.

$m$ - Mass of the sailboat, measured in kilograms.

$a$ - Net acceleration of the sailboat, measured in meters per square second.

If we know that $F = 3.90\times 10^{3}\,N$, $f = 4.35\times 10^{3}\,N$ and $m = 1250\,kg$, then the net acceleration of the sailboat is:

$a = \frac{F-f}{m}$

$a = \frac{3.90\times 10^{3}\,N-4.35\times 10^{3}\,N}{1250\,kg}$

$a = -\frac{9}{25}\,\frac{m}{s^{2}}$

If sailboat decelerates uniformly, then we can get the final velocity of the boat by using this equation of motion:

$v = v_{o}+a\cdot t$ (Eq. 2)

Where:

$v_{o}$, $v$ - Initial and final velocities of the sailboat, measured in meters per second.

$t$ - Time, measured in seconds.

If we get that $v_{o} = 11\,\frac{m}{s}$, $a = -\frac{9}{25}\,\frac{m}{s^{2}}$ and $t = 15\,s$, then the final velocity of the sailboat is:

$v = 11\,\frac{m}{s} +\left(-\frac{9}{25}\,\frac{m}{s^{2}} \right)\cdot (15\,s)$

$v = 5.6\,\frac{m}{s}$

The velocity of the boat 15 seconds later is 5.6 meters per second.