Physics, 03.11.2020 16:20 matiasjjj14

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

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Answer from: perezz97

The final velocity of the sailboat 15 seconds later is 5.6 m/s.

Given the following data:

Mass of sailboat = 1250 kgInitial velocity = 11 m/sForward force = $3.90 \times 10^3$ NewtonResistive drag force = $4.35 \times 10^3$ NewtonTime = 15 seconds.

To find the final velocity of the sailboat 15 seconds later:

First of all, we would determine the net force acting on the sailboat.

$Net\;force = Forward\;force - Resistive\;force\\\\Net\;force = 3.90 \times 10^3 - 4.35 \times 10^3$

Net force = $-4.35 \times 10^3$ Newton.

Next, we would calculate the acceleration of the sailboat by applying Newton's Second Law of Motion:

$Acceleration = \frac{Net\;force}{Mass} \\\\Acceleration = \frac{-0.45 \times 10^3}{1250}$

Acceleration = -0.36 m/s^2

To calculate the final velocity of the boat 15 seconds later, we would use the first equation of motion:

$V = U + at\\\\V = 11 + (-0.36)15\\\\V = 11 - 5.4$

Final velocity, V = 5.6 m/s

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Answer from: yasameenakbari

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:

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

- Drag force of the water, measured in newtons.

- Mass of the sailboat, measured in kilograms.

- 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:

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

- 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.

Answer from: Quest

increase masses, decrease separation

Answer from: Quest

the total distance that martina travels that day is

(distance from home to mom's office) plus (distance from mom's office to school)

plus (distance from school to library) plus (distance from library to friend's house)

plus (distance from friend's house to martina's house).

(we don't know what any of these distance actually is.)

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