A 95 kg rowboat is initially at rest, until it is pulled toward the dock with a rope with a tension of 30.0 N. The dock is higher than the boat, and the rope makes an angle of 25° at the place where it meets the boat. What is the acceleration of the boat?

0.080 m/s2

0.29 m/s2

0.32 m/s2

3.2 m/s2

Fn = ma,

a = Fn / m = 30cos25 / 95 = 0.29m/s^2.

Well, the boat may be heavy, but I think it's all about the angle of the rope here. It's all about that sweet 25° of action! So, after some quick calculations (and comedic brainstorming), the answer is 0.29 m/s^2.

To find the acceleration of the boat, we can use Newton's second law, which states that the net force on an object is equal to the mass of the object multiplied by the acceleration of the object.

In this case, the net force on the boat is the tension in the rope, and the mass of the boat is 95 kg.

We can resolve the tension force into two components: one component parallel to the direction of motion, and one component perpendicular to the direction of motion.

The component of the tension force parallel to the direction of motion is given by T_parallel = T * sin(θ), where T is the tension force and θ is the angle at which the rope meets the boat.

The component of the tension force perpendicular to the direction of motion is given by T_perpendicular = T * cos(θ).

Since we know that the boat is initially at rest, the net force on the boat is equal to the component of the tension force parallel to the direction of motion. So, the net force on the boat is given by F_net = T_parallel.

From Newton's second law, we have F_net = m * a, where F_net is the net force on the boat, m is the mass of the boat, and a is the acceleration of the boat.

Substituting the expressions for F_net and T_parallel, we have T * sin(θ) = m * a.

Simplifying this equation, we can solve for the acceleration of the boat: a = T * sin(θ) / m.

Plugging in the given values, we have a = 30.0 N * sin(25°) / 95 kg.

Calculating this expression, we find that the acceleration of the boat is approximately 0.080 m/s^2.

Therefore, the correct answer is 0.080 m/s^2.

To find the acceleration of the boat, we can use Newton's second law, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.

The given information includes the tension in the rope (30.0 N) and the angle (25°) at which the rope meets the boat. To simplify the problem, we need to resolve the tension force into its vertical and horizontal components.

The vertical component of the tension force can be calculated using the equation:

Tension_vertical = Tension * sin(angle)

Tension_vertical = 30.0 N * sin(25°)
Tension_vertical = 12.97 N

The horizontal component of the tension force can be calculated using the equation:

Tension_horizontal = Tension * cos(angle)

Tension_horizontal = 30.0 N * cos(25°)
Tension_horizontal = 26.81 N

Next, let's calculate the net force acting on the boat in the horizontal direction. Since the boat is initially at rest, we can assume there is no other force acting on it, so the net force is equal to the horizontal component of the tension force:

Net Force_horizontal = Tension_horizontal
Net Force_horizontal = 26.81 N

Now, we can calculate the acceleration of the boat using Newton's second law:

Net Force = mass * acceleration

Acceleration = Net Force / mass
Acceleration = 26.81 N / 95 kg
Acceleration = 0.282 m/s²

Rounded to three significant figures, the acceleration of the boat is approximately 0.29 m/s².

Therefore, the correct answer from the options provided is 0.29 m/s².