Physics, 30.10.2019 06:31 cdizzlesmith2511

Aboat moves through the water with two forces acting on it. one is a 2000 n forward push by the water on the propeller, and the other is an 1800 n resistive force due to the water around the bow. a) what is the acceleration of the 1000 kg boat? (10 points) b) if it starts from rest, how far will the boat move in 10 sec.? (10 points) c) what will its velocity be at the end of that time?

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Answer from: Admin

a). a=0.2 $\frac{m}{s^{2} }$

b). x=10m

c). v=2 $\frac{m}{s}$

Explanation:

Force forward push = 2000N

Resistive force due water=1800N

Net force=2000N-1800N

Fn=200N

m=1000kg

t=10sec

a).

$F_{n} =m*a\\a=\frac{F_{n}}{m}\\ a=\frac{200N}{1000kg}\\ a=0.2 \frac{m}{s^{2} }$

b).

$x_{f}=x_{o}+ v_{o}*t+\frac{1}{2}*a*t^{2}\\x_{o}=0\\v_{o}=0\\x_{f}=\frac{1}{2}*a*t^{2}\\x_{f}=\frac{1}{2}*0.2*(10)^{2} \\x_{f}=10m$

c).

$v_{f}=v_{o}+a*t\\v_{o}=0\\v_{f}=a*t\\v_{f}=0.2\frac{m}{s^{2} }*10s\\v_{f}=2\frac{m}{s}$

Answer from: gonzaloc

The following parameters were calculated from a boat moving through the water with two forces acting on it:

a. Acceleration = 0.2 m/s^2

b. Distance, S = 10 meters.

c. Final velocity, V = 2 m/s.

Given the following data:

Forward push force = 2000 NewtonResistive force = 1800 Newton.Mass of boat = 1000 kg

a. To find the acceleration of the 1000 kg boat, we we would apply Newton's Second Law of Motion:

Mathematically, Newton's Second Law of Motion is given by this formula;

$Acceleration = \frac{Net\; force}{Mass}$

First of all, we would determine the net force acting on the boat:

$Net \; force = Forward \;push\;force - Resistive\; force\\\\Net \; force = 2000 - 1800$

Net force = 200 Newton

Substituting the values, we have;

$Acceleration = \frac{200}{1000}$

Acceleration = 0.2 m/s^2

b. If the boat starts from rest, how far (distance) will it move in 10 seconds:

To find the distance covered by the boat, we would use the second equation of motion.

Note: Initial velocity = 0 m/s (since the boat starts from rest).

$S = ut + \frac{1}{2} at^2\\\\S = 0(10) + \frac{1}{2} (0.2)(10^2)\\\\S = 0 + 0.1(100)$

Distance, S = 10 meters.

c. To find the boat's velocity at the end of that time (10 seconds), we would use the first equation of motion:

$V = U + at\\\\V = 0 + 0.2(10)$

Final velocity, V = 2 m/s

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