The accelerating force of the wind on a small 200-kg sailboat is 707 N north. If the drag of the keel is 500 N acting west, what is the acceleration of the boat?
A) 1.5 m/s2 35 degrees north of east
B) 2.5 m/s2 55 degrees north of west
C) 3.0 m/s2 35 degrees north of east
D) 4.3 m/s2 55 degrees north of west
E) 1.5 m/s2 55 degrees north of west
To determine the acceleration of the boat, we first need to find the net force acting on the boat. We can do this by using the Pythagorean theorem to find the magnitude of the net force and inverse tangent to find the direction.
First, let's find the magnitude of the net force.
F_net = sqrt(F_wind^2 + F_drag^2) = sqrt(707^2 + 500^2) = 867 N
Next, let's find the direction. We can do this by finding the angle θ between the net force and the east direction using inverse tangent (arctan).
θ = arctan(F_wind / F_drag) = arctan(707 / 500) = 54.7 degrees
Finally, we can find the acceleration by dividing the net force by the mass of the boat.
a = F_net / m = 867 N / 200 kg = 4.335 m/s^2
Therefore, the acceleration of the boat is approximately 4.3 m/s^2, 55 degrees north of west, which is option D.
To find the acceleration of the boat, we need to consider the net force acting on it. The net force is the vector sum of the accelerating force of the wind and the drag force of the keel.
Given:
Wind force = 707 N north
Drag force of keel = 500 N west
To find the net force, we need to resolve both forces into their x and y components.
Wind force:
The wind force is acting north, so its y-component is 707 N * sin(90°) = 707 N.
The x-component of the wind force is 0 N since it is acting purely in the vertical direction.
Drag force:
The drag force is acting west, so its x-component is -500 N.
The y-component of the drag force is 0 N since it is acting purely in the horizontal direction.
Now, let's find the net force in the x and y directions:
Net force in the x direction:
Since the wind force and the drag force are the only forces acting in the x direction, the net force in the x direction is the sum of their x-components.
Net force in the x direction = 0 N + (-500 N) = -500 N.
Net force in the y direction:
Since the wind force and the drag force are acting in opposite directions in the y direction, we subtract their y-components to find the net force.
Net force in the y direction = 707 N - 0 N = 707 N.
Now, we can calculate the resultant force:
Resultant force = √((Net force in the x direction)^2 + (Net force in the y direction)^2)
Resultant force = √((-500 N)^2 + (707 N)^2)
Resultant force = √(250000 N^2 + 499849 N^2)
Resultant force ≈ √749849 N^2 ≈ 866 N
The acceleration is given by Newton's second law, F = ma. Rearranging the equation, we have a = F/m.
Acceleration = Resultant force / Mass of the boat = 866 N / 200 kg ≈ 4.33 m/s^2
The direction of acceleration can be determined using the angles provided in the options. Comparing this with the given options:
A) 1.5 m/s^2, 35 degrees north of east
B) 2.5 m/s^2, 55 degrees north of west
C) 3.0 m/s^2, 35 degrees north of east
D) 4.3 m/s^2, 55 degrees north of west
E) 1.5 m/s^2, 55 degrees north of west
The correct answer is D) 4.3 m/s^2, 55 degrees north of west.
To find the acceleration of the boat, we need to consider the net force acting on it. The sailboat experiences two forces: the accelerating force of the wind and the drag force of the keel.
First, let's break down the given information:
- Accelerating force of the wind: 707 N north
- Drag force of the keel: 500 N west
Since the forces are acting in different directions, we need to resolve them into their x and y components. Let's consider the x-axis as east-west and the y-axis as north-south.
The force of the wind has a magnitude of 707 N and acts directly north. To resolve it into x and y components, we can use trigonometry. Let's call the angle between the force and the positive x-axis as θ.
Using the sine and cosine functions, we can find the x and y components of the wind force:
Wind force in the x-direction = 707 N * sin(θ)
Wind force in the y-direction = 707 N * cos(θ)
Next, let's consider the drag force of the keel, which has a magnitude of 500 N acting west. Since it is acting in the opposite direction of the x-axis, the drag force in the x-direction will simply be -500 N, while the y-component is zero.
Now, we can find the net force acting on the boat:
Net force in the x-direction = Wind force in the x-direction + Drag force in the x-direction
Net force in the y-direction = Wind force in the y-direction + Drag force in the y-direction
Finally, we can use Newton's second law, F = ma, to find the acceleration of the boat. The net force will be divided by the mass of the sailboat (200 kg) to find the acceleration.
a = Net force / Mass
Now let's calculate the net force and find the resulting acceleration:
Net force in the x-direction = 707 N * sin(θ) - 500 N
Net force in the y-direction = 707 N * cos(θ)
a = √((Net force in the x-direction)^2 + (Net force in the y-direction)^2) / Mass
To determine the direction of the acceleration, we can use the tangent function:
θ = atan(Net force in the y-direction / Net force in the x-direction)
Now, using the given angles in the answer choices, we can calculate the values and compare them to the options provided.